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\title{hw7-solutions}
\begin{document}
\centerline{\Large Math 171 Homework 7}
\centerline{\small (due May 20)}
\vskip .2in
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 45.2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb{45.2}
\bi
\itema
Give an example or a subset of $\R$ which is connected but not compact.
\itemb
Give an example of a subset of $\R$ which is compact but not connected.
\itemc
Characterize the compact connected subsets of $\R$.
\ei
\sol
\bi
\itema
$\R$ is connected by Theorem 45.7 and not compact by Theorem 43.9.
\itemb
$\{0, 1\}$ is not connected because the point $\{0\}$ is open and closed in
$\{0, 1\}$ and compact because it is finite.
\itemc
\begin{claim}
The compact connected subsets of $\R$ are
\bi
\item the empty set,
\item singleton sets $\{x\}$, $x \in \R$ and
\item closed intervals $[a, b]$, $a, b \in \R$, $a < b$.
\ei
\end{claim}
\begin{proof}
By Corollary 45.4, the connected subsets of $\R$ are the empty set,
the singleton sets, bounded intervals (open, closed and half-open),
rays ($[a, \infty)$ and $(-\infty, a]$) and $\R$ itself.
By Theorem 43.9 the compact subsets of $\R$ are the closed bounded sets.
Open intervals are not closed. Rays and $\R$ are not bounded. Hence,
we get the list in the claim.
\end{proof}
\ei
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 45.5
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb{45.5}
Let $X$ be a connected subset of a metric space $M$. Prove that $\overline X$
is connected. Is $\mathring X$ necessarily connected?
\sol
Assume that $\overline X$ is not connected. We will show that $X$ is also
not connected. Write $\overline X = U \cup V$ where $U$ and $V$ are disjoint
non-empty open subsets of $\overline X$. Then $X = (X \cap U) \cup (X \cap V)$
with $X \cap U$ and $X \cap V$ disjoint open in $X$. To show that $X$
is not connected it suffices to show that $X \cap U$ and $X \cap V$ are
non-empty. Assume that $X \cap U$ is empty. Then $X = X \cap V$, so
$X \subset V$.
We show that this implies that $\bar X = V$ contradicting our assumptions
that $U$ in nonempty. Indeed, let $x$ be a limit point of $X$ and let
\sequence x be a sequence of elements of $X$ converging to $x$. Then
each $x_n$ is contained in $V$. Since $V$ is closed in $\bar X$,
and $x$ is a limit point of $V$, $x \in V$.
Next we give an example when $X$ is connected and $\mathring X$ is not
connected.
Let
\[
X = ([-2, 0] \times [-2, 0]) \cup ([0, 2] \times [0, 2]).
\]
We show that $\mathring X$ is not connected. Because for every $\e > 0$,
the point $(-\e/2, \e/2)$ is an element of $B_{\e}((0, 0))$ and
not of $X$, $(0, 0) \notin \mathring X$.
Let $f: \R^2 \ra \R$ be given by $f(x, y) = x + y$.
Since $(0, 0) \notin \mathring X$, $f(x, y) \neq 0$ for every
$(x, y) \in \mathring X$.
Since
$f$ is continuous, the set $U := f^{-1}((0, \infty))$ is open
in $\R^2$. Therefore, $U \cap \mathring X$ is open in $\mathring X$.
The set $U \cap \mathring X$ is nonempty because it contains
the point $(1, 1)$. The set $U \cap \mathring X$ is not all of
$\mathring X$ because it does not contain the point $(-1, -1) \in \mathring X$.
Finally, $U \cap \mathring X$ is closed in $\mathring X$ because
$U \cap \mathring X = f^{-1}([0, \infty)) \cap \mathring X$.
Thus, $\mathring X$ is not connected.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 45.7ab
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb{45.7}
\bi
\itema
Show, by example, that unions and intersections of connected sets are not
necessarily connected.
\itemb
Prove that if $X$ and $Y$ are connected subsets of $\R$, then
$X \cap Y$ is connected.
