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\title{18.901 Problem Set 2}
\author{Chris Laas}
\date{March 17, 2000}

\begin{document}
\maketitle




\section*{Problem 1: 18.13}

Let us assume that there exist distinct continuous extensions $g_1,
g_2: \bar{A} \to Y$ of $f: A \to Y$.  Then there exists some $x \in
\bar{A}$ such that $g_1(x) \neq g_2(x)$.  Then, since $Y$ is
Hausdorff, there exist neighborhoods $U_1$ and $U_2$ of $g_1(x)$ and
$g_2(x)$, respectively, that are disjoint.  Since $U_1$ and $U_2$ are
open and $g_1$ and $g_2$ are continuous, $V_1 = g_1^{-1}(U_1)$ and
$V_2 = g_2^{-1}(U_2)$ are open.  Thus, $V_1 \cap V_2$ is open, and
$x \in V_1 \cap V_2$ (since $g_1(x) \in U_1$ and $g_2(x) \in U_2$).
Since $x \in \bar{A}$, $x \in V_1 \cap V_2$, and $V_1 \cap V_2$ is
open, $V_1 \cap V_2$ intersects $A$.

Thus, there exists $y \in V_1 \cap V_2 \cap A$.  Since
$y \in V_1 \cap V_2$, $g_1(y) \in U_1$ and $g_2(y) \in U_2$.  But
since $y \in A$, $g_1(y) = f(y) = g_2(y)$.  Thus
$f(y) \in U_1 \cap U_2$.  But this is a contradiction, since $U_1$ and
$U_2$ are disjoint.

Thus, the original assumption is false: there exists no pair of
distinct continuous extensions $g_1, g_2: \bar{A} \to Y$ of
$f: A \to Y$.  Thus, $g$ is unique.




\section*{Problem 2: 20.6}
\begin{enumerate}

\item%(a)

Take the point $y = (y_1, y_2, \dots)$ where
$$y_n = x_n + \epsilon \left(1 - \frac{1}{n}\right)$$

Then since
$y_n \in (x_n - \epsilon, x_n + \epsilon)$, $y \in U(x, \epsilon)$.
But $\bar{d}(x_n, y_n) = \epsilon \left(1 - \frac{1}{n}\right)$; thus
$\sup \{\bar{d}(x_n, y_n)\} = \epsilon$.  Thus
$y \notin B_{\bar{\rho}}(x, \epsilon)$, and thus
$U(x, \epsilon) \neq B_{\bar{\rho}}(x, \epsilon)$.


\item%(b)

Take $y \in U(x, \epsilon)$ as defined above, and take any
$\delta > 0$.  Then set $z = (z_1, z_2, \dots)$ by
$$z_n = y_n + \frac{\delta}{2}
      = x_n + \epsilon \left(1 - \frac{1}{n}\right) + \frac{\delta}{2}$$
Then, since
$$\bar{\rho}(y, z) = \sup \left\{\left|y_n - \left(y_n+\frac{\delta}{2}\right)\right|\right\}
                  = \frac{\delta}{2}$$
$z \in B_{\bar{\rho}}(y,\delta)$.  But, since there exists
$N \in \Z_+$ such that $\frac{\delta}{2} > \frac{1}{N}$,
$z_N > x + \epsilon$, and thus $z \notin U(x, \epsilon)$.

Thus $B_{\bar{\rho}}(y,\delta) \not\subset U(x,\epsilon)$.  Thus there
exists $y \in U(x, \epsilon)$ such that for all $\delta > 0$,
$B_{\bar{\rho}}(y,\delta) \not\subset U(x,\epsilon)$.  Thus
$U(x, \epsilon)$ is not open in the uniform topology.

\end{enumerate}




\section*{Problem 3: 20.8}
\def\l2{$\ell^2$}
\def\Bl2{B_{\ell^2}}
\def\Brho{B_{\bar{\rho}}}
\begin{enumerate}

\setcounter{enumi}{1}
\item%(b)

First, we will show that the box topology is not a subset of the
\l2-topology.  Take

$$ U = \(\prod_{n \in \Z_+} \(-\frac{1}{n},\frac{1}{n}\)\) \cap \R^\infty $$

Since $U$ is a basis element of the box topology, it is clearly open
in the box topology.

