path: root/lacey/hw6.tex

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{\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