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\def\choice#1{\left\{ \begin{array}{ll} #1 \end{array} \right.}

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\title{18.901 Problem Set 1}
\author{Chris Laas}
\date{February 25, 2000}

\begin{document}
\maketitle




\section*{Problem 1: 3.13}

Given an ordered set $A$ with the least upper bound property, we must
prove that $A$ has the greatest lower bound property.

To begin, take any nonempty subset $A_0$ of $A$ that is bounded below.
Let $B_0$ be the set of all lower bounds of $A_0$.  Since $A_0$ is
bounded below, $B_0$ is nonempty.  Since $b \leq a$ for all $a \in
A_0$ and $b \in B_0$, every element of $A_0$ is an upper bound of
$B_0$.  Thus, $B_0$ is bounded above.  Therefore, by the least upper
bound property, $B_0$ has a least upper bound $u \in A$.  Then,
$u \leq a$ for any $a \in A_0$, and $u \geq b$ for any $b \in B_0$.

Thus, since $u$ is a lower bound of $A_0$, and $u$ is greater than any
other lower bound of $A_0$, $u$ is the greatest lower bound of $A_0$.

Thus, any nonempty subset of $A$ that is bounded below has a greatest
lower bound, and thus $A$ has the greatest lower bound property.




\section*{Problem 2: 7.8}

First, we will construct a countably infinite subset
$U \subset X^\omega$ such that for all $x \in U$, $x_1 = 0$.
Define a map $u: \Z_+ \to X^\omega$ by $u(n) = x$ such that
$$
  x_i = \choice{0 & i \leq n \\
                1 & i >    n   }
$$

Then simply let $U$ be the image of $u$: $U = u(\Z_+)$.

Next, we define the map
$\sigma: X^\omega \to \{ x \in X^\omega | x_1 = 1 \}$ by
$\sigma(x) = x'$ such that
$$
  x'_i = \choice{
                 1       & i = 1 \\
                 x_{i-1} & i \neq 1 \\
                }
$$

Using these constructions, we will show that there is an injection
from the set ${\cal B}$ of all countable subsets of $X^\omega$ to the
set ${\cal C}$ of all countably infinite subsets of $X^\omega$.  To
construct this injection $\tau: {\cal B} \to {\cal C}$, take any element
$B \in {\cal B}$.  If $B$ is a finite subset of $X^\omega$, define
$\tau(B) = \sigma(B) \cup U$.  Else, if $B$ is a countably infinite
subset of $X^\omega$, define $\tau(B) = \sigma(B)$.  In compact form:
$$
  \tau(B) = \choice{
                    \sigma(B) \cup U & \mbox{if $B$ is finite} \\
                    \sigma(B)        & \mbox{if $B$ is infinite} \\
                   }
$$

Since the result of $\tau$ is always countably infinite,
$\tau({\cal B}) \subset {\cal C}$.  Since the output of $\sigma$ and any
element of $U$ always differ in the first coordinate, it is clear that
$\tau$ is injective.

Now, define an injection $\alpha: {\cal C} \to (X^\omega)^\omega$ as
follows:  for any $C \in {\cal C}$, there is a bijection
$\eta_C: \Z_+ \to C$, since $C$ is countably infinite.  Define
$\alpha(C)_n = \eta_C(n)$.  Note that this definition makes use of the
axiom of choice.

We already know that there exists a bijection
$\gamma: \Z_+ \to \Z_+ \times \Z_+$.

Hence, define the injection $\beta: (X^\omega)^\omega \to X^\omega$
by $\beta(y)_n = (y_{\gamma(n)_1})_{\gamma(n)_2}$.  Intuitively, this
function ``unrolls'' the two-dimensional elements of
$(X^\omega)^\omega$ into the linear elements of $X^\omega$.

Putting it all together, the composite function
$\beta \circ \alpha \circ \tau$ defines an injection from ${\cal B}$
to $X^\omega$.

The map from $x \in X^\omega$ to the set containing only $x$,
$\{x\} \in {\cal B}$, is clearly an injective map from $X^\omega$ to
${\cal B}$.

Hence, since there exists an injection from ${\cal B}$ to $X^\omega$
and an injection from $X^\omega$ to ${\cal B}$, ${\cal B}$ and
$X^\omega$ have the same cardinality.




\section*{Problem 3: 13.7, 17.16}

\subsection*{13.7}

I provide the answer both in intuitive graphical format, and in
complete tabular format.
\vspace{1cm}

\def\rsupset{\rotatebox[origin=c]{-45}{$\subset$}}
\def\lsupset{\rotatebox[origin=c]{ 45}{$\supset$}}

\begin{tabular}{ccccc}
${\cal T}_3$ &          &              &          & ${\cal T}_5$ \\
             & \rsupset &              & \lsupset &              \\
             &          & ${\cal T}_1$ &          &              \\
             & \lsupset &              & \rsupset &              \\
${\cal T}_2$ &          &              &          & ${\cal T}_4$ \\
\end{tabular}


\begin{tabular}{ccc}
${\cal T}_1$ & is coarser than      & ${\cal T}_2$ \\
             & is finer than        & ${\cal T}_3$ \\
             & is coarser than      & ${\cal T}_4$ \\
             & is finer than        & ${\cal T}_5$ \\\hline
${\cal T}_2$ & is finer than        & ${\cal T}_3$ \\
             & is not comparable to & ${\cal T}_4$ \\
             & is finer than        & ${\cal T}_5$ \\\hline
${\cal T}_3$ & is coarser than      & ${\cal T}_4$ \\
             & is not comparable to & ${\cal T}_5$ \\\hline
${\cal T}_4$ & is finer than        & ${\cal T}_5$ \\
\end{tabular}


\subsection*{17.16(a)}

\begin{tabular}{ll}
Under the standard topology ${\cal T}_1$,
& the closure of $K$ is $K \cup \{ 0 \}$. \\

Under the topology of $\R_K$, ${\cal T}_2$,
& the closure of $K$ is $K$. \\

Under the finite complement topology ${\cal T}_3$,
& the closure of $K$ is $\R$. \\

Under the upper limit topology ${\cal T}_4$,
& the closure of $K$ is $K$. \\

Under the topology having all sets $(-\infty, a)$ as basis, ${\cal T}_5$, 
& the closure of $K$ is $[0,+\infty)$.
\end{tabular}


\subsection*{17.16(b)}

\begin{tabular}{lll}
${\cal T}_1$ & is Hausdorff     & and satisfies the $T_1$ axiom. \\
${\cal T}_2$ & is Hausdorff     & and satisfies the $T_1$ axiom. \\
${\cal T}_3$ & is not Hausdorff & but satisfies the $T_1$ axiom. \\
${\cal T}_4$ & is Hausdorff     & and satisfies the $T_1$ axiom. \\
${\cal T}_5$ & is not Hausdorff & and does not satisfy the $T_1$ axiom. \\
\end{tabular}




\section*{Problem 4: 17.18}

\begin{eqnarray*}
\bar{A} & = & A \cup \{ 0 \times 1 \} \\
\bar{B} & = & B \cup \{ 1 \times 0 \} \\
\bar{C} & = & C \cup \{ x \times 1 | 0 \leq x < 1 \} \cup \{ 1 \times 0 \} \\
\bar{D} & = & D \cup \{ x \times 0 | 0 <    x \leq 1 \}
                \cup \{ x \times 1 | 0 \leq x <    1 \} \\
\bar{E} & = & E \cup \left\{ \frac{1}{2} \times 0 \right\}
                \cup \left\{ \frac{1}{2} \times 1 \right\} \\
\end{eqnarray*}



\end{document}