Now take the point $y \in U$ such that $y = (0,0,0,\dots)$.  For any
$\delta > 0$, there is $N \in \Z_+$ such that
$\frac{1}{N} < \frac{\delta}{2}$.  Let $z = (z_1,z_2,\dots)$ be
defined such that

$$ z_n = \choice{\frac{\delta}{2}, & n = N \\
                 0               , & n \neq N} $$

Then $d(y,z) = \frac{\delta}{2}$, so $z \in \Bl2(y,\delta)$; but,
since $z_N = \frac{\delta}{2} > \frac{1}{N}$, $z \notin U$.

Thus $\Bl2(y,\delta) \not\subset U$ for all $\delta > 0$.  Thus $U$ is
not open in the \l2-topology, but is open in the box topology.

Thus, the \l2-topology is distinct from the box topology on
$\R^\infty$.

Next, we will show that the \l2-topology is not a subset of the
uniform topology.  Take $y \in \R^\infty$ as above:
$y = (0,0,0,\dots)$.  Then take $U = \Bl2(y,\frac{1}{2})$.  Since $U$
is a basis element of the \l2-topology, it is clearly open in the
\l2-topology.

Now, for any $\delta > 0$, there is $N \in \Z_+$ such that
$\frac{1}{N} < \delta$.  Let $z = (z_1,z_2,\dots)$ such that

$$ z_n = \choice{
                 \frac{\delta}{2}, & n \leq N^2 \\
                 0               , & n >    N^2 \\
                 } $$

Then $z \in \R^\infty$ and, since
$\sup \{|z_n - y_n|\} = \sup \{z_n\} = \frac{\delta}{2}$,
$z \in \Brho(y,\delta)$.

But, since

$$ d(y,z) = \sqrt{\sum_{n \in \Z_+} z_n^2}
          = \sqrt{N^2 \(\frac{\delta}{2}\)^2}
          = N \(\frac{\delta}{2}\)
          > \frac{1}{2} $$

$z \notin \Bl2(y,\frac{1}{2}) = U$.

Hence $\Brho(y,\delta) \not\subset \Bl2(y,\frac{1}{2})$ for all
$\delta > 0$.  Thus $\Bl2(y,\frac{1}{2})$ is not open in the uniform
topology, but is open in the \l2-topology.

Thus the uniform topology is distinct from the \l2-topology.

Lastly, we will show that the uniform topology is not a subset of the
product topology.  Again, take $y$ as above.  Then take

\begin{eqnarray*}
U & = & \Brho\(y,\frac{1}{2}\) \\
  & = & \left\{ x \in \R^\infty
                | \sup \{\bar{d}(0,x_n)\} < \frac{1}{2} \right\} \\
  & = & \(\(-1/2,1/2\) \times \{0\} \times \{0\} \times \cdots\) \cup \\
     && \(\(-1/2,1/2\) \times \(-1/2,1/2\) \times \{0\} \times \{0\} \times \cdots\) \cup \\
     && \(\(-1/2,1/2\) \times \(-1/2,1/2\) \times \(-1/2,1/2\) \times \{0\} \times \{0\} \times \cdots\) \cup \cdots \\
  & = & \(\prod_{n \in \Z_+} \(-\frac{1}{2},\frac{1}{2}\)\) \cap \R^\infty \\
\end{eqnarray*}

Since $U$ is a basis element of the uniform topology, it is clearly
open in the uniform topology.

Now take any basis element $B$ of the product topology which contains
$y$.  Then, by definition,
$B = \(\prod_{n \in \Z_+} V_n\) \cap \R^\infty$
such that $V_n$ is open in $\R$ and all but a finite number of the
$V_n$ are equal to $\R$.  Thus there is some $N \in Z_+$ such that
$V_N = \R$.

