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Diffstat (limited to 'lacey')
| -rw-r--r-- | lacey/hw1.tex | 368 | ||||
| -rw-r--r-- | lacey/hw2.tex | 294 | ||||
| -rw-r--r-- | lacey/hw3.tex | 339 | ||||
| -rw-r--r-- | lacey/hw4i.tex | 138 | ||||
| -rw-r--r-- | lacey/hw4ii.tex | 110 | ||||
| -rw-r--r-- | lacey/hw5.tex | 130 | ||||
| -rw-r--r-- | lacey/hw6.tex | 187 | ||||
| -rw-r--r-- | lacey/hw7.tex | 153 | ||||
| -rw-r--r-- | lacey/hw8.tex | 190 |
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diff --git a/lacey/hw1.tex b/lacey/hw1.tex new file mode 100644 index 0000000..338a531 --- /dev/null +++ b/lacey/hw1.tex @@ -0,0 +1,368 @@ +\font\bigbf=cmbx12 at 24pt +\newfam\bbold +\font\bbten=msbm10 +\font\bbsev=msbm7 +\font\bbfiv=msbm5 +\textfont\bbold=\bbten +\scriptfont\bbold=\bbsev +\scriptscriptfont\bbold=\bbfiv +\def\bb#1{{\fam\bbold #1}} + +\newcount\pno +\def\tf{\advance\pno by 1\smallskip\bgroup\item{(\number\pno)}} +\def\endtf{\egroup\medskip} + +\newcount\qno +\def\question#1{\ifnum\qno=0\else\vfil\eject\fi\bigskip\pno=0\advance\qno by 1\noindent{\bf Question \number\qno.} {\it #1}} + +\def\proof{{\it Proof.}\quad} +\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt +depth 0pt{\parfillskip0pt\medskip}} + +{\bigbf\centerline{Holden Rohrer}\bigskip\centerline{MATH 4317\qquad HOMEWORK \#1}} + +\question{For each item, if it is a true statement, say so. +Otherwise, give a counterexample.} + +\tf For sets $A_1,A_2,\ldots,A_n,$ +$\left(\bigcup_{k=1}^n A_k\right)^c = \bigcap_{k=1}^n A_k.$ +\endtf + +False. +With $U = \{1\},$ and $A_1 = \emptyset,$ (where $n=1$), +the union is $\emptyset,$ so the complement is $\{1\},$ which is not +equal to the intersection $\emptyset.$ + +\tf For sets $A_1,A_2,\ldots,A_n,$ +$\left(\bigcup_{k=1}^\infty A_k\right)^c = \bigcap_{k=1}^\infty A_k^c.$ +\endtf + +True. + +\tf An infinite sequence of decreasing closed intervals $I_1\supset I_2 +\supset I_3 \supset \cdots$ has a non-empty intersection. +\endtf + +True, when defined on the reals. + +\tf Every convergent sequence of rationals has a rational limit. +\endtf + +False. Let $a_n = 1 + {1\over a_{n-1}}$ with $a_0 = 1.$ +This is a rational expression but converges to the irrational $\phi.$ + +Also see the Babylonian method. + +\tf A finite set can contain its supremum, but not its infimum. +\endtf + +False. The set $\{0\}$ has lower bound $m=0.$ +This is an infimum because all other lower bounds $w\leq 0=m$ by +definition. +This is in the set. + +\tf An unbounded set does not have a supremum. +\endtf + +False. A set can be unbounded because it lacks a lower bound like +$\{-2^n : n\in\bb N\},$ but this set has supremum $-1.$o + +\tf Two real numbers $a$ and $b$ satisfy $a < b$ if for all $\epsilon > +0,$ $a \leq b + \epsilon.$ +\endtf + +{\let\to\Rightarrow +False. Let $a = b.$\par $0 < \epsilon \to 0\leq\epsilon\to a\leq a ++\epsilon\to a \leq b + \epsilon.$ +} + +\tf Let $f: X\to Y$ be a function, and $A,$ $B$ be sets in the codomain +$Y.$ Then, $f^{-1}(A\cap B) = f^{-1}(A)\cap f^{-1}(B).$ +\endtf + +\proof +Let $A,B\in Y.$ +Let $s\in f^{-1}(A)\cap f^{-1}(B).$ +$f(s) \in A$ and $f(s)\in B,$ so $f(s) in A\cap B,$ and $s\in +f^{-1}(A\cap B).$ + +Now, to show the other direction, let $t\in f^{-1}(A\cap B).$ +Then, $f(t) \in A\cap B,$ so $f(t) \in A$ and $f(t)\in B,$ so $t\in +f^{-1}(A),$ and $t\in f^{-1}(B),$ and $t\in f^{-1}(A)\cap f^{-1}(B).$ + +We have shown these two sets are equal. +\endproof + +\tf Let $f: X\to Y$ be a function, and $A,$ $B$ be sets in the codomain +$Y.$ Then, $f^{-1}(A\cup B) = f^{-1}(A)\cup f^{-1}(B).$ +\endtf + +\proof +Let $s\in f^{-1}(A\cup B).$ Then, $f(s)\in A\cup B.$ +WLOG, let $f(s)\in A,$ so $s\in f^{-1}(A)\cup f^{-1}(B).$ + +The other direction is $s\in f^{-1}(A)\cup f^{-1}(B),$ and WLOG, $s\in +f^{-1}(A).$ +Then, $f(s)\in A,$ so $f(s)\in A\cup B,$ and $s\in f^{-1}(A\cup B).$ + +We have shown these two sets are equal. +\endproof + +\question{(Exercise 1.2.8) Here are two important definitions related to +a function $f: A\to B.$ The function $f$ is one-to-one (1-1, injective) +if $a_1\neq a_2$ in $A$ implies that $f(a_1)\neq f(a_2)$ in $B.$ The +function $f$ is onto (surjective) if, given any $b\in B,$ it is possible +to find an element $a\in A$ for which $f(a) = b.$ Give an example of +each, or state that the request is impossible.} + +I have assumed $0\in\bb N.$ + +\tf +$f: \bb N\to\bb N$ that is one-to-one but not onto. +\endtf + +$f(x) = 2x.$ + +\tf +$f: \bb N\to\bb N$ that is not one-to-one but is onto. +\endtf + +$f(x) = \left\lfloor {x\over 2}\right\rfloor.$ + +\tf +$f: \bb N\to\bb Z$ that is one-to-one and onto. +\endtf + +$f(x) = \left\lfloor {x+1\over 2}\right\rfloor (-1)^x.$ + +\question{(Exercise 1.2.10) Decide which of the following are true +statements. Provide a short justification for those that are valid and a +counterexample for those that are not:} + +\tf Two real numbers satisfy $a<b$ if and only if $a < b+\epsilon$ for +every $\epsilon>0.$ +\endtf + +False. Where $a=b,$ $\epsilon>0$ implies $a < b+\epsilon.$ + +\tf Two real numbers satisfy $a < b$ if $a<b+\epsilon$ for every +$\epsilon > 0.$ +\endtf + +False. Where $a=b,$ $\epsilon>0$ implies $a < b+\epsilon.$ + +\tf Two real numbers satisfy $a\leq b$ if and only if $a<b+\epsilon$ for +every $\epsilon>0.$ +\endtf + +\proof + +$(\Rightarrow)$ + +Let $a\leq b.$ If $\epsilon > 0,$ then $a < b + \epsilon.$ + +$(\Leftarrow)$ + +We will show the contrapositive. +Let $a > b.$ If $\epsilon > 0,$ then $a > b + \epsilon,$ or $a \geq b + +\epsilon.$ + +\endproof + +\question{(Exercise 1.3.2) Give an example of each of the following, or +state that the request is impossible.} + +\tf +A set $B$ with $\inf B\geq\sup B.$ +\endtf + +Let $B = \{0\}.$ $\inf B = \sup B = 0,$ so $\inf B\geq\sup B.$ + +\tf +A finite set that contains its infimum but not its supremum. +\endtf + +This is impossible. +All finite nonempty sets include their infimum and their supremum. + +\tf +A bounded subset of $\bb Q$ that contains its supremum but not its +infimum. +\endtf + +$A = \{2^{-n} : n\in\bb N\}$ is bounded (for all $a\in A,$ $|a|\leq 1$) +and it contains its supremum $1$ but not its infimum $0.$ + +\question{(1.3.7) Prove that if $a$ is an upper bound for $A,$ and if +$a$ is also an element of $A,$ then it must be that $a=\sup A.$} + +\proof +Let $a$ be an upper bound for $A.$ That is, for all $b\in A,$ $a\geq b.$ +Also let $a\in A.$ +Therefore all upper bounds $m$ must satisfy $m\geq a.$ + +These are the conditions of supremum, so $a$ is a supremum. +\endproof + +\question{(1.3.6) Given sets $A$ and $B,$ define $A+B = \{a+b:a\in +A,b\in B\}.$ Follow these steps to prove that if $A$ and $B$ are +nonempty and bounded above then $\sup(A+B)=\sup A + \sup B.$} + +\tf +Let $s = \sup A$ and $t = \sup B.$ Show $s+t$ is an upper bound for +$A+B.$ +\endtf +\proof +Let $a+b\in A+B.$ From the construction of $A+B,$ $a\in A$ and $b\in B.$ + +From definition of supremum, $a\leq s$ and $b\leq t.$ Therefore, +$a+b\leq s+t,$ giving $s+t$ is an upper bound for all elements of $A+B.$ +\endproof + +\tf +Now let $u$ be an arbitrary upper bound for $A+B,$ and temporarily fix +$a\in A.$ Show $t\leq u-a.$ +\endtf +\proof +Let $a+b\in A+B$ with fixed $a\in A.$ + +If $u$ is an upper bound for $A+B,$ then $a+b \leq u.$ + +As $t$ is the least upper bound of $B,$ for all $\epsilon > 0,$ we have +for some $b\in B,$ that $t-\epsilon < b.$ +Thus, $a+t-\epsilon < a+b \leq u.$ +$\epsilon$ may be reduced, and we obtain $a+t \leq u$ or $t\leq u-a.$ +\endproof + +\tf +Finally, show $\sup(A+B) = s+t.$ +\endtf +\proof + +Let $\epsilon > 0.$ +From definition of supremum, for some $a\in A,$ we have +$s-\epsilon < a.$ +From the last proof, $s-\epsilon+t < a+t \leq u.$ +Since $s+t-\epsilon < u$ for all $\epsilon > 0,$ $s+t\leq u.$ + +\endproof + +\tf +Construct another proof of this same fact using Lemma 1.3.8. +\endtf +\proof +Let $\epsilon > 0.