Let $(X, d)$ be any metric space, and let $E \subset X$ such that
$$
E \subset \cup_{i = 1}^k B \left( x_i, \epsilon \right),
$$
for some points $x_1, \ldots, x_k \in X$ and for some $\epsilon > 0$. Then how to show that
$$
\overline{E} \subset \cup_{i = 1}^k B \left( x_i, 2\epsilon \right)?
$$
My Attempt:
Let $p$ be any point of $\overline{E}$. Suppose that $d \left( p, x_i \right) \geq 2 \epsilon$ for each $i = 1, \ldots, k$. Then since, for any point $x \in E$, there is a point $x_{i_0}$ such that $d \left( x, x_{i_0} \right) < \epsilon$, therefore we obtain
$$
d (p, x) \geq d \left( p, x_{i_0} \right) – d\left( x_{i_0}, x \right) \geq 2 \epsilon – \epsilon = \epsilon.
$$
So for any real number $\delta$ such that $0 < \delta < \epsilon$, we have
$$
B (p, \delta) \cap E = \emptyset,
$$
which contradicts the assumption that $p \in \overline{E}$. Hence our supposition is wrong and thus there exists an $i = i_1$ such that $d \left( p, x_{i_1} \right) < 2 \epsilon$.
Is this proof correct and clear enough? Or, are there any errors?
Best Answer
Yes, this is correct, but a direct proof is more elegant: Let $x \in \overline{E}$ and $y \in E$ such that $d(x, y) < \varepsilon$. Furthermore let $j \in \lbrace 1, ..., k \rbrace$ such that $y \in B(x_j, \varepsilon)$. Then: $$ d(x, x_j) \leq d(x, y) + d(y, x_j) < \varepsilon + \varepsilon = 2\varepsilon $$ Therefore $x \in B(x_j, 2 \varepsilon) \subseteq \displaystyle \bigcup_{i = 1}^k B(x_i, 2 \varepsilon)$.