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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 |