$ + +From lemma 1.3.8, there exists $a\in A$ and $b\in B$ such that +$s-\epsilon/2 < a$ and $t-\epsilon/2 < b.$ +Since we know $s+t$ is an upper bound for $A+B,$ +and we may now show $s+t-\epsilon < a+b$ for all $a+b\in A+B.$ +The lemma tells us this makes $s+t$ a supremum for $A+B.$ +\endproof + +\question{(1.3.10, Cut Property) The Cut Property of the real numbers is +the following. If $A$ and $B$ are nonempty, disjoint sets with $A\cup +B=\bb R$ and $a<b$ for all $a\in A$ and $b\in B,$ then there exists $c\in\bb +R$ such that $x\leq c$ whenever $x\in A$ and $y\geq c$ whenever $y\in +B.$} + +\tf +Use the Axiom of Completeness to prove the Cut Property. +\endtf +\proof +Let $A$ and $B$ be sets satisfying the properties above. +We will find $s$ such that for all $a\in A,$ and for all $b\in B,$ +$a\leq s\leq b.$ + +All $b\in B$ are upper bounds for $A,$ so the axiom of completeness +tells us $A$ has a least upper bound we'll call $s = \sup A.$ +Since $A\cup B\in\bb R$ and $s\in\bb R,$ but $A\cap B = \emptyset$ (by +disjointedness), either $s\in A$ or $s\in B.$ + +WLOG, let $s\in A$ (we might just as easily have defined $s$ as the +greatest lower bound of $B$). +By construction, $s$ is an upper bound for $A,$ so for all $a\in A,$ $a +\leq s,$ and since $s\in A,$ from our definition of $B,$ for all $b\in B,$ +$s < b,$ giving us our cut. +\endproof + +\tf +Show that the implication goes the other way; that is, assume $\bb R$ +possesses the Cut Property and let $E$ be a nonempty set that is bounded +above. Prove $\sup E$ exists. +\endtf +\proof +Let $A = \bigcup_{e\in E} \{x\in\bb R: x \leq e\}.$ +Let $B$ be the complement of $A.$ +This creates a disjoint set with $A\cup B = \bb R.$ + +$E$ is bounded above, so we have $b$ such that for all $e\in E,$ $e < +b.$ +This makes $B$ nonempty, and $A$ is trivially nonempty if $E$ is +nonempty. + +We also have that all $b\in B$ and $a\in a$ satisfy $b>a$ because all +$b\in B$ satisfy $b>e$ for a some $e\in E,$ but $a \leq e$ in that same +case. +The cut property gives us $c$ between $A$ and $B$ such that $a\leq c\leq +b,$ and since $B$ contains all possible upper bounds for $A,$ this is a +least upper bound. + +\endproof + +\tf +The punchline of parts $(1)$ and $(2)$ is that the Cut Property could be +used in place of the Axiom of Completeness as the fundamental axiom that +distinguishes th ereal numbers from the rational numbers. To drive this +point home, give a concrete example showing that the Cut Property is not +a valid statement when $\bb R$ is replaced by $\bb Q.$ +\endtf +\proof +Let $A = \{x\in\bb Q : x<\sqrt 2\}$ and $B = \{x\in\bb Q : \sqrt 2\leq +x\}.$ +$A\cup B = \bb Q,$ and $A\cap B = \emptyset,$ and for all $a\in A$ and +$b\in B,$ $a<\sqrt 2\leq b,$ so this satisfies all of the conditions. + +However, the only number which is between $A$ and $B$ is $\sqrt 2,$ +and any nearby rational $r$ satisfies either $r>\sqrt 2$ (putting it in +$B$), or $r<\sqrt 2$ (putting it in $A$), telling us $\bb Q$ doesn't +hold the cut property. +\endproof + +\question{(1.3.11) Decide if the following statements about suprema and +infima are true or false. Give a short proof for those that are true. +For any that are false, supply an example where the claim in question +does not appear to hold.} + +\tf +If $A$ and $B$ are nonempty, bounded, and satisfy $A\subset B,$ then +$\sup A\leq\sup B.$ +\endtf +\proof +For all $a\in A,$ $a\in B,$ so by definition of supremum, $a\leq\sup B.$ +Therefore, $\sup B$ is an upper bound for $A,$ and $\sup A < \sup B,$ +again by definition of the supremum. +\endproof + +\tf +If $\sup A<\inf B$ for sets $A$ and $B,$ then there exists a $c\in\bb R$ +satisfying $a<c<b$ for all $a\in A$ and $b\in B.$ +\endtf +\proof +Let $\sup A<\inf B.$ Let $a\in A$ and $b\in B.$ Let $c = {\sup A + \inf +B\over 2}.$ + +From definitions of supremum and infimum, $a \leq \sup A < c < \inf B \leq +b,$ so $a < c < b.$ +\endproof + +\tf +If there exists $c\in\bb R$ satisfying $a<c<b$ for all $a\in A$ and +$b\in B,$ then $\sup A < \inf B.$ +\endtf + +False. Let $B = \{2^{-n} : n\in\bb N\}$ and $A = \{-b : b\in B\}.$ + +For all $a\in A$ and $b\in B,$ $a < 0 < b,$ but $\sup A = \inf B = 0.$ + +\bye diff --git a/lacey/hw2.tex b/lacey/hw2.tex new file mode 100644 index 0000000..c553b8c --- /dev/null +++ b/lacey/hw2.tex @@ -0,0 +1,294 @@ +\font\bigbf=cmbx12 at 24pt +\newfam\bbold +\font\bbten=msbm10 +\font\bbsev=msbm7 +\font\bbfiv=msbm5 +\textfont\bbold=\bbten +\scriptfont\bbold=\bbsev +\scriptscriptfont\bbold=\bbfiv +\def\bb#1{{\fam\bbold #1}} + +\newcount\pno +\def\tf{\advance\pno by 1\smallskip\bgroup\item{(\number\pno)}} +\def\endtf{\egroup\medskip} + +\newcount\qno +\long\def\question#1{\ifnum\qno=0\else\vfil\eject\fi\bigskip\pno=0\advance\qno by 1\noindent{\bf Question \number\qno.} {\it #1}} + +\def\proof{\medskip\noindent{\it Proof.}\quad} +\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt +depth 0pt{\parfillskip0pt\medskip}} + +\let\bu\bullet + +{\bigbf\centerline{Holden Rohrer}\bigskip\centerline{MATH 4317\qquad HOMEWORK \#2}} + +\noindent{\bf Definition.} For any set $A,$ finite or infinite, define +the power set of $A$ to be the collection of all subsets of $A.$ That +is, +$${\cal P}(A) = \{A' : A'\subseteq A\}.$$ + +\noindent{\bf Definition.} A set $A\subseteq\bb R$ is dense if for all +$a\neq b\in\bb R,$ there is an element $c\in A$ between $a$ and $b.$ + +\question{For each item, if it is a true statement, say so. Otherwise, +give a counterexample.} + +\tf +A decreasing nested sequence of bounded intervals has non empty +intersection. +\endtf + +False. +This fails if they are bounded and open; +for example, $A_n = (0,{1\over n}].$ + +\tf +A decreasing nested sequence of closed intervals has non-empty +intersection. +\endtf + +True. + +\tf +For reals $a < b,$ we have $\sup Q\cap [a,b] = b.$ +\endtf + +True. + +\tf +For reals $a < b,$ we have $\sup Q^c\cap [a,b] = b.$ +\endtf + +True. + +\tf +The rationals are dense. +\endtf + +True. + +\tf +The subset of the rationals $\{p/2^r | p\in\bb Z, +r\in\{100,101,\ldots\}\}$ are dense. +\endtf + +True. % i feel like i want to write a proof. + +\tf +The subset of the rationals $\{p/2^r | p\in\bb Z, +r\in\{1,2,\ldots,100\}\}$ are dense. +\endtf + +False. Between 0 and $1/2^{100},$ there is no number in this set. + +\tf +The set $\{\pi r|r\in\bb Q\}$ is dense. +\endtf + +True. + +\question{Prove that there is a positive number $s$ such that $s^2 = 3.$ +\item{$\bu$} This depends upon the Supremum Axiom of course, applied to +a good set $S.$ +\item{$\bu$} Letting $\sigma = \sup S,$ there are two cases which need +to lead to a contradiction: $\sigma^2 < 3$ and $\sigma^2 > 3.$ +\item{$\bu$} You could find it useful to note that $1.7^2<3,$ but only +just barely. (But it is not necessary, either) +} + +\proof +Let $S = \{s\in\bb R|s^2 < 3\}.$ + +2 is an upper bound of $S.$ + +For the sake of contradiction, let $s\in S$ s.t. $s>2.$ +Then, $s^2>4>3,$ contradicting our definition of $S.$ + +By the supremum axiom, this upper-bounded set must have a supremum +$\sigma = \sup S.$ + +If $\sigma^2 < 3,$ +we must have $s>0\in\bb R$ such that $\sigma^2<s^2<3,$ (by density of +the reals) meaning $s\in S$ and $s>\sigma,$ so $\sigma$ is not the +supremum. % TODO: check if we can sorta assume continuity of s^2 like +% this (probably ask Lacey) + +If $\sigma^2 > 3,$ choose $s>0$ such that $3 < s^2 < \sigma^2,$ and +let $\epsilon = \sigma-s > 0.$ +From a property of the supremum, we must have $a\in S$ such that +$a>\sigma-\epsilon=s.$ +We already know $s^2 > 3,$ so $a^2 > 3$, contradicting our construction. + +We have thus determined $\sigma^2 = 3,$ proving there is such a positive +real number. +\endproof + +\question{Show that these statements are equivalent: for a sequence of +reals $\{x_n\},$ +\item{$(1)$} The sequence converges to $x$: for all $\epsilon > 0,$ +there is a $N_\epsilon > 1$ so that for all $n > N_\epsilon,$ +$|x_n-x|<\epsilon.