Exercise 8.2.8

Question

Let $(X,d)$ be a metric space.
\begin{enumerate}[label=(\alph*)] 
\item Verify that a typical $\epsilon$-neighborhood $V_\epsilon(x)$ is an open set. Is the set
$$C_\epsilon(x) = \{y \in X : d(x,y) \leq \epsilon\}$$
a closed set?
\item Show that a set $E \subseteq X$ is open if and only if its complement is closed.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item For any point $y \in V_\epsilon(x)$, define $a = \epsilon - d(x,y)$; by the triangle inequality $V_a(y) \subseteq V_\epsilon(x)$.

Consider any limit point $y$ of $C_\epsilon(x)$. For any $a > 0$, $\exists z \in C_\epsilon$ where
$$d(x, y) \leq d(x, z) + d(z, y) < \epsilon + a$$
therefore $d(x,y) \leq \epsilon$ and so $C_\epsilon(x)$ is closed.

\item The proof is identical to that of Theorem 3.2.13, which is this statement in the special case for $\mathbf{R}$ with the usual metric.
 \end{enumerate}

Exercise 8.2.8

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Exercise 8.2.8

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