Thus, if we define $z = (z_1,z_2,\dots)$ by

$$ z_n = \choice{
                 1, & n =    N \\
                 0, & n \neq N \\
                 } $$

then $z \in \R^\infty$ and $z_n \in V_n$; thus $z \in B$.  But, since
$\bar{\rho}(y,z) = 1$, $z \notin \Brho(y,\frac{1}{2}) = U$.

Thus, for any basis element $B$ of the product topology which contains
$y$, $B \not\subset U$.  Thus, $U$ is not open in the product
topology, but is open in the uniform topology.

Thus, the product topology is distincy from the uniform topology.

Putting all these results together with the known fact that

$$ \text{box topology}
   \supset \text{\l2-topology}
   \supset \text{uniform topology}
   \supset \text{product topology} $$

we obtain the final result for these topologies on $\R^\infty$:

$$ \text{box topology}
   > \text{\l2-topology}
   > \text{uniform topology}
   > \text{product topology} $$



\item%(c)

The final result for $H$ will be:

$$ \text{box topology}
   > \text{\l2-topology}
   = \text{uniform topology}
   = \text{product topology} $$

First, we will show that the box topology is not a subset of the
\l2-topology.  Take

$$ U = \prod_{n \in \Z_+} \[0,\frac{1}{n}\)
     = \[0,1\) \times \[0,\frac{1}{2}\) \times \[0,\frac{1}{3}\) \times \cdots
$$

Since

$$ U = \( \prod_{n \in \Z_+} \(-1,\frac{1}{n}\) \) \cap H $$

$U$ is open in the box topology on $H$.

Take $y = (0,0,0,\dots)$.  Clearly $y \in U$.  Now, for any
$\delta > 0$, there is $N \in \Z_+$ such that $\frac{1}{N} < \delta$.
Then let $z = (z_1, z_2, \dots)$ such that

$$ z_n = \choice{
                 \frac{1}{N}, & n =    N \\
                 0          , & n \neq N \\
                 } $$

Then $d(y,z) = \frac{1}{N} < \delta$, so $z \in \Bl2(y,\delta)$; but
clearly $z \notin U$, since $\frac{1}{N} \notin \[0,\frac{1}{N}\)$.

Thus $\Bl2(y,\delta) \not\subset U$ for all $\delta > 0$.  Thus $U$ is
not open in the \l2-topology, but is open in the box topology.

Thus the \l2-topology is distinct from the box topology.

Next, we will show that the \l2-topology {\em is} a subset of the
uniform topology on H.  Take any basis element of the \l2-topology,
i.e. $U = \Bl2(x,\epsilon)$ for any $x \in H$ and $\epsilon > 0$.

Then take any $y \in U$, and define $\gamma = \epsilon - d(x, y)$.
Now, there exists some $N \in \Z_+$ such that
$\frac{1}{N} < \frac{\gamma^2}{2}$.  Finally, define
$\delta = \frac{1}{N}$.  Then

$$ \Brho(y,\delta) = \left\{ z \in H |
                             \sup \{|z_n - y_n|\} < \frac{1}{N} \right\} $$

Take any $z \in \Brho(y, \delta)$.  Then, since
$|z_n - y_n| < \frac{1}{N}$ for all $n$, we can say that

$$ \sum_{n=1}^{N} (z_n - y_n)^2 < \sum_{n=1}^{N} \frac{1}{N^2} = \frac{1}{N} $$

And, since $0 \leq y_n, z_n \leq \frac{1}{n}$,
$|z_n - y_n| \leq \frac{1}{n}$.  Thus

$$ \sum_{n=N+1}^{\infty} (z_n - y_n)^2
     \leq \sum_{n=N+1}^{\infty} \frac{1}{n^2}
     \leq \int_{N}^{\infty} \frac{1}{x^2} dx
     = \frac{1}{N} $$

