8 homeworks from analysisHEADmaster

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+\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
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+++ 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