Exercise 8.2.14

Question

\begin{enumerate}[label=(\alph*)] 
\item Give the details for why we know there exists a point $x_2 \in V_{\epsilon_1}(x_1) \cap O_2$ and an $\epsilon_2 > 0$ satisfying $\epsilon_2 < \epsilon_1 / 2$ with $V_{\epsilon_2}(x_2)$ contained in $O_2$ and
$$\overline{V_{\epsilon_2}(x_2)} \subseteq V_{\epsilon_1}(x_1)$$
\item Proceed along this line and use the completeness of $(X, d)$ to produce a single point $x \in O_n$ for every $n \in \mathbf{N}$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item It's clear from the definition that, for a dense set $S$, any point $x \in X$, and any $\epsilon > 0$, $V_\epsilon(x) \cap S \neq \emptyset$. Since $O_2$ is dense, there must be some $x_2 \in V_{\epsilon_1 / 2}(x_1) \cap O_2$. Now since $O_2$ is open, we have that for some $\epsilon_a$, $V_{\epsilon_a}(x_2) \subseteq O_2$. We can also find $\epsilon_b$ so that $\overline{V_{\epsilon_b}(x_2)} \subseteq V_{\epsilon_1}(x_1)$. To do this, recall from Exercise 8.2.12 (a) that if $x \in \overline{V_{\epsilon_b}(x_2)}$, then $d(x, x_2) \leq \epsilon_b$. By choosing any $\epsilon_b < \epsilon/2$, we have $$d(x, x_1) \leq d(x, x_2) + d(x_2, x_1) < \epsilon/2 + \epsilon/2 = \epsilon$$ and hence $x \in V_{\epsilon_1}(x_1)$. Finally, set $\epsilon_2 = \min\{\epsilon_a, \epsilon_b \} $, so that $\epsilon_2$ satisifies both properties. \item We can repeat this process to define $\epsilon_n > 0$ and $x_n$ for $n > 1$, satisfying $\epsilon_n < \epsilon_{n-1} / 2$, $V_{\epsilon_n}(x_n) \subseteq O_n$, and $\overline{V_{\epsilon_n}(x_n)} \subseteq V_{\epsilon_{n-1}}$. Note that $(\epsilon_n) \to 0$, and that $\sum^\infty_{i=n} \epsilon_i < 2\epsilon_n$. These two facts imply that $(x_n)$ is a Cauchy sequence, and by completeness $(x_n)$ converges to some $x \in X$. Theorem 3.2.5 is still true for arbitrary metric spaces, and the proof can be reused (replacing the absolute value function with a general distance metric). Let $N \in \mathbf{N}$ be arbitrary. Noting that for $n \geq N + 1$, $x_n \in V_{\epsilon_{N+1}}(x_{N+1})$, we have $x \in \overline{V_{\epsilon_{N+1}}} \subseteq V_{\epsilon_N}$. Hence, $x \in O_n$ for any $n \in \mathbf{N}$. \end{enumerate}

Exercise 8.2.14

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Exercise 8.2.14

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