Putting these two results together, we find that

\begin{eqnarray*}
d(y,z) & = & \sqrt{\sum_{n=1}^{\infty} (z_n - y_n)^2} \\
       & = & \sqrt{\sum_{n=1}^{N} (z_n - y_n)^2
               + \sum_{n=N+1}^{\infty} (z_n - y_n)^2} \\
       & < & \sqrt{\frac{1}{N} + \frac{1}{N}} \\
       & = & \sqrt{\frac{2}{N}} \\
       & < & \sqrt{2 \( \frac{\gamma^2}{2} \)} \\
       & = & \gamma \\
\end{eqnarray*}

Therefore, using the triangle inequality, we have

$$ d(x, z) \leq d(x,y) + d(y,z)
           <    d(x,y) + \gamma
           =    d(x,y) + (\epsilon - d(x,y))
           =    \epsilon $$

Therefore, we can finally say that $z \in \Bl2(x,\epsilon)$.
Thus, $\Brho(y,\delta) \subset \Bl2(x,\epsilon)$.

Thus, for any $y \in U$, there exists a basis element $B$ of the
uniform topology such that $y \in B \subset U$.  Thus $U$ is open in
the uniform topology.

Thus any basis element of the \l2-topology is open in the uniform
topology; thus, the \l2-topology is a subset of the uniform topology
on $H$.  Since we already knew that the uniform topology is a subset
of the \l2-topology, we now have that the two topologies are the
same.

Now, finally, we will show that the uniform topology is a subset of
the product topology on $H$.  Take any basis element of the uniform
topology, i.e. $U = \Brho(x,\epsilon)$, for any $x \in H$ and
$\epsilon > 0$.

Then take any $y \in U$.  Define $a_n = \epsilon - |y_n - x_n|$; we
know that $a_n > 0$ for all $n$.

Take $N \in \Z_+$ such that $\frac{1}{N} < \epsilon$.  Then let

$$ V_n = \choice{
                 \( y_n - \frac{a_n}{2}, y_n + \frac{a_n}{2} \)
                       \cap \[0,\frac{1}{n}\] ,
                            & 1 \leq n \leq N \\
                 \[0,\frac{1}{n}\] , & n > N \\
                 } $$

and let $V = \prod V_n$.  Since $V = \(\prod V_n' \) \cap H$ for

$$ V_n' = \choice{
                  \( y_n - \frac{a_n}{2}, y_n + \frac{a_n}{2} \) ,
                       & 1 \leq n \leq N \\
                  \R , & n > N \\
                 } $$

and thus the $V_n'$ are open in $\R$ and equal to $\R$ for all but
finitely many $n$, $V$ is a basis element of the product topology on
$H$.

Now, take any $z \in V$.  Then

\begin{eqnarray*}
\sup \{|z_n - x_n|\} & \leq & \max \{ |z_n-x_n| \text{ for } 1 \leq n \leq N,
                                      \sup \{ 1/n \} \text{ for } n > N \} \\
    & \leq & \max \{ \frac{\epsilon+|y_n-x_n|}{2} \text{ for } 1 \leq n \leq N,
                     \frac{1}{N+1} \} \\
    & <    & \max \{ \frac{\epsilon+\epsilon}{2}, \frac{1}{N+1} \} \\
    & =    & \max \{ \epsilon, \frac{1}{N+1} \} \\
    & =    & \epsilon
\end{eqnarray*}

Thus, $z \in \Brho(x, \epsilon)$; and thus, we have
$V \subset \Brho(x, \epsilon) = U$.

Thus, for any $y \in U$, there is a basis element $V$ of the product
topology such that $y \in V \subset U$.  Thus $U$ is open in the
product topology.

Thus any basis element of the uniform topology is open in the product
topology; thus, the uniform topology is a subset of the product
topology on $H$.  Since we already knew that the product topology is a
subset of the uniform topology, we now have that the two topologies
are the same.

