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\bf{Question}

\quad Show from the definition of {\it open set} in $\br^n$ that
if $A$ and $B$ are open sets then $A\cap B$ and $A\cup B$ are open
sets.  Show the same results hold with {\it open} replaced by {\it
closed}.  Deduce that the intersection and union of a finite
collection of open (closed) sets is open (closed).
\smallskip
Show from the definition that the {\it union} of an infinite
collection of open sets is necessarily open.  By considering the
open intervals $(-{1\over n},{1\over n})$ in $\br$ or otherwise,
show that the {\it intersection} of an infinite collection of open
sets need not be open.
\smallskip
Show that the {\it intersection} of an infinite collection of
closed sets is necessarily closed, and give an example of an
infinite collection of closed sets whose {\it union} is not
closed.




\bf{Answer}

$A$, $B$ open: Let $p \in A \cap B$.

Since $A$, $B$ are open there exist $\alpha$, $\beta>0$ such that
$B_{\alpha}(p) \subset A$, $B_{\beta}(p) \in B$.

Let $\epsilon=\textrm{min}\{ \alpha, \beta \}$: Then
$B_{\epsilon}(p)\subset A\cap B$. Thus $A\cap B$ open.

Let $q \in A \cup B$. If $q \in A$ there exists $\alpha>0$ such
that $B_{\alpha}(q) \subset A \subseteq A \cup B$. If $q \in B$
there exists $\beta >0$ such that $B_{\beta}(q) \subset B
\subseteq A \cup B$. Thus $A \cup B$ open.

\begin{eqnarray*}
A, B \textrm{closed} & \Leftrightarrow & \Re^n \backslash A, \Re^n
\backslash B \textrm{ open (by definition)}\\ & \Rightarrow &
(\Re^n \backslash A) \cap (\Re^n \backslash B) \textrm{ open by
the above}\\ & \Rightarrow & \Re^n \backslash (A \cup B) \textrm{
open}\\ & \Rightarrow & A \cap B \textrm{ closed (by definition)}.
\end{eqnarray*}
Likewise for $A \cap B$, using $\Re^n \backslash (A \cap B) =
(\Re^n \backslash A) \cup (\Re^n \backslash B).$

Let $\{ C_{\lambda} \}_{\lambda\in\Lambda}$ be a collection of
open sets, and suppose $p \in\bigcup_{\lambda\in\Lambda}
U_{\lambda} = W$, say.

This means there is at least one $\lambda \in Lambda$ with $p\in
U_{\lambda}$. Since $U_{\lambda}$ is open there exists $\epsilon
>0$ with $B_{\epsilon}(p) \subset U_{\lambda} \subseteq W$. Hence
$W$ is open.

If $I_n = (-\frac{1}{n}, \frac{1}{n})$ then $\bigcap_n=1^\infty
I_n$ consists of $\{ 0 \}$ only: this is not open in $\Re$.

If $\{ C_{\lambda} \}_{\lambda\in\Lambda}$ is a collection of
closed sets, then $\{ \Re^n \backslash C_{\lambda}
\}_{\lambda\in\Lambda}$ is a collection of open sets, with $\Re^n
\backslash \bigcap_{\lambda\in\Lambda} C_{\lambda} =
U_{\lambda\in\Lambda} (\Re^n - C_{\lambda})$; then use above
result.

If $J_n=\Re \backslash I_n = (-\infty, -frac{1}{n}] \cup [
\frac{1}{n}, \infty)$ which is closed in $\Re$ then
$\bigcup_{n=1}^{\infty} = \Re^n \backslash \{ 0 \}$ which is not
closed in $\Re$.


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