$ +\item{$(2)$} For all $\epsilon > 0,$ there is a $N_\epsilon > 1$ so that +for all $n>N_\epsilon,$ $|x_n-x|<100\epsilon.$ +\item{$(3)$} For all $0<\epsilon<10^{-10}$ there is a $N_\epsilon > 1$ +so that for all $n>N_\epsilon,$ $|x_n-x|<\epsilon.$ +\item{$(4)$} For all $\epsilon > 0$ there is a $N_\epsilon > 10^{10}$ so +that for all $n>N_\epsilon,$ $|x_n-x|<\epsilon.$ +} + +\proof +We will prove these in a loop. + +\smallskip +$(1)\Rightarrow(2)$ +\smallskip + +For all $\epsilon>0,$ there is a $N_\epsilon>1$ so that for all +$n>N_\epsilon,$ $|x_n-x|<\epsilon<100\epsilon,$ satisfying our +condition. + +\smallskip +$(2)\Rightarrow(3)$ +\smallskip + +We will show for all $0<\epsilon<10^{-10},$ there is a $N_\epsilon>1$ so +that for all $n>N_\epsilon,$ $|x_n-x|<\epsilon.$ + +Select some $0<\epsilon/100<10^{-12}$ and $N_{\epsilon}$ satisfying +$(2)$. +Then, for all $n>N_\epsilon,$ $|x_n-x|<100(\epsilon/100)=\epsilon,$ +satisfying $(3).$ + +\smallskip +$(3)\Rightarrow(4)$ +\smallskip + +Let $\epsilon>0.$ +Account for where $\epsilon > 10^{-11},$ we take instead $N_\epsilon>1$ +such that for all $n>N_\epsilon,$ $|x_n-x| < \min\{10^{-11},\epsilon\} +\leq \epsilon.$ + +If $N_\epsilon<10^{10},$ the result implies +$M_\epsilon>10^{10}>N_\epsilon$ such that for all +$n>M_\epsilon>N_\epsilon,$ $|x_n-x|<\epsilon.$ + +\smallskip +$(4)\Rightarrow(1)$ +\smallskip + +Let $\epsilon > 0.$ We have $N_\epsilon > 10^{10} > 1$ such that for all +$n>N_\epsilon,$ $|x_n-x|<\epsilon.$ +This satisfies $(1).$ + +\endproof + +\question{Let $A$ be a nonempty set of positive real numbers, that is +bounded above. Set $B = \{1/a : a\in A\}.$ Prove that $B$ is bounded +below, and that +$$\inf B = {1\over\sup A}.$$ +} + +\proof + +All members of $A$ are positive, so $1/a>0$ for all $a\in A,$ giving a +lower bound for $B.$ +By the supremum axiom, there is an infimum $b$ for $B.$ + +We know that for all $a\in A,$ $b\leq 1/a,$ and for all $\epsilon>0$ and +some respective $c\in A,$ that $1/c<b+\epsilon.$ + +$1/b\geq a,$ so we know $b\geq\sup A.$ +Then, $$c > {1\over b+\epsilon} = {1\over b} + {\epsilon\over +b(b+\epsilon)},$$ +of which the error can be reduced to tell us that ${1\over b} = \sup A.$ + +\endproof + +\question{Let $A = \{a_n|n\in\bb N\}$ and $B = \{b_n|n\in\bb N\}$ be two +bounded sets in $\bb R.$ Show that +$$\inf_m a_m + \inf_n b_n \leq \inf_n(a_n + b_n).$$ +$$\inf_m a_m + \inf_n b_n = \inf_{m,n} a_m + b_n.$$ +} + +\proof +We will show the equality first. +Let $a$ and $b$ refer to the infima of $A$ and $B$ respectively. +Let $\epsilon > 0.$ +From our definition of infima, we have $m,n\in\bb N$ such that +$a_n < a + \epsilon$ and $b_m < b + \epsilon.$ +Thus, $a_n+b_m < a+b+2\epsilon.$ +And for all $o,p\in\bb N,$ +we have $a \leq a_o$ and $b \leq b_p,$ so $a+b\leq a_o+b_p,$ +meaning our infimum $c$ satisfies $a+b\leq c<a+b+2\epsilon,$ so $c = +a+b.$ + +The inequality is then trivially implied since $\{a_n+b_n:n\in\bb +N\}\subseteq \{a_n+b_m:n,m\in\bb N\},$ so the infimum of the former is +less than or equal to all elements of the latter, and thus forms a lower +bound less than or equal to the infimum of the latter. +\endproof + +\question{This concerns the power set, defined above. +\item{$(1)$} What is the power set of the empty set, ${\cal +P}(\emptyset)?$ +\item{$(2)$} Find a formula for the cardinality of ${\cal +P}(\{1,2,\ldots,n\})$ for $n=1,2,3,\ldots$ Prove the formula, using +mathematical induction. + +\noindent Recall that the mathematical induction is a two step procedure, +requiring +\item{$(1)$} Establish the proposition in a base case, $n=1$ in this +case. +\item{$(2)$} Assuming the proposition true for integer $n\geq 1,$ show +that it is true for $n+1.$ +} + +\proof +${\cal P}(\emptyset) = \{\emptyset\}.$ + +This is the $n=0$ case of ${\cal P}(\{1,2,\ldots,n\}),$ and it has +cardinality $1 = 2^0.$ + +To show the inductive case, assume that $|{\cal P}(\{1,2,\ldots,n\})|=2^n.$ + +${\cal P}(\{1,2,\ldots,n,n+1\}) = {\cal P}(\{1,2,\ldots,n\})\cup\{x\cup\{n+1\} : +x\in{\cal P}(\{1,2,\ldots,n\})\},$ +so its cardinality is $2^{n+1},$ showing the inductive case. + +These sets are disjoint because none of the left ones and all of the +right ones have the element $n+1,$ and each have cardinality $2^n,$ and +this enumerates all subsets of $\{1,2,\ldots,n\}$ which are subsets of +$\{1,2,\ldots,n,n+1\},$ and the subsets which include $n+1.$ + +\endproof + +\question{Find bijections for the following functions. +\item{$(2)$} Find a bijection between $(0,1)$ and $[0,1).$ +\item{$(3)$} Find a bijection between $(0,1)$ and $(0,1)\cup\bb N.$ +} + +\tf (Hilbert Hotel) Find a bijection between $\bb Z$ and $\{1/2\}\cup\bb +Z.$ +\endtf + +Let $f: \bb Z\to\{1/2\}\cup\bb Z.$ + +For $z>1,$ $f(z) = z-1.$ +For $z=0,$ $f(z) = 1/2.$ +For $z<0,$ $f(z) = z.$ + +\tf Find a bijection between $(0,1)$ and $[0,1).$ +\endtf +Let $f: [0,1)\to(0,1)$ +where $f(0) = 1/2,$ and where $n\in\bb N^{\geq 2},$ +$f(1/n)={1\over n+1},$ and $f(x) = x$ otherwise. + +\tf Find a bijection between $(0,1)$ and $(0,1)\cup\bb N.$ +\endtf +Let $f: (0,1)\cup\bb N\to(0,1).$ + +If $n\in\bb N,$ $f(n) = {1\over 2^{n+1}},$ and for $m,l\in\bb N,$ +$f({1\over 2^m3^l}) = {1\over 2^m3^{l+1}},$ and $f(x) = x$ otherwise. + +\bye diff --git a/lacey/hw3.tex b/lacey/hw3.tex new file mode 100644 index 0000000..93a275a --- /dev/null +++ b/lacey/hw3.tex @@ -0,0 +1,339 @@ +\font\bigbf=cmbx12 at 24pt +\newfam\bbold +\font\bbten=msbm10 +\font\bbsev=msbm7 +\font\bbfiv=msbm5 +\textfont\bbold=\bbten +\scriptfont\bbold=\bbsev +\scriptscriptfont\bbold=\bbfiv +\def\bb#1{{\fam\bbold #1}} +\input color + +\newcount\pno +\def\tf{\advance\pno by 1\smallskip\bgroup\item{(\number\pno)}} +\def\endtf{\egroup\medskip} +\def\false{\centerline{TRUE\qquad{\color{blue} FALSE}}} +\def\true{\centerline{{\color{blue} TRUE}\qquad FALSE}} + +\newcount\qno +\long\def\question#1{\ifnum\qno=0\else\vfil\eject\fi\bigskip\pno=0\advance\qno by 1\noindent{\bf Question \number\qno.} {\it #1}} + +\def\proof{\medskip\noindent{\it Proof.}\quad} +\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt +depth 0pt{\parfillskip0pt\medskip}} + +\let\bu\bullet + +{\bigbf\centerline{Holden Rohrer}\bigskip\centerline{MATH 4317\qquad HOMEWORK \#3}} + +\question{Answer True or False. No justification is needed.} + +\tf +A sequence of reals that is bounded above converges. +\endtf + +\false + +\tf +A monotone sequence of reals converges. +\endtf + +\false + +\tf +A sequence of reals that is bounded above and is monotone converges. +\endtf + +\false + +\tf +If $\{x_n\}$ is a sequence, and $p_1=2, p_2=3, p_3=5,\ldots$ are the +primes listed in increasing order, then $\{x_{p_k}\}$ is a subsequence. +\endtf + +\true + +\tf +If $\{x_n\}$ is a sequence which converges, and +$p_1=2,p_2=3,p_3=5,\ldots$ are the primes listed in increasing order, +then $\{x_{p_k}\}$ converges. +\endtf + +\true + +\tf +If $\{x_n\}$ is a sequnce, then $\{x_{n+(-1)^n}\}$ is a subsequence. +\endtf + +\false + +\tf +The rationalso are closed under addition, multiplication, and division. +And a sequence of rationals that is convergent converges to a rational. +\endtf + +\false + +\tf +The supremum of the empty set is $\sup\emptyset = -\infty.$ +\endtf + +\true + +\tf +If non-empty set $S$ has supremum $\pi,$ then there are infinitely many +elements $s\in S$ with $\pi-1/100<s\leq\pi.$ +\endtf + +\false + +\question{Show directly, from the definition, and using elementary +methods that the three sequences below are Cauchy.} + +\item{(1)} $\{1/n : n\in\bb N\}$ + +\proof +Let $\epsilon > 0.$ +Choose $N \in\bb N$ such that $N > 1/\epsilon > 0.$ + +Let $m,n > N.$ +$$0 < 1/m, 1/n < 1/N < \epsilon.$$ + +Subtracting, +$$-\epsilon < 1/n - 1/m < \epsilon \Longrightarrow |1/n-1/m| < +\epsilon,$$ +giving us that the sequence is Cauchy. +\endproof + +\item{(2)} $\{{3n-1\over 2n+5} : n\in\bb N\}$ + +\proof +$$a_n := {3n-1\over 2n+5} = {3n+7.5-8.5\over 2n+5} = {3\over 2} - +{8.5\over 2n+5}.