In conclusion, we can sum up the relations between these four
topologies on $H$ thus:

$$ \text{box topology}
   > \text{\l2-topology}
   = \text{uniform topology}
   = \text{product topology} $$

\end{enumerate}




\section*{Problem 4: 24.8}
\begin{enumerate}

\item%(a)

Yes.  Given a product $X = \prod_{\alpha \in J} X_\alpha$ of
path-connected spaces $X_\alpha$, and two points $x, y \in X$, we can
construct $f: [0,1] \to X$ thus:

For every $X_\alpha$, there exists a continuous map
$f_\alpha: [a_\alpha, b_\alpha] \to X_\alpha$ such that
$f_\alpha(a_\alpha) = x_\alpha$ and $f_\alpha(b_\alpha) = y_\alpha$.

Define $f'_\alpha: [0,1] \to X_\alpha$ by
$f'_\alpha(t) = f_\alpha((1-t) a_\alpha + t b_\alpha)$.
By composition, $f'_\alpha$ is continuous, and clearly
$f'_\alpha(0) = x_\alpha$ and $f'_\alpha(1) = y_\alpha$.

Now, define $f: [0,1] \to X$ by $(f(t))_\alpha = f_\alpha(t)$.  Then,
since products of continuous functions are continuous, $f$ is a
continuous map.  And, clearly, $f$ is a path in $X$ from $x$ to $y$.

Thus, $X$ is path-connected.


\item%(b)

No.  Take $A \subset \R^2$ as the toplogist's sine curve:

$$ A = \left\{\left. x \times \sin\frac{1}{x} \right| 0 < x \leq 1 \right\} $$

Take any two points $x,y \in A$; we may take $x_1 < y_1$ without loss
of generality.  Then, we may define $f: [x_1, y_1] \to A$ by
$f(t) = t \times \sin\frac{1}{t}$.  Then $f$ is a path in $A$ from $x$
to $y$.  Thus, $A$ is path-connected.

However, as shown in Example 7 on page 157 of Munkres, $\bar{A}$ is
not path-connected.


\item%(c)

Yes.  Take any $x,y \in f(X)$; choose some $x' \in f^{-1}(x)$
$y' \in f^{-1}(y)$.  Since $X$ is path-connected, there exists some
path $g: [a,b] \to X$ from $x'$ to $y'$.

Construct the function $h: [a,b] \to Y$ by $h(t) = f(g(t))$.  Since it
is a composition of continuous functions, $h$ is continuous; and,
clearly, $h(a) = f(g(a)) = f(x') = x$ and $h(b) = f(g(b)) = f(y') =
y$.  Finally, for any $t \in [a,b]$, $g(t) \in X$; thus,
$h(t) = f(g(t)) \in f(X)$.  Thus, $h$ is a path in $f(X)$ from $x$ to
$y$.

Thus, $f(X)$ is path-connected.


\item%(d)
\def\Aax{A_{\alpha_x}}
\def\Aay{A_{\alpha_y}}

Yes.  Take any $x, y \in \bigcup A_\alpha$.  Then there are $\Aax$ and
$\Aay$ such that $x \in \Aax$ and $y \in \Aay$.  Also, since
$\bigcap A_\alpha$ is not empty, there exists
$z \in \bigcap A_\alpha$; thus $z \in \Aax$ and $z \in \Aay$.

Since $\Aax$ and $\Aay$ are path-connected, there exists a path
$f:[a,b] \to \Aax$ from $x$ to $z$ and a path $g:[c,d] \to \Aay$ from
$z$ to $y$.  Construct the function $h: [a, b+d-c] \to \bigcup A_\alpha$
by defining $h(t) = f(t)$ for $t \in [a,b]$ and
$h(t) = g(t-b+c)$ for $t \in [b,b+d-c]$.  By the pasting lemma, $h$ is
continuous; and clearly $h(a) = x$ and $h(b+d-c) = y$.  Thus $h$ is a
path in $\bigcup A_\alpha$ from $x$ to $y$.

Thus $\bigcup A_\alpha$ is path-connected.


\end{enumerate}




\end{document}
