path: root/lacey/hw8.tex

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\def\clos{\mathop{\rm closure}}

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\long\def\question#1{\ifnum\qno=0\else\vfil\eject\fi\bigskip\advance\qno by 1\noindent{\bf Question \number\qno.} {\it #1}}

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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