$$ + +Let $\epsilon > 0.$ +We choose $N = {4.25\over\epsilon}-2.5,$ and for all $n,m>N,$ +$$|a_n-a_m| = \left|\left({3\over 2} - {8.5\over 2n+5}\right) - +\left({3\over 2} - {8.5\over 2m+5}\right)\right| = \left|{8.5\over 2n+5} +- {8.5\over 2m+5}\right| < {8.5\over 2N+5} = \epsilon,$$ +with the inequality found by maximizing the absolute value variable (we +might also show this inequality by starting from $n,m > N$ and +rearranging). +%TODO: may be worth improving + +We have shown the sequence is Cauchy. +\endproof + +\item{(3)} Let $\{s_n\}$ be any sequence of digits with +$s_n\in\{0,1,2\}$ for all integers $n.$ Show that the sequence $\{x_n\}$ +is Cauchy, where $x_n = \sum_{m=1}^n {s_m\over 3^m}.$ + +\proof +Let $\epsilon > 0.$ +Pick $N$ such that $3^{-N} < \epsilon.$ +We will show that $|x_n-x_m| < \epsilon$ for all $n,l > N.$ +$$x_n = \sum_{m=1}^n {s_m\over 3^m} = \sum_{m=1}^N {s_m\over 3^m} + +\sum_{m=N}^n {s_m\over 3^m} \leq \sum_{m=1}^N {s_m\over 3^m} + +\sum_{m=1}^N {2\over 3^m} = \sum_{m=1}^N {s_m\over 3^m} + 3^{-N} - +3^{-n} \leq \sum_{m=1}^N {s_m\over 3^m} + 3^{-N},$$ +and, eliminating the second term, we also get $x_n \geq \sum_{m=1}^N.$ +Taking these together, +$|x_m-x_n| \leq 3^{-N} = \epsilon.$ + +\endproof + +\question{} + +\item{(1)} Show that the sequence below converges + +$$\sqrt 2, \sqrt{2+\sqrt 2}, \sqrt{2 + \sqrt{2 + \sqrt 2}}, \ldots$$ + +\proof +Let $a_1 = \sqrt 2$ and for $n\geq 1,$ that $a_{n+1} = \sqrt{2+a_n}.$ + +We will show by induction that it is bounded by 2 and, later, that it is +monotonically increasing. + +Clearly, $\sqrt 2 < 2,$ so $a_1 < 2.$ +For $k\geq 1,$ we assume $a_k < 2,$ then $2+a_k < 4,$ so $\sqrt{2+a_k} < +2,$ or in other words, $a_{k+1} < 2.$ +We have shown the sequence is bounded above. + +$0 < a_k < 2,$ so $a_k-2 < 0$ and $a_k+1 > 0.$ +Therefore, +$$(a_k-2)(a_k+1) = a_k^2 - a_k - 2 < 0,$$ +Rearranging, and noting that both sides are positive, so the square root +function can be used and is monotonically increasing so as not to change +the direction of the inequality, +$$a_k < \sqrt{a_k+2} = a_{k+1},$$ showing that our sequence is +strictly monotonically increasing, so the set is bounded below by the +first element. + +By the Monotone Convergence Theorem, the sequence converges. +\endproof + +\item{(2)} Does the sequence below converge? If so, to what? + +$$\sqrt 2, \sqrt{2\sqrt 2}, \sqrt{2\sqrt{2\sqrt 2}}, \ldots$$ + +\proof +Yes, this sequence converges to 2 (heuristically, $\sqrt{2x} = x$ has +roots 0 and 2, and all elements are positive) + +We will show it is monotone, bounded by 2, and that the limit of the set +cannot be less than 2. +Let $x_1 = \sqrt 2$ and $x_n = \sqrt{2x_{n-1}}$ otherwise. + +$x_1 = \sqrt 2 < 2,$ and assuming $x_k < 2,$ then $2x_k < 4,$ and +$\sqrt{2x_k} = x_{k+1} < 2.$ +This shows the sequence is bounded by 2. + +All elements are positive. +Specifically, $x_1 = \sqrt 2 > 0,$ and $\sqrt{2x_n}>0$ where $x_n>0,$ giving +the inductive step. +We have also already shown $2 > x_n,$ so $2x_n > x_n^2,$ so finally +$\sqrt{2x_n} = x_{n+1} > x_n,$ giving monotonicity. + +To show the limit cannot be less than 2, assume for the sake of +contradiction that the limit is $1.5 < 2-\mu < 2$ (or, phrased +otherwise, $0 < \mu < .5$) + +By the definition of convergence, we may find $x_n > 2-1.1\mu.$ +We will show $\sqrt{2x_n} = x_{n+1} > 2-\mu.$ +Note that $\mu < .9,$ so $.9-\mu > 0,$ and $\mu > 0.$ +Therefore, +$$(.9-\mu)\mu = .9\mu-\mu^2 > 0 \Rightarrow 2-1.1\mu > 2-2\mu+\mu^2/2 +\Rightarrow 2(2-1.1\mu) > (2-\mu)^2 \Rightarrow \sqrt{2(2-1.1\mu)} = +x_{n+1} > 2-\mu.$$ +we have thus shown that the sequence may exceed the limit, instructing +us that the limit is 2. + +\endproof + +\question{} + +\item{(1)} In Section 1.4, we used the Axiom of Completeness (AoC) to +prove the Archimedean Property of $\bb R$ (Theorem 1.4.2). Show that the +Monotone Convergence Theorem can also be used to prove the Archimedean +Property without making any use of AoC. + +\proof + +The sequence $a_n = n$ is monotone. + +We will show the Archimedean Property by contradiction. +Let there be $x\in\bb R$ such that there is no $n\in\bb N$ satisfying +$n>x.$ +Therefore, $a_n$ is bounded and converges to some number $a < n$ by the +Monotone Convergence Theorem. +Let $0 < \epsilon < 1/2.$ + +From the definition of convergence, there exists $N$ such that for all +$m\geq N,$ that $|a-a_N| < \epsilon.$ +Rewriting, $a_N-\epsilon < a < a_N+\epsilon,$ and from our inequality +about $\epsilon,$ +$$a < a_N+1/2 \to a < a_{N+1}-1/2,$$ +which contradicts our convergence statement. + +\endproof + +\item{(2)} Use the Monotone Convergence Theorem to supply a proof for +the Nested Interval Property (Theorem 1.4.1) that doesn't make use of +AoC. These two results suggest that we could have used the Monotone +Convergence Theorem in place of AoC as our starting axiom for building a +proper theory of the real numbers. + +\proof +Let $I_i = [a_i,b_i]$ be a sequence of nonempty closed intervals such +that $$I_1\supseteq I_2 \supseteq I_3 \supseteq \cdots.$$ +$a_i$ is a weakly monotone increasing sequence because if $a_{i+1}<a_i,$ +then $a_{i+1}\in I_{i+1},$ but $a_{i+1}\not\in I_i,$ violating the +interval condition. +It is clearly bounded by $b_1$ because if $a_i > b_1,$ then +$I_i\not\subseteq I_1,$ and $I_1$ is nonempty. + +The Monotone Convergence Theorem tells us that $a_i$ converges to some +$x\geq a_i$ because $a_i$ is strictly increasing. + +For the sake of contradiction, let there be $b_k < x.$ +Let $\epsilon = x-b_k.$ +Because $a_i$ converges to $x,$ there must be for some $i>k,$ an $a_i > +x-\epsilon = b_k,$ violating the definitions of our intervals. +We have thus obtained $x\in [a_i, b_i],$ for all $i\in\bb N,$ and +therefore in the intersection of these intervals. +\endproof + +\question{Exercise 2.4.7 (Limit Superior). Let $(a_n)$ be a bounded +sequence.} + +\item{a)} Prove that the sequence defined by $y_n = \sup\{a_k : k\geq +n\}$ converges. + +\proof +$y_n \geq y_{n+1},$ because $\{a_k : k\geq n\}\supseteq\{a_k : k \geq +n+1\},$ so any upper bound for the former set is an upper bound for the +latter. +It is monotone and bounded by whatever bound $(a_n)$ is bounded by, so +it must converge by the Monotone Convergence Theorem. +\endproof + +\item{b)} The limit superior of $(a_n)$ or $\limsup a_n,$ is defined +by $\limsup a_n = \lim y_n,$ where $y_n$ is the sequence from part +$(a)$ of this exercise. Provide a reasonable definition for $\liminf +a_n$ and briefly explain why it always exists for any bounded sequence. + +\proof +We can define $$\liminf a_n := -\limsup (-a_n).$$ +If $|a_n| < b$ for some $b,$ then $|-a_n| = |a_n| < b$ too. +Clearly, this exists because $(-a_n)$ is also a bounded sequence where +$\limsup$ exists. + +Also, this is equivalent to the more intuitive definition +$$\liminf a_n := \lim(\inf\{a_k : k\geq n\})$$ because the lower bound of that set maps +onto the upper bound of its negation. +This would also exist because the infimum would be monotonically +increasing and bounded in the same way. +\endproof + +\item{c)} Prove that $\liminf a_n \leq \limsup a_n$ for every bounded +sequence, and give an example of a sequence for which the inequality is +strict. + +\proof +For all sets of bounded numbers $S_k = \{a_k : k\geq n\},$ +we know $\inf S_k \leq \sup S_k$ (the lower bound must be below the +upper bound). +By the Nested Interval Property, and because the infimum and supremum +should create a set of nested intervals (they are increasing and +decreasing respectively), we should have some $x\in [\liminf a_n, +\limsup a_n],$ meaning $\liminf a_n \leq \limsup a_n.$ +The inequality is strict for the sequence $(0)_{n\in\bb N}.$ +\endproof + +\item{d)} Show that $\liminf a_n = \limsup a_n$ if and only if $\lim +a_n$ exists. In this case, all three share the same value. + +\proof +$(\Rightarrow)$ +Let $x := \liminf a_n = \limsup a_n.