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\bf{Question}

\quad  Let $[a,b]$ be a closed interval in $\br$, and let $A$ be a
given subset of $\br$ such that $a\in A$ and $b\notin A$. The
purpose of this exercise is to show that $A$ cannot be both open
and closed (so the only sets in $\br$ which are both open and
closed are the empty set $\emptyset$ and the real line $\br$
itself).
\smallskip
Let $V=A\cap[a,b]$, and let $\xi$ be the {\it supremum} (least
upper bound) of $V$.  Clearly $a&lt;\xi&lt;b$ since $a\in V$ but
$b\notin V$.  Does $\xi$ belong to $A$ or not ?  Show that if $A$
is open then $\xi$ does not belong to $A$, and if $A$ is closed
then $\xi$ does not belong to the complement of $A$.  Therefore
$A$ cannot be both open and closed.



\bf{Answer}

Suppose $\xi\in A$. Since $A$ is open, there exists $\epsilon &gt;0$
with $(\xi -\epsilon, \xi + \epsilon) \subset A$ and with $\xi +
\epsilon &lt; b$, so $\xi + \epsilon \in V$ and of course $\xi +
\epsilon &gt; \xi$.

This contradicts $\xi$ being an upper bound for $V$.

Let $B = \Re \backslash A$. Suppose $\xi \in B$, since $B$ is open
, there exists $\epsilon &gt; 0$ with $(\xi -\epsilon, \xi +
\epsilon) \subset B$ and with $\xi-\epsilon&gt;a$, so $\xi-\epsilon$
is an upper bound for $V$ but $\xi - \epsilon &lt; \xi$.

This contradicts $\xi$ being the least upper bound for $V$. So
$\xi$ can belong to neither $A$ nor $B=\Re \backslash A$;
impossible, so $A$ cannot be both open and closed.


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