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\begin{flushright}
A. Miller \\
M551 \\
 Spring 2001\\
Exam 1\\
\end{flushright}


\prob Let $X$ be a topological space and $A\subseteq X$.  Prove
that for every $x\in X$
\par $x\in\overline{A}$ iff
\par for all $U$ an open neighborhood of
$x$ we have that $U\cap A\not=\emptyset$.


\prob Let $X$ and $Y$ be topological spaces and $f:X\to Y$ a function.
Assume that for every $A\subseteq X$ that $f(\overline{A})\subseteq
\overline{f(A)}$.
\par Prove that $f$ is continuous.

\prob % 2-2
Suppose $f:X\to Y$ is a function and $A$ and $B$ are subsets of
$Y$.
\par Prove that $f^{-1}(A-B)=f^{-1}(A)-f^{-1}(B)$.

\prob % 13-5
Suppose ${\cal B}$ is a basis for a topology $\tau$ on $X$.
\par Prove that $\tau$ is the intersection of all topologies $\tau^\prime$
on $X$ such that ${\cal B}\subseteq \tau^\prime$.

\prob Let $X$ be a Hausdorff space and $A\subseteq X$.
\par Prove that
the set of limit points of $A$ is a closed subset of $X$.

\begin{center}
  Answers
\end{center}

1. Thm 17.5a, p.96

2. Thm 18.1, p.104 (2)$=>$(3)$=>$(1)

3. Exer 2d, p.20

4. Exer 5, p.83

5. $A^\prime$ denotes the limit points of $A$. Suppose $p\in X$ is not in
$A^\prime$.  Then there exists an open neighborhood $U$ of $p$ such that
there are no $q\in A\cap U$ which are different from $p$.

Claim. $U\cap A^\prime=\emptyset$

Suppose $q\in U$.  If $q=p$, then we already know that $q\notin A^\prime$.
If $q\not=p$, then $q\in V=U-\{p\}$.  But $V$ is an open set, since
in a Hausdorff space, the set $\{p\}$ is closed and so
$V=U\cap(X-\{p\})$ is the intersection of two open sets and hence is open.
Since $A\cap V=\emptyset$ we have that $q\notin A^\prime$ (in fact, it
is not even in the closure of $A$).

The claim proves that for any $p\in X-A^\prime$ there
is an open neighborhood $U$ of $p$ such that $U\cap A^\prime=\emptyset$
It follows that
$$X-A^\prime=\bigcup\{U^{\rm open}\subseteq X\;:\; U\cap A^\prime=
\emptyset\}$$
and so the complement of $A^\prime $ is the union of open sets hence
open and so $A^\prime$ is
closed.





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