$ +Let $\epsilon > 0.$ + +Using the definition of convergence, for some $N,$ +$$|\inf\{a_k : k \geq N\} - x| < \epsilon \Rightarrow x > \inf\{a_k : +k\geq N\}-\epsilon \Rightarrow x > a_j-\epsilon,$$ +for all $j\geq N.$ +Similarly, we can obtain $a_j+\epsilon > x.$ +This means for all $j>N,$ the sequence converges as $|a_j-x| < +\epsilon,$ so $\lim a_n = x$ exists. + +$(\Leftarrow)$ +Let $\lim a_n = x$ exist. +Let $\epsilon > 0.$ +We must have $N$ such that for all $n > N,$ $|a_n - x| < \epsilon.$ +That means $$\limsup a_n \leq \sup\{a_n : n > N\} < x+\epsilon,$$ and a +similar lower limit for $\liminf.$ +$|\limsup a_n - \liminf a_n| < 2\epsilon,$ and by the "give me some +room" principle, $\limsup a_n = \liminf a_n = x.$ +\endproof + +\bye diff --git a/lacey/hw4i.tex b/lacey/hw4i.tex new file mode 100644 index 0000000..767c62b --- /dev/null +++ b/lacey/hw4i.tex @@ -0,0 +1,138 @@ +\font\bigbf=cmbx12 at 24pt +\newfam\bbold +\font\bbten=msbm10 +\font\bbsev=msbm7 +\font\bbfiv=msbm5 +\textfont\bbold=\bbten +\scriptfont\bbold=\bbsev +\scriptscriptfont\bbold=\bbfiv +\def\bb#1{{\fam\bbold #1}} +\input color + +\newcount\pno +\def\tf{\advance\pno by 1\smallskip\bgroup\item{(\number\pno)}} +\def\endtf{\egroup\medskip} +\def\false{\centerline{TRUE\qquad{\color{blue} FALSE}}} +\def\true{\centerline{{\color{blue} TRUE}\qquad FALSE}} + +\newcount\qno +\long\def\question#1{\ifnum\qno=0\else\vfil\eject\fi\bigskip\pno=0\advance\qno by 1\noindent{\bf Question \number\qno.} {\it #1}} + +\def\intro#1{\medskip\noindent{\it #1.}\quad} +\def\proof{\intro{Proof}} +\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt +depth 0pt{\parfillskip0pt\medskip}} + +\let\bu\bullet + +{\bigbf\centerline{Holden Rohrer}\bigskip\centerline{MATH 4317\qquad +HOMEWORK \#4, Part I}} + +\question{Answer True or False. No justification is needed.} + +\tf +If $\lim a_n = 0$ and $\{b_n\}$ is a bounded sequence, then $a_nb_n$ is +a convergent sequence. +\endtf + +\true + +\tf +If $\limsup_n a_n = 1$ and $\rho>1,$ then there is an $N>1$ such that +$a_n<\rho$ for all $n>N.$ +\endtf + +\true + +\tf +If $(a_n)$ is a Cauchy sequence, then $|a_n|$ is a Cauchy sequence. +\endtf + +\true + +\tf +If $(a_n)$ is a Cauchy sequence, then $\lfloor a_n\rfloor$ is a Cauchy +sequence. +\endtf + +\false + +\tf +If $(a_n)$ is a Cauchy sequence, then $\arctan(a_n)$ is a Cauchy +sequence. +\endtf + +\true + +\question{(Exercise 2.6.5) Consider the following (invented) definition: +a sequence $(s_n)$ is pseudo-Cauchy if, for all $\epsilon>0,$ there +exists an $N$ such that if $n>N,$ then $|s_n-s_{n+1}|<\epsilon.$ Supply +a proof for the valid statements and a counterexample for the false +statements.} + +\item{a.} Pseudo-Cauchy sequences are bounded. + +\intro{Disproof} +False. $s_n = \log n$ is clearly unbounded, but for all $\epsilon > 0,$ +we can choose some $N > 1/\epsilon.$ + +For all $n > N > 1/\epsilon,$ (or $\epsilon > 1/n$), starting from a +result of the definition of the exponential, +$$e^{1/n} > 1+1/n = (n+1)/n \Longrightarrow +\epsilon > 1/n > \log(n+1) - \log(n) = |s_n-s_{n+1}|.$$ + +\endproof + +\item{b.} Bounded Pseudo-Cauchy sequences are convergent. + +\intro{Disproof} +Consider $s_n = \log n,$ but modulated such that it ``bounces off'' the +bounds of the interval $(0,1),$ or more directly, +$s_n = |\log n\bmod 2-1|.$ +\endproof + +\item{c.} If $(x_n)$ and $(y_n)$ are pseudo-Cauchy, then $(x_n+y_n)$ is +Pseudo-Cauchy as well. + +\proof +Let $\epsilon > 0.$ We must have $N$ and $M$ such that for all $n > +\max\{N,M\},$ for both sequences, $|x_n-x_{n+1}|<\epsilon/2$ and +$|y_n-y_{n+1}|<\epsilon/2.$ +By the triangle inequality, +$$|x_n+y_n-x_{n+1}-y_{n+1}|<|x_n-x_{n+1}|+|y_n-y_{n+1}|<\epsilon.$$ +\endproof + +\question{Give an example of each or explain why the request is +impossible referencing the proper theorem(s).} + +\item{a.} Two series $\sum x_n$ and $\sum y_n$ that both diverge but +where $\sum x_ny_n$ converges. + +\intro{Example.} +Let $x_n = 1/n$ and $y_n = -1.$ +\endproof + +\item{b.} A convergent series $\sum x_n$ and a bounded sequence $(y_n)$ +such that $\sum x_n y_n$ diverges. + +\intro{Example.} +Let $x_n = (-1)^n 1/n$ and $y_n = (-1)^n.$ +\endproof + +\item{c.} Two sequences $(x_n)$ and $(y_n)$ where $x_n$ and $(x_n+y_n)$ +both converge but $y_n$ diverges. + +\intro{Disproof.} +If $x_n$ and $(x_n+y_n)$ converge, their difference should also +converge, their difference being $x_n+y_n-x_n = y_n.$ +This is implied from the Algebraic Limit Theorem for Series. +\endproof + +\item{d.} A sequence $(x_n)$ satisfying $0<x_n\leq 1/n$ where $\sum_n +(-1)^n x_n$ diverges. + +\intro{Disproof.} +This converges by the Alternating Series Test. +\endproof + +\bye diff --git a/lacey/hw4ii.tex b/lacey/hw4ii.tex new file mode 100644 index 0000000..e69f9da --- /dev/null +++ b/lacey/hw4ii.tex @@ -0,0 +1,110 @@ +\font\bigbf=cmbx12 at 24pt +\newfam\bbold +\font\bbten=msbm10 +\font\bbsev=msbm7 +\font\bbfiv=msbm5 +\textfont\bbold=\bbten +\scriptfont\bbold=\bbsev +\scriptscriptfont\bbold=\bbfiv +\def\bb#1{{\fam\bbold #1}} +\input color + +\newcount\pno +\def\tf{\advance\pno by 1\smallskip\bgroup\item{(\number\pno)}} +\def\endtf{\egroup\medskip} +\def\false{\centerline{TRUE\qquad{\color{blue} FALSE}}} +\def\true{\centerline{{\color{blue} TRUE}\qquad FALSE}} + +\newcount\qno +\long\def\question#1{\ifnum\qno=0\else\vfil\eject\fi\bigskip\pno=0\advance\qno by 1\noindent{\bf Question \number\qno.} {\it #1}} + +\def\intro#1{\medskip\noindent{\it #1.}\quad} +\def\proof{\intro{Proof}} +\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt +depth 0pt{\parfillskip0pt\medskip}} + +\let\bu\bullet + +{\bigbf\centerline{Holden Rohrer}\bigskip\centerline{MATH 4317\qquad +HOMEWORK \#4, Part II}} + +\question{Give an example of each of the following, or argue that such a +request is impossible.} + +\item{a.} A sequence that has a subsequence that is bounded but contains +no subsequence that converges. + +\intro{Disproof} +This is impossible. +Let $(a_n)$ be a sequence bounded by $M.$ + +We will show by contradiction that for any $\epsilon > 0,$ there must be +a susbequence $(a_{\phi(n)})$ such that $|a_n-l|<\epsilon$ for some $l.$ + +First, we break the interval $[-M,M] \supseteq \{a_n\}$ into the +partition $$[-M,-M+\epsilon)\cup +[-M+\epsilon,-M+2\epsilon)\cup\cdots\cup [M-\epsilon,M] = +S_1\cup S_2\cup\cdots\cup S_n.$$ +For contradiction, we assume that there is no subsequence meeting the +Cauchy criterion for $\epsilon.$ + +This means $(a_n)\cap S_i$ is finite for all $S_i$ because otherwise +these would be subsequences satisfying the Cauchy Criterion for +$\epsilon.$ +$$S_1\cup\ldots\cup S_n = [-M,M],$$ +so $[-M,M]\cap (a_n) = (a_n)$ has a length less than or equal to the sum +of lengths across $S_i\cap (a_n),$ which makes it finite, but the +sequence is infinite, giving a contradiction. + +\endproof + +\vfil\eject +\item{b.} A sequence that does not contain $0$ or $1$ as a term but +contains subsequences converging to each of these values. + +\proof +Let $$a_n = 1/2 + (-1)^n{n\over 2(n+1)}.$$ +\endproof + +\vfil\eject +\item{c.} A sequence that contains subsequences converging to every +point in the infinite set $\{1, 1/2, 1/3, 1/4, 1/5,\ldots\},$ and no +subsequences converging to points outside of this set. + +\intro{Disproof} +Such a sequence would necessarily contain a subsequence converging to +$0.$ + +Let $\epsilon > 0.$ +We can easily choose some element in the set $0<s = 1/n < \epsilon/2.$ +Then, because we have a subsequence converging to this point, for some +$N,$ we get $$|a_N - s| < \epsilon/2 \Longrightarrow s-\epsilon/2 < a_N +< s+\epsilon/2.$$ +We already know $0 < s < \epsilon/2,$ so $|s-\epsilon/2| < \epsilon/2 < +|s+\epsilon/2| < \epsilon,$ and $|a_N| < +\max\{|s-\epsilon/2|,|s+\epsilon/2|\} < \epsilon,$ showing that we must +have a subsequence converging to $0.$ +\endproof + +\vfil\eject +\item{d.