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Homework 5 [adapted from latex HW of Colin McKinney]
\\ \\
\noindent \begin{bf}5.6A\end{bf} Show that if $A \subset B$, then
$\overline{A} \subset \overline{B}$.  \\  \noindent Take $x \in
\overline{A}$.  By Theorem 17.5, every neighborhood $U$ of $x$ intersects
$A$.  Since $A \subset B$, $U$ also intersects $B$, which by Theorem 17.5
implies that $x \in \overline{B}$. 
\\ \\ \noindent \begin{bf}
5.6B\end{bf}
Show $\overline{A \cup B} = \overline{A} \cup \overline{B}$.  \\ \\
\noindent 
$A \subset \overline{A}$ and $B \subset \overline{B}$.  Hence, 
$\overline{A} \cup \overline{B}$ is a closed set containing $A \cup B$.  
Since $\overline{A \cup B}$ is the smallest closed set containing $A \cup 
B$,  $\overline{A \cup B} \subset \overline{A} \cup \overline{B}$.
$\overline{A} \cup \overline{B} \subset \overline{A \cup B}$ by 5.6C.
\\ \\ \noindent \begin{bf}
5.6C\end{bf} 
Show $\cup
\overline{A_\alpha} \subset \overline{\cup A_\alpha}$; give an example
where equality fails.  
\\ \\ \noindent Fix $x \in \cup
\overline{A_\alpha}$.  This implies that $x \in \overline{A_{\alpha_0}}$ for
some $\alpha_0$.  By Theorem 17.5, this implies that every neighborhood $U$
of $x$ intersects $A_{\alpha_0}$.  Since $A_{\alpha_0} \subset
\cup A_\alpha$, we know that every neighborhood $U$ of $x$ intersects
$\cup A_\alpha$.  By Theorem 17.5, then, $x \in \overline{\cup A_\alpha}$.  
Hence $\cup \overline{A_\alpha} \subset \overline{\cup A_\alpha}$.  \\ \\
Specific counter-example where equality fails:  $\cup_{q \in Q} \overline{\{q\}} = \cup_{q \in Q} 
\{q\} = Q$, but $\overline{\cup_{q \in Q} \{q\}} = \overline{Q} = R$
\\ \\ \noindent 
\begin{bf}8\end{bf}
$A \cap B \subset A$.  Hence $\overline{A \cap B} \subset \overline{A}$.  
Similarly $\overline{A \cap B} \subset \overline{B}$.  Hence $\overline{A 
\cap B} \subset \overline{A} \cap \overline{B}$
\\ \\
Alternatively, $\overline{A \cap B}$ is the smallest closed set containing 
$A \cap B$. $A 
\subset \overline{A}$ and $B
\subset \overline{B}$.  Hence $A \cap B \subset \overline{A} \cap 
\overline{B}$.  Thus, $ \overline{A} \cap
\overline{B}$ is a closed set containing $A \cap B$.  Hence $\overline{A 
\cap B}\subset \overline{A} \cap
\overline{B}$.
\\ \\
Specific counter-example where equality fails:  $\overline{Q} \cap \overline{Q^c} = R \cap R = R$, but
$\overline{Q \cap Q^c} = \overline{\emptyset} = \emptyset$.
\\ \\
$\cap A_\alpha \subset A_{\alpha_0}$.  Hence $\overline{\cap A_\alpha} 
\subset \overline{ A_{\alpha_0}}$ for all $\alpha_0$.  Hence 
$\overline{\cap A_\alpha}
\subset \cap \overline{ A_{\alpha}}$
\\ \\
 Take $x \in \overline{ A} - \overline{B} $.  Take $U$ open such that $x \in U$.  $x \not\in
\overline{B}$ implies there exists an open set $V$ such the $x \in V$ and $V \cap B = \emptyset$.  $x
\in U \cap V$, $U \cap V$ is open, $x \in \overline{ A}$ implies there exists a $y \in (U \cap V) \cap
A$.
  Suppose $y \in B$.  Then $y \in B \cap V$, a contradiction.  Hence $y \in A - B$.  Thus $y \in U
\cap (A - B)$.  Hence $x \in \overline{A - B}$ and $\overline{A} - \overline{B} \subset\overline{A -
B}$.
 \\ \\
Specific counter-example where equality fails:  
$\overline{Q} - \overline{Q^c} = R - R = \emptyset$, but
$\overline{Q - Q^c} = \overline{Q} = R$ or
$\overline{[0, 1]} - \overline{(0, 1)} = [0, 1] - [0, 1] = \emptyset$, but
$\overline{[0, 1] - (0, 1)} = \overline{\{0, 1\}} = \{0, 1\}$.
\\ \\ \noindent 
\begin{bf}5.11\end{bf} 
Show that the product of two
Hausdorff spaces is Hausdorff. \\ \\ \noindent Let $A$ and $B$ be two
Hausdorff spaces.  
Take $x_1 \times
y_1$, $x_2 \times
y_2$ in $A \times B$ such that $x_1 \times y_1 \not= x_2 \times
y_2$.  Then either $x_1 \not= x_2$ or $y_1 \not= y_2$.  Without loss of 
generality assume  $x_1 \not= x_2$.
Since $A$ is Hausdorff, there exists $U_1$, $U_2$ 
such that $U_1$ is a neighborhood
of $x_1$ in $A$ and $U_2$ is a neighborhood of $x_2$ in $A$, where $U_1$
and $U_2$ are disjoint.  
Then $U_1 \times B$ will be a neighborhood of $x_1 \times
y_1$ in $A \times B$, and $U_2 \times B$ will be a neighborhood of $x_2
\times y_2$ in $A \times B$.  Since $U_1$ and $U_2$ are disjoint, 
it follows that $U_1 \times B$ and $U_2
\times B$ are disjoint.  Since we have found disjoint neighborhoods for
two arbitrary points in $A \times B$, by the definition of Hausdorff, $A
\times B$ is Hausdorff. \\ 

\noindent \begin{bf}5.12\end{bf} Show that a subspace of a Hausdorff
spaces is Hausdorff. \\ \\ \noindent Let $X$ be a Hausdorff space, and let
$Y$ be a subspace of $X$.  Let $x_1$ and $x_2$ be elements of $Y$ such 
that $x_1 \not= x_2$.  Since $X$ is Hausdorff, there exist
disjoint neighborhoods $U_1$ and $U_2$ in X of $x_1$ and $x_2$,
respectively.  Hence a set containing $x_1$ in $Y$ is $V_1 = U_1 \cap Y$,
which is open in $Y$ by definition of the subspace topology on $Y$.  Thus
$V_1$ is a neighborhood of $x_1$ in $Y$.  Similarly, a set containing
$x_2$ in $Y$ is $V_2 = U_2 \cap Y$, which is open in $Y$ by the definition
of the subspace topology on $Y$.  Thus $V_2$ is a neighborhood of $x_2$
in $Y$.  Now since $V_1 \subset U_1$ and $V_2 \subset U_2$, and $U_1$ and
$U_2$ are disjoint, it follows that $V_1$ and $V_2$ are disjoint.  Thus, 
$Y$ is Hausdorff. 
\\ \\ \noindent 

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\begin{bf}5.13\end{bf} 
Show
that $X$ is Hausdorff if and only if the diagonal $\Delta = \{ x \times x
\; | \; x \in X \}$ is closed in $X \times X$. \\ \\ \noindent $(1)
\Rightarrow (2)$ Given $X$ is Hausdorff, show $\Delta$ is closed in $X
\times X$. 