\ei
\sol
\bi
\itema
Consider connected singleton sets $\{0\}$ and $\{1\}$. Their union
$\{0, 1\}$ is not connected.
\itemb
By Theorem 45.3, it suffices to show that whenever $a, b \in X \cap Y$ with
$a < b$ then $[a, b] \subset X$. Assume that $a, b \in X \cap Y$.
Since $a, b \in X$ and $X$ is connected, by Theorem 45.3,
$[a, b] \subset X$. Similarly, $[a, b] \subset Y$. Thus,
$[a, b] \subset X \cap Y$, as desired.
\ei
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 46.2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb{46.2}
Give an example of a complete metric space that is not compact.
\sol
$\R$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 46.3
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb{46.3}
Given an example of a connected metric space that is not complete.
\sol
$\ (0, 1)$ is connected by Corollary 45.4, but not complete because the Cauchy
sequence $\{1/n\}_{n > 1}$ does not converge in $(0, 1)$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 46.5
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb{46.5}
Let $M$ be a metric space.
\bi
\itema
Prove that if $C$ is a complete subset of $M$, then $C$ is closed.
\itemb
Prove that if $M$ is complete, then every closed subset of $M$ is complete.
\ei
\sol
\bi
\itema
Let $x$ be limit point of $C$ in $M$.
Let \sequence x be a sequence of elements of $C$ converging to an element
$x$ of $M$. By Theorem 46.2, \sequence x is a Cauchy sequence in $M$.
Since each $x_n$ is an element of $C$, \sequence x is a Cauchy sequence
in $C$. Since $C$ is complete, \sequence x converges to some $y \in C$
as a sequence in $C$. Thus, \sequence x converges to both $x$ and $y$ as
a sequence in $M$, so $x = y \in C$. Thus, $C$ is contains all of its limit
points.
\itemb
Assume $M$ is complete and let $C$ be a closed subset of $M$. Let
\sequence x be a Cauchy sequence in $C$. Since $M$ is complete,
\sequence x converges to some $x \in M$. Since $C$ is closed
$x \in C$. Thus, every Cauchy sequence in $C$ converges to an element of $C$.
\ei
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 1
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb{1}
Let $X$ be any two element set, for instance $\{1, 2\}$, endowed with the
discrete metric. Prove that a metric space $M$ is connected if and only if
continuous functions $f: M \ra X$ are constant.
\sol
Assume that there exists an non-constant function $f: M \ra X$.
Since the singleton sets $\{1\}$ and $\{2\}$ are open in the discrete metric
and $f$ is continuous it follows that $f^{-1}(\{1\})$ and $f^{-1}(\{2\})$ are
open subsets of $M$. Since $X = \{1\} \cup \{2\}$ it follows that
\[
M = f^{-1}(\{1\}) \cup f^{-1}(\{2\}).
\]
Since $f$ is non-constant, both $f^{-1}(\{1\})$ and $f^{-1}(\{2\})$ and
are non-empty. Thus, $M$ can be written as a union of two disjoint nonempty
open sets $f^{-1}(\{1\})$ and $f^{-1}(\{2\})$, so $M$ is not connected.
Assume $M$ is not connected. Say $M = U \cup V$ with $U$ and $V$ disjoint
nonempty open subsets of $M$. Define $f: M \ra X$ by
\[
f(x) \manylinedefinition{
1 & \text{if $x \in U$,} \\
2 & \text{if $x \in V$.}
}
\]
Then $f$ is continuous because the preimage of every open subset of $X$
is open in $M$. Indeed, $X$ has 4 subsets: $\emptyset$, $\{1\}$, $\{2\}$, $X$,
each of them open in $X$. The preimage of each of the 4 subsets of $X$
is open in $M$: $f^{-1}(\emptyset) = \emptyset$, $f^{-1}(\{1\}) = U$,
$f^{-1}(\{2\}) = V$, $f^{-1}(X) = M$.
\end{document}