} A sequence that contains subsequences converging to every +point in the infinite set below, and no subsequences converging to +points outside of this set. + +$$\{2^{-m} + 2^{-n} : 1\leq m < n\} = \bigcup_{m=1}^\infty \{2^{-m} + +2^{-m-1}, 2^{-m}+2^{-m-2},\ldots\}.$$ + +\intro{Disproof} +For a very similar reason as (c), we cannot have such a sequence +containing such subsequences without containing their limit point $0$ +(and possibly others). + +We will dispense with the long $\epsilon$ proof and simply note that we +must have some $m$ such that $2^{-m+1}<\epsilon/4,$ and +$2^{-m}+2^{-{m+1}} = 3(2^{-m+1})<3\epsilon/4,$ and choosing a point +$a_N$ from +our sequence within $\epsilon/4$ because of the subsequence already +established, we again get an $a_N<\epsilon,$ showing it converges to +$0,$ which is outside the set. + +\bye diff --git a/lacey/hw5.tex b/lacey/hw5.tex new file mode 100644 index 0000000..08b8638 --- /dev/null +++ b/lacey/hw5.tex @@ -0,0 +1,130 @@ +\font\bigbf=cmbx12 at 24pt +\newfam\bbold +\font\bbten=msbm10 +\font\bbsev=msbm7 +\font\bbfiv=msbm5 +\textfont\bbold=\bbten +\scriptfont\bbold=\bbsev +\scriptscriptfont\bbold=\bbfiv +\def\bb#1{{\fam\bbold #1}} +\input color + +\newcount\pno +\def\tf{\advance\pno by 1\smallskip\bgroup\item{(\number\pno)}} +\def\endtf{\egroup\medskip} +\def\absolute{\centerline{{\color{blue} Absolute}\qquad Conditional\qquad Divergent}} +\def\conditional{\centerline{Absolute\qquad{\color{blue} Conditional}\qquad Divergent}} +\def\divergent{\centerline{Absolute\qquad Conditional\qquad{\color{blue} Divergent}}} + +\newcount\qno +\long\def\question#1{\ifnum\qno=0\else\vfil\eject\fi\bigskip\pno=0\advance\qno by 1\noindent{\bf Question \number\qno.} {\it #1}} + +\def\intro#1{\medskip\noindent{\it #1.}\quad} +\def\proof{\intro{Proof}} +\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt +depth 0pt{\parfillskip0pt\medskip}} + +\let\bu\bullet + +{\bigbf\centerline{Holden Rohrer}\bigskip\centerline{MATH 4317\qquad +HOMEWORK \#5}} + +\question{Are the series absolutely convergent, conditionally +convergent, or divergent? No proof required. (Cauchy Criteria can help +on some of these.)} + +\tf +$$\sum_n {\sin n\over n^2}$$ +\endtf + +\absolute + +\tf +$$\sum_n {(-1)^n\over n\log n}$$ +\endtf + +\conditional + +\tf +$${3\over 4}-{4\over 6}+{5\over 8}-{6\over 10}+{7\over 12}-\cdots$$ +\endtf + +\divergent + +\tf +$$\sum_n {(-1)^n\over n(\log n)^2}$$ +\endtf + +\conditional + +\question{Show that the series $$\sum_n {(-1)^{n+1}\over n} = 1-{1\over +2}+{1\over 3}-{1\over 4}+\cdots$$ is convergent. (This is the simplest +instance of the Alternating Series Test (Thm 2.7.7). But you can't just +cite that, since the Theorem is not proved in the text. You can follow +any of the proof strategies outlined in Exercise 2.7.1. Or come up with +your own, tailored to this particular example.} + +\proof +We start by stating $$s_n = 1-{1\over 2}+{1\over 3}-{1\over 4}+\cdots\pm +{1\over n}.$$ +Next, we calculate, for $n\geq 2k,$ $$s_n-s_{2k} = {1\over 2k+1} - +{1\over 2k+2} +\cdots \pm {1\over n} = {1\over (2k+1)(2k+2)} + \cdots +\pm {1\over n} > -{1\over n} > -{1\over 2k}.$$ +By similar reasoning, $s_n-s_{2k+1}<{1\over 2k}.$ + +Therefore, $s_{2k}-{1\over 2k} < s_n < s_{2k+1}+{1\over 2k},$ +and since $|s_{2k+1}+{1\over 2k}-s_{2k}+{1\over 2k}| \leq {3\over 2k},$ +we can show for both that +$$|s_n-s_{2k}| \leq {3\over 2k}\qquad |s_n-s_{2k+1}| \leq +{3\over 2k}.$$ +This means the sequence is Cauchy and thus convergent. + +\endproof + +\question{(2.7.9, Ratio Test) Given a series $\sum_n a_n,$ with $a_n +\neq 0,$ the Ratio Test states that if +$$\limsup_n \left|{a_n\over a_{n+1}}\right| = r < 1,$$ +then the series $$\sum_n a_n$$ converges absolutely. Take these steps to +verify this statement.} + +\item{a.} Show that for $r<s<1,$ there is $N$ so that for all $n\geq +N,$ we have +$$\left| {a_n\over a_{n+1}} \right| < s.$$ + +\proof +The definition of $\limsup$ tells us that for some $N,$ that for all +$n\geq N,$ +$$\left| {a_{n+1}\over a_n} \right| \leq r < s.$$ +\endproof + +\item{b.} Conclude that for $n > N,$ that $|a_n| < |a_N|s^{n-N}.$ + +\proof +We will show this by induction. + +Note that $$\left|{a_{n+1}\over a_n}\right| = {|a_{n+1}|\over |a_n|}.$$ +(This can be shown by a sort of casework: the signs of $a_n$ and +$a_{n+1}$ can be independently flipped without affecting the expression, +and the positive case is trivial) + +Using the same value of $N$ as in the previous question, we show the +base case for $n = N+1,$ +$${|a_{N+1}|\over |a_N|} < s \Longrightarrow |a_n| < s|a_N| +\Longrightarrow |a_n| < |a_N|s^{n-N}.$$ +For the inductive step, we assume for $n > N,$ that $|a_n| < +s^{n-N}|a_N|,$ and then use ${|a_{n+1}|\over |a_n|} < s$ to show $|a_{n+1}| +< s|a_n| < s^{n-N+1}|a_N|,$ completing the inductive step. +\endproof + +\item{c.} Conclude that $\sum_n a_n$ converges absolutely. + +\proof +Taking $N$ again as in previous questions, +$$\sum_{n>N} |a_n| < \sum_{n>N} |a_N|s^{n-N} = {|a_N|\over 1-s},$$ and +since the sequence is positive, the partial series are monotone +increasing, giving convergence by the Monotone Convergence Theorem. + +The geometric sum is from the book. +\endproof + +\bye diff --git a/lacey/hw6.tex b/lacey/hw6.tex new file mode 100644 index 0000000..7b941ca --- /dev/null +++ b/lacey/hw6.tex @@ -0,0 +1,187 @@ +\font\bigbf=cmbx12 at 24pt +\newfam\bbold +\font\bbten=msbm10 +\font\bbsev=msbm7 +\font\bbfiv=msbm5 +\textfont\bbold=\bbten +\scriptfont\bbold=\bbsev +\scriptscriptfont\bbold=\bbfiv +\def\bb#1{{\fam\bbold #1}} +\input color + +\newcount\pno +\def\tf{\advance\pno by 1\smallskip\bgroup\item{(\number\pno)}\par} +\def\endtf{\egroup\medskip} +\def\false{\centerline{TRUE\qquad{\color{blue} FALSE}}} +\def\true{\centerline{{\color{blue} TRUE}\qquad FALSE}} + +\newcount\qno +\long\def\question#1{\ifnum\qno=0\else\vfil\eject\fi\bigskip\pno=0\advance\qno by 1\noindent{\bf Question \number\qno.} {\it #1}} + +\def\intro#1{\medskip\noindent{\it #1.}\quad} +\def\proof{\intro{Proof}} +\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt +depth 0pt{\parfillskip0pt\medskip}} + +\let\bu\bullet + +{\bigbf\centerline{Holden Rohrer}\bigskip\centerline{MATH 4317\qquad +HOMEWORK \#6}} + +\question{(3 points each) Answer True or False. No justification +needed.} + +\item{a)} A set can be both open and closed. +\tf +\true +\endtf + +\item{b)} For a set $A\subset\bb R,$ the set of limit points of $A$ is closed. +\tf +\true +\endtf + +\item{c)} The Cantor set is uncountable. +\tf +\true +\endtf + +\item{d)} Make a `thin' Cantor set by starting with $[0,1],$ remove the middle +$2/3.$ From the two intervals remaining, remove the middle $3/4.$ From +the four remaining intervals the middle $4/5,$ and so on. The remaining +set is uncountable. +\tf +\true +\endtf + +\item{e)} If $\sum_n a_n$ converges absolutely, then $\sum_n a_n^2$ converges +absolutely. +\tf +\true +\endtf + +\item{f)} A countable subset of $\bb R$ must have a limit point. +\tf +\false +\endtf + +\item{g)} An uncountable subset of $\bb R$ must have a limit point. +\tf +\true +\endtf + +\question{Decide whether the following statements are true or false. +Provide counterexamples for those that are false, and supply proofs for +those that are true.} + +\item{a)} An open set that contains every rational number must +necessarily be all of $\bb R.$ + +\intro{Disproof} +False. Let us have the bijection $\phi:\bb N\to\bb Q.