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I'm not even convinced that this is true, and I'm hoping that someone can help me to see why it is true.

I am attempting to prove the following:

Consider a metric space $F$ and a set $E$ that is dense in $F$. Show that the set of all open balls in E, centered at $e \in E$ and with rational radii, say $\{B(e, r_i)\}$, are a base for $F$.

To show that $\{B(e, r_i)\}$ is a base for $F$, I'd need to show that for any open set $T \subset F$ and element $t \in T$, there exists a ball $B(e, r_t) \in \{B(e, r_i)\}$ such that the following is true:

$$t \in B(e, r_t) \subset T$$

But, I don't even believe that this is true. If all of the balls in $\{B(e, r_i)\}$ must be centered at a fixed point $e$, then how can one of them always be contained in an arbitrary, open subset of $F$? The picture below is meant to illustrate my confusion.

on-pasta
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    "...centered at $e\in E$..." Shouldn't that be red as "...centered at some $e\in E$..."? – drhab Oct 13 '18 at 06:53
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    @drhab It seems so, but no. In the question I've got, it says "centered at $e \in E$", which I've taken to mean a fixed point $e$. – on-pasta Oct 13 '18 at 06:55
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    There is some controversy about the interpretation of the statement (e.g. I think it means we should take balls around all points $e \in E$, not just a fixed one, and in that thinking, I am not alone). So do you actually want to ask (i) whether you are correct that your interpretation makes the statement false, or (ii) which interpretation is the correct one? – Adayah Oct 13 '18 at 10:54

3 Answers3

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The correct statement is: if $E$ is dense in $F$, then the set

$$\mathcal{B} = \{B(x,r): x \in E, r \in \mathbb{Q}, r > 0\}$$ is a base for $F$.

So the radii are not a fixed point, but are members of $E$ in general. So they are centered at some $e \in E$..

See the implication 7 -> 1 in this answer for details of the proof itself.

Henno Brandsma
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Minimal counterexample. Take $E = F = \{a,b\}$ equipped with the discrete metric: $$ d(x,y) = \begin{cases} 0 & \text{if $x = y$} \\ 1 & \text{if $x \not= y$} \end{cases} $$ The topology defined by $d$ is the discrete topology. However, if you take only one $e \in E$, you only get two open balls, $\{e\}$ and $F$ and these two balls do not form a basis of the discrete topology.

That being said, I fully subscribe to @drhab comment: I would understand the statement as "for some $e \in E$".

J.-E. Pin
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You are correct! The statement, as you write, is untrue.

Every metric space is Hausdorff. Therefore, for $e$ and any other point $p$, there exist disjoint open sets $E$ and $P$ such that $e \in E, p \in P$. In particular, $e \not \in P$. However, any union of open sets, all containing $e$, contains $e$.

Therefore P is not the union of open sets containing $e$, so the balls around $e$ are not a base for the metric space (containing at least 2 points).

user156213
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