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\begin{center}
\textbf{Mathematics 468} \\
\textbf{Homework 1 solutions}
\end{center}
\vspace{-0.05in}

1. Is the set 
\[
\{(p,q) \in \mathbf{R}^{2} | p,q \in \mathbf{Q} \}
\]
countable?

\textsc{Answer:} Yes.  Since the set of rationals $\mathbf{Q}$ is
countable, you can list them: $p_{1}, p_{2}, p_{3}, \dots$.  Now you
can write down all of the ordered pairs of rationals in an array:
\[
\begin{array}{ccccc}
(p_{1}, p_{1}) &  (p_{1}, p_{2}) &  (p_{1}, p_{3}) & (p_{1}, p_{4}) & \dots \\
(p_{2}, p_{1}) &  (p_{2}, p_{2}) &  (p_{2}, p_{3}) & (p_{2}, p_{4}) & \dots \\
(p_{3}, p_{1}) &  (p_{3}, p_{2}) &  (p_{3}, p_{3}) & (p_{3}, p_{4}) & \dots \\
(p_{4}, p_{1}) &  (p_{4}, p_{2}) &  (p_{4}, p_{3}) & (p_{4}, p_{4}) & \dots \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array}
\]
Now the set of all of these ordered pairs is countable: order them
according to this chart:
\[
\begin{array}{ccccc}
{1} & {3} & {6} & {10} & {\dots} \\
{2} & {5} & {9} & {14} & {\dots} \\
{4} & {8} & {13} & {19} & {\dots} \\
{7} & {12} & {18} & {25} & {\dots} \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array}
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\]

\myline 

2. If $A_{1}$, $A_{2}$, $A_{3}$, \dots are each countable, is their
union?

\textsc{Answer:} Yes, for pretty much the same reason as in the first
problem.  Since each $A_{j}$ is countable, you can list its elements:
\[
A_{j} = \{a_{1j}, a_{2j}, a_{3j}, \dots \}.
\]
Now the union $\displaystyle \bigcup_{j=1}^{\infty} A_{j}$ consists of
the elements in the following array (I've put the elements of
$A_{j}$ in the $j$th column):
\[
\begin{array}{cccc}
a_{11} & a_{12} & a_{13} & \dots \\
a_{21} & a_{22} & a_{23} & \dots \\
a_{31} & a_{32} & a_{33} & \dots \\
\vdots & \vdots & \vdots & \ddots
\end{array}
\]
Now cross out any repetitions, and count the remaining elements as in
the first problem.

\myline 

3. What is the smallest closed subset of $\mathbf{R}$ which contains
$\mathbf{Q}$?

\textsc{Answer:} $\mathbf{R}$ itself is the smallest closed subset
which contains $\mathbf{Q}$.  Suppose that $A$ is a subset of
$\mathbf{R}$ which contains $\mathbf{Q}$; I claim that if 
$A \neq \mathbf{R}$, then $A$ cannot be closed.

Since $A \neq \mathbf{R}$, then there is some point $x \in A^{c}$.
For every $\varepsilon > 0$, the ball $B_{\varepsilon} (x)$ must
contain at least one rational number (in fact, it contains countably
many rationals, but that's not important for this problem).
Therefore, this ball does not lie completely inside of $A^{c}$, and
therefore $A^{c}$ is not open.  By the definition of ``closed,'' $A$
is not closed.

So if $A$ is any proper subset of $\mathbf{R}$ which contains
$\mathbf{Q}$, then $A$ is not closed; hence $\mathbf{R}$ is the only
closed subset of $\mathbf{R}$ which contains $\mathbf{Q}$.
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