$ +$$\bigcup_{n=0}^\infty (\phi(n)-2^{-n},\phi(n)+2^n)$$ +is a union of arbitrary (countably many) open intervals, and it must +contain every rational number, but their length is clearly less than the +length of the reals (the sum of their lengths, which is greater than the +length of their union, is 4). +\endproof + +\item{b)} The Nested Interval Property remains true if the term ``closed +interval'' is replaced by ``closed set.'' + +\proof{Disproof} +False. The integers are a closed set, but the decreasing subsets +$A_n = \bb N\setminus\{1,\ldots, n\}$ will not limit to one number. +\endproof + +\item{c)} Every nonempty open set contains a rational number. + +\proof +True. Let $x\in A$ where $A$ is an open set. For some $\epsilon>0,$ +$V_\epsilon(x)\subseteq A,$ and there must be $y\in\bb Q\cap +V_\epsilon(x)$ by density of the rationals in the reals. +$y\in A,$ so every nonempty open set contains a rational number. +\endproof + +\item{d)} Every bounded infinite closed set contains a rational number. + +\intro{Disproof} +$\{(1-2^{-n})\sqrt2 : n\in\bb N\}\cup \{\sqrt2\}$ contains its limit +point ($\sqrt2$), and it is infinite and bounded by $\sqrt 2,$ but it +does not contain a rational number since a rational times an irrational +is irrational. +\endproof + +\question{Given $A\subset\bb R,$ let $L$ be the set of all limit points +of $A.$} + +\item{a)} Show that the set $L$ is closed. + +\proof +Let $l$ be a limit point of $L.$ +We will show $l$ is a limit point of $A,$ and thus $l\in L.$ + +Let $\epsilon > 0.$ + +By definition of limit point, $V_{\epsilon/2}(l)\cap L$ isn't empty. +We choose $x$ from that set. +Again, by definition of limit point, $V_{\epsilon/2}(x)\cap A$ isn't +empty. +We have thus shown, by the triangle inequality, that $V_\epsilon(l)\cap +A$ isn't empty, so $l\in L,$ meaning $L$ contains all of its limit +points and is therefore closed. +\endproof + +\item{b)} Argue that if $x$ is a limit point of $A\cup L,$ then $x$ is a +limit point of $A.$ + +\proof +Let $l$ be a limit point of $A\cup L.$ + +Then, for all $\epsilon > 0,$ there must be a distinct $x\in +V_\epsilon(l)\cap(A\cup L).$ +If $x\in A,$ then we have shown that $x$ is a limit point of $A.$ +If $x\in L,$ then we can reuse the theorem in the last answer, so $x$ is +a limit point of $A.$ +\endproof + +\question{A set $A$ is called an $F_\sigma$ set if it can be written as +the countable union of closed sets. A set $B$ is called a $G_\delta$ set +if it can be written as the countable intersection of open sets.} + +\item{(1)} Show that a closed interval $[a,b]$ is a $G_\delta$ set. + +\proof +Let $A_n = (a-1/n,b+1/n).$ +All of these intervals contain $[a,b],$ but they do not contain any +element greater than $b$ or less than $a,$ so their intersection is +$[a,b],$ proving that it is a $G_\delta$ set. +\endproof + +\item{(2)} Show that the half-open interval $(a,b]$ is both a $G_\delta$ +and an $F_\sigma$ set. + +\proof +Let $A_n = (a,b+1/n).$ +The intersection of these sets is $(a,b],$ so it is a $G_\delta$ set. + +Let $B_n = [a+1/n,b].$ +The union of these sets is $(a,b],$ so it is a $F_\sigma$ set. +\endproof + +\item{(3)} Show that $\bb Q$ is an $F_\sigma$ set, and the set of +irrationals $\bb I$ forms a $G_\delta$ set. (We will see in section 3.5 +that $\bb Q$ is not a $G_\delta$ set, nor is $\bb I$ an $F_\sigma$ +set.) + +\proof +We know $\bb Q$ is countable, so we take the union of $\{q\}$ over every +$q\in\bb Q,$ generating a countable union, so $\bb Q$ is an $F_\sigma$ +set. + +The complement of a closed set is open, so $\bb R\setminus\{q\}$ is an +open set, and the intersection of these must exclude every rational +number, constructing the set of irrationals as a $G_\delta$ set. +\endproof + +\bye diff --git a/lacey/hw7.tex b/lacey/hw7.tex new file mode 100644 index 0000000..a846350 --- /dev/null +++ b/lacey/hw7.tex @@ -0,0 +1,153 @@ +\font\bigbf=cmbx12 at 24pt +\newfam\bbold +\font\bbten=msbm10 +\font\bbsev=msbm7 +\font\bbfiv=msbm5 +\textfont\bbold=\bbten +\scriptfont\bbold=\bbsev +\scriptscriptfont\bbold=\bbfiv +\def\bb#1{{\fam\bbold #1}} +\input color + +\newcount\pno +\def\tf{\advance\pno by 1\smallskip\bgroup\item{(\number\pno)}\par} +\def\endtf{\egroup\medskip} +\def\false{\centerline{TRUE\qquad{\color{blue} FALSE}}} +\def\true{\centerline{{\color{blue} TRUE}\qquad FALSE}} + +\newcount\qno +\long\def\question#1{\ifnum\qno=0\else\vfil\eject\fi\bigskip\pno=0\advance\qno by 1\noindent{\bf Question \number\qno.} {\it #1}} + +\def\intro#1{\medskip\noindent{\it #1.}\quad} +\def\proof{\intro{Proof}} +\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt +depth 0pt{\parfillskip0pt\medskip}} + +\let\bu\bullet + +{\bigbf\centerline{Holden Rohrer}\bigskip\centerline{MATH 4317\qquad +HOMEWORK \#7}} + +\question{(3 points each) Answer True or False. No justification +needed.} + +\item{a)} The arbitrary intersection of compact sets is compact. +\tf +\true +\endtf + +\item{b)} The arbitrary union of compact sets is compact. +\tf +\false +\endtf + +\item{c)} Let $A$ be arbitrary, and let $K$ be compact. Then, the +intersection $A\cap K$ is compact. +\tf +\false % open set \subset closed bounded set +\endtf + +\item{d)} If $F_1\supset F_2\supset F_3\supset F_4\supset\cdots$ is a +nested sequence of nonempty closed sets, then the intersection $\cap_n +F_n$ is not empty. +\tf +\false +\endtf + +\item{e)} The intersection of a perfect set and a compact set is +perfect. +\tf +\false +\endtf + +\item{f)} The rationals contain a perfect set. +\tf +\true % empty set +\endtf + +\question{(3.3.10) Let $K = [a,b]$ be a closed interval. Let ${\cal O} = +\{O_\lambda | \lambda\in \Lambda\}$ be an open cover for $[a,b].$ Show +that there is a finite subcover using this strategy: define $S$ to be +the set of all $x\in [a,b]$ such that $[a,x]$ has a finite subcover from +${\cal O}.$} + +\item{a)} Argue that $S$ is nonempty and bounded, and thus $s = \sup S$ +exists. + +\proof +Let $x = a.$ +We know that $[a,a] = \{a\}\subseteq [a,b],$ so there must exist, in +${\cal O},$ an $O_\lambda$ such that $a\in O_\lambda.$ +Taking $\{O_\lambda\}$ as the subcover, we have shown that $S$ is +nonempty. + +From the definition, $S\subseteq [a,b],$ so $|S| \leq \max\{|a|,|b|\},$ +making $S$ bounded. +Therefore, $s = \sup S$ exists. +\endproof + +\item{b)} Show that $s = b.$ Explain why this implies ${\cal O}$ admits +a finite subcover. That is, every closed interval is compact. + +\proof +From the definition $S\subseteq [a,b],$ and for all $y\in[a,b],$ we know +$y \leq b,$ so $s\leq b.$ + +For the sake of contradiction, assume $s < b.$ +From the open cover, we know there must be an open interval +$V_\epsilon(s)$ containing $s,$ with some size $\epsilon,$ and from the +definition of supremum, there must be $x\in [s-\epsilon,s)\cap S,$ so +by taking the finite cover of $x$ and appending $V_\epsilon(s),$ we +construct a finite cover which includes $y\in (s,s+\epsilon)\cap[a,b],$ +and this is nonempty since $s < b,$ and $y > s,$ giving us a +contradiction of our assumption about the supremum. + +This conclusively tells us that $s = b,$ so an open cover of $[a,b]$ +must admit a finite subcover. +This is the definition of compactness. +\endproof + +\item{c)} Use the fact that closed intervals are compact to conclude +that a set $F$ that is closed and bounded is compact. + +\proof +Let ${\cal O}$ be an open cover for $F.$ +We are given that $F$ is bounded by some $M,$ so we construct +$${\cal O}\cup\left\{\left(\bigcup_{\lambda\in\Lambda} +O_\lambda\right)^c\right\},$$ +The appended set is an open set because the complement of a closed set +is an open set. +We thus have an open cover for $[-M,M]\supseteq F,$ +and we have shown that $[-M, M]$ admits a finite subcover, so taking the +obtained finite subcover sans the appended set, we have a finite +subcover for $F.$ + +\endproof + +\question{We proved: the countable intersection of compact nonempty +$K_1\supset K_2\supset \cdots$ is not empty. We did so using a +diagonalization and the limit definition of compact sets. Prove this +theorem using the open cover definition of a compact set. (Assume +otherwise, and construct an open cover of $K_1$ without finite +subcover.)} + +\proof +Assume for the sake of contradiction that the intersection of these sets +is empty. +Since $K_i$ is closed (assuming we are working in a Hausdorff space), +$K_i^c$ is open, and $C = \{K_2^c,K_3^c,\ldots\}$ is an open cover, +since $\bigcup_{2\leq i} K_i^c = \left(\bigcap_{2\leq i} K_i\right)^c = +X,$ where $X$ is the space. + +This set is therefore an open cover of $K_1,$ but any finite subcover +${\cal O}$ must omit, for some $N,$ all $K_n$ where $n>N.