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I'm working on this problem from Munkres:

Show that every metrizable space with a countable dense subset has a countable basis.

Here's my attempt at a proof.

Let $X$ be a metrizable space with metric $d$ and a countable dense subset $A$.

Claim: $\mathscr{B} = \{B_{(a,1/n)} = \{x : d(a,x) < \epsilon = 1/n\} \forall a\in A$} is a countable basis for $X$.

This set is countable since it is in bijection with $A\times \mathbb{N}$ which is countable as a countable product of countable sets.

Now I want to prove that $\mathscr{B}$ is indeed a basis.

$1)$ Let $y \in X$ then $y \in \overline A$ (by the definition of dense) which implies $\exists$ $B_{(y,1/n)}$ s.t. $\exists$ $y'\in A$ s.t. $y' \in B_{(y,1/n)}$. But then since $y$ is within $\epsilon = 1/n$ for some $n\in\mathbb{N}$ of $y'$, $\exists$ $B_{(y',1/n)}$ which is an element of $\mathscr{B}$. Therefore since $y$ was arbitrary we know that there is a basis element containing every element of $X$.

$2)$ Now I want to show that for an element $x$ of the intersection of two arbitrary basis elements that there is a third basis element which contains $x$ that is a subset of the intersection. This is where I get stuck, any help would be appreciated.

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Take two basic open sets, $B_1 = B(a_1, \frac{1}{n_1})$ and $B_2 =B(a_2, \frac{1}{n_2})$ where $a_1, a_2 \in A$. Suppose $x \in B_1 \cap B_2$. Since $B_1 \cap B_2$ is open, there exists an $n$ such that $B(x, \frac{1}{n}) \subseteq B_1 \cap B_2$. Since $A$ is dense, there exists $a_3 \in A \cap B(x, \frac{1}{2n})$. Then $x \in B(a_3, \frac{1}{2n}) \subseteq B(x, \frac{1}{n}) \subseteq B_1 \cap B_2$.

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  • $\begingroup$ I'm a little bit confused about the $\frac{1}{2n}$ part, how does this guarantee that $B(a_3,\frac{1}{2n}) \subseteq B(x, \frac{1}{n})$. Certainly if $a_3$ were more than half of the way towards the edge of $B(x, \frac{1}{n})$ then $B(a_3,\frac{1}{2n}$ would not be a subset. $\endgroup$ – EgoKilla May 9 '14 at 4:58
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    $\begingroup$ @EgoKilla $a_3 \in A \cap B(x,\frac{1}{2n})$. Use triangle inequality, if $y \in B(a_3, \frac{1}{2n})$, then $d(x,y) \leq d(x, a_3) + d(a_3, y) = \frac{1}{2n} + \frac{1}{2n}$ $\endgroup$ – William May 9 '14 at 5:35
  • $\begingroup$ Wonderful thank you for elaborating. $\endgroup$ – EgoKilla May 9 '14 at 5:43

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