$ +Let $x\in K_n.$ +For $m\leq n,$ $K_n\subseteq K_m,$ so $x\in K_m,$ and $x\not\in K_m^c,$ +so $x\not\in \bigcup_{o\in\cal O} o.$ +But $x\in K_1,$ contradicting the existence of a finite subcover from +this open cover. + +Therefore, $K_1$ is not compact, but that is a contradiction. +\endproof + +\bye diff --git a/lacey/hw8.tex b/lacey/hw8.tex new file mode 100644 index 0000000..adb730e --- /dev/null +++ b/lacey/hw8.tex @@ -0,0 +1,190 @@ +\font\bigbf=cmbx12 at 24pt +\newfam\bbold +\font\bbten=msbm10 +\font\bbsev=msbm7 +\font\bbfiv=msbm5 +\textfont\bbold=\bbten +\scriptfont\bbold=\bbsev +\scriptscriptfont\bbold=\bbfiv +\def\bb#1{{\fam\bbold #1}} +\input color + +\def\tf{\smallskip\bgroup\par} +\def\endtf{\egroup\medskip} +\def\false{\centerline{TRUE\qquad{\color{blue} FALSE}}} +\def\true{\centerline{{\color{blue} TRUE}\qquad FALSE}} +\def\clos{\mathop{\rm closure}} + +\newcount\qno +\long\def\question#1{\ifnum\qno=0\else\vfil\eject\fi\bigskip\advance\qno by 1\noindent{\bf Question \number\qno.} {\it #1}} + +\def\intro#1{\medskip\noindent{\it #1.}\quad} +\def\proof{\intro{Proof}} +\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt +depth 0pt{\parfillskip0pt\medskip}} + +\let\bu\bullet + +{\bigbf\centerline{Holden Rohrer}\bigskip\centerline{MATH 4317\qquad +HOMEWORK \#8}} + +\question{A set $E$ is totally disconnected if, given any two distinct +points $x,y\in E,$ there exists sets $A$ and $B$ with $x\in A,$ $y\in +B,$ $E=A\cup B,$ and $\clos(A)\cap B = +A\cap\clos(B) = \emptyset.$ The rationals are a basic example +of a totally disconnected set. + +Show that the Cantor set $C$ is totally disconnected. You can use either +strategy below: +{\leftskip.5in +\item{$\bu$} As suggested in Exercise 3.4.8: $C = \bigcap_n C_n$ where +$C_n$ is a union of $2^n$ closed intervals, of length $3^{-n}.$ And any +two distinct intervals are a distance of at least $3^{-n}$ apart. +\item{$\bu$} The ternary expansion property: +$$C = \left\{\sum_{n=1}^\infty s_n 3^{-n} : s_n = 0,2\right\}.$$ +Given $x\neq y\in C,$ use the first ternary digit in which they differ +to construct the separating sets. +}} + +{\bf Lemma. \it If $A\subseteq B,$ then $\clos(A)\subseteq\clos(B).$} +\proof +Any sequence in the smaller set $A$ is contained in the larger set $B,$ +so the limit points of $A$ are in the limit points of $B,$ or in other +words, $\clos(A)\subseteq\clos(B).$ +\endproof + +\medskip +\proof +Let $x\neq y\in C,$ differing on the $k^{\rm th}$ ternary digit of their +respective expansions. + +To define $C_k$ as above, we take $\Gamma_k$ to be numbers that +terminate in $k$ ternary digits or less (i.e. all digits after $k$ are +0) and only have the digits 0 and 2. +$C_k$ is then the union of $[\gamma,\gamma+3^{-k}]$ for all +$\gamma\in\Gamma_k\cap[0,1).$ + +The numbers $x$ and $y$ must differ by at least $2(3^{-k}),$ so +we choose $\gamma$ such that we may take $x\in A=[\gamma,\gamma+3^{-k}]$ +and $B=C_k\setminus A.$ +Clearly, $A\cup B = C_k.$ +And $\clos(A)=A,$ and $A\cap B = \emptyset.$ +Finally, $(\gamma-3^{-k},\gamma+2(3)^{-k})\not\subseteq B,$ (the lower +ends of these intervals are numbers which will have a 1 as a ternary +digit because $...0222...2+1 = ...1000...$) so $A$ is not in $\clos(B),$ +which means $A\cap\clos(B) = \emptyset.$ + +Now, $C\subseteq C_k$ from the definition of $C,$ so the set $A\cap +C\subseteq A$ and the set $B\cap C\subseteq B.$ +By the lemma, these sets are also disjoint, and +$(A\cap C)\cup((C_k\setminus A)\cap C) = C.$ + +Therefore, the Cantor set is totally disconnected. +\endproof + +\question{Let $\{r_1, r_2, r_3,\ldots\}$ be an enumeration of the +rational numbers. Define $G = \bigcup_{n=1}^\infty +(r_n-2^{-n},r_n+2^{-n}).$ Thus, $G$ is an open set that contains all +rationals. Let $F = G^c.$} + +\item{(1)} Argue that $F$ is a closed, nonempty set consisting only of +irrational numbers. + +\proof +$G$ is open and contains all rationals, so $G^c=F$ must be closed and +contain no rationals (i.e. only contain irrationals). +The remaining task is to show that $F$ is nonempty + +We will show that $F\cap[0,3]$ is nonempty (which implies that $F$ is +nonempty). +We note that $F\cap[0,3]$ is the intersection of compact sets +$$F_n = [0,3]\setminus\bigcup_{k=1}^n +(r_k-2^{-k},r_k+2^{-k}) = [0,3]\cap\bigcap_{k=1}^n +(r_k-2^{-k},r_k+2^{-k})^c.$$ +They are compact because the complement of an open interval is closed +and the intersection of a compact and closed set is compact. + +The intersection of nonempty compact sets is nonempty, so we will show +that each $F_n$ is nonempty and thus show that $F$ is nonempty. + +We start with a definition: the length of an interval $[a,b]$ or $(a,b)$ +is $b-a,$ and the length of a finite union of disjoint intervals is the +sum of the lengths of those intervals (note that this is consistent: +breaking up an interval into two smaller intervals $[a,c]$ and $(c,b]$ +gives length $b-c+c-a=b-a,$ implying the general finite case). +All such lengths are nonnegative. + +If the length of such a set is positive, that means it contains an +interval with positive length, and $b-a>0$ implies $b>a$ and $[a,b]$ is +nonempty, so that set is also nonempty. + +Let us have a set $S$ that is a finite union of disjoint intervals, and +thus has a length. +First, +$$\mathop{\rm len}((a,b)\cap S)+\mathop{\rm len}((a,b)\setminus S) = +\mathop{\rm len}((a,b)),$$ +implying $\mathop{\rm len}(S\cap(a,b))\leq\mathop{\rm len}((a,b)).$ +$$\mathop{\rm len}(S\setminus(a,b))+\mathop{\rm len}(S\cap(a,b)) = +\mathop{\rm len}(S),$$ +and substituting our previous identity, $$\mathop{\rm len}(S) = +\mathop{\rm len}(S\setminus(a,b))+\mathop{\rm len}(S\cap(a,b))\leq +\mathop{\rm len}(S\setminus(a,b))+\mathop{\rm len}((a,b)),$$ +and rearranging, +$$\mathop{\rm len}(S\setminus(a,b)) \geq \mathop{\rm len}(S) - +\mathop{\rm len}((a,b)).$$ +% show that len(such a set \ (a,b)) >= len(such a set) - len((a,b)) + +We proceed by induction to show that $\mathop{\rm len}(F_k)\geq +2^{1-k}$ and thus that $F_k$ is nonempty. + +The length of $F_0 = [0,3]$ is clearly $3\geq 2^{1}$, so $F_0$ is +nonempty + +Taking the inductive hypothesis, $\mathop{\rm len}(F_k)\geq 2^{1-k},$ +we compute $$\mathop{\rm len}(F_{k+1}) = \mathop{\rm len}(F_k\setminus +(r_{k+1}-2^{-(k+1)},r_{k+1}+2^{-(k+1)})) \geq \mathop{\rm +len}(F_k)-2^{-k} = 2^{1-k}-2^{-k} = 2^{-k}.$$ +(from the earlier result). + +We have finally shown that $F_k$ is nonempty and that $F$ is nonempty. +\endproof + +\medskip +\item{(2)} Is $F$ totally disconnected? + +\proof +Yes. Between any two irrational $x,y\in F,$ there is some $r\in\bb Q$ +such that $x<r<y,$ determining our partition of $F$ into $L = \{f<r|f\in +F\}$ and $R = \{f>r|f\in F\}.$ + +We use the Lemma from question 1 here, that $A\subseteq B$ implies +$\clos(A)\subseteq\clos(B).$ + +$\clos(L)\subseteq\clos((-\infty,r))=(-\infty,r].$ +Similarly, $\clos(R)\subseteq\clos((r,\infty))=[r,\infty).$ + +$$\clos(A)\cap B = A\cap\clos(B)\subseteq\{r\},$$ +but $r\not\in\clos(F).$ +And the sets $L,R$ are subsets of $F,$ so +$\clos(L),\clos(R)\subseteq\clos(F),$ so $r$ is not in $\clos(B)$ or +$\clos(A).$ + +We have shown that $F$ is totally disconnected. +\endproof + +\medskip +\item{(3)} As presented, the set $F$ could have isolated points. (How?) +Modify the construction of $G$ so that $F$ is perfect. (A perfect set +is closed with no isolated points.) + +\proof +Trivially, if we let $G = \bigcup_{k=1}^\infty (r_k-1,r_k+1) = \bb R,$ +we get $F = G^c = \emptyset,$ which is perfect. +However, if we want to make $F$ nonempty, we must do more work. + +Let $I$ be the isolated points of $F$ or $F\setminus\clos(F).$ +Set $G' = G\cup I,$ and now $(G')^c = F\setminus I = \clos(F),$ is +perfect. +\endproof + +\bye |