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Let $d$ be a metric on a set $X$, and let $$ B=\{B(p,e) = \{y\in X \mid d(p, y)<\epsilon \}\text{for every $p\in X$ and every $\epsilon>0$}\} $$ For $B$ to be the basis of a topology on $X$, then $\emptyset\in B$, but I don't see how this can be guaranteed since $\epsilon>0$.

Thanks so much!

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It is not true that the empty set must be in the basis. – Jonas Meyer Nov 16 '12 at 6:15
Every open set must be a union of basis elements, so doesn't that imply that the empty set must be in the basis? – William Stagner Nov 16 '12 at 6:22
$\varnothing$ is the union of the empty set of elements of the base. That is, if $\mathscr{B}$ is the base, $\varnothing$ is one of the subsets of $\mathscr{B}$, and the open set $\varnothing=\bigcup\varnothing$. – Brian M. Scott Nov 16 '12 at 6:24
Oh, okay, that makes sense. Thank you very much! – William Stagner Nov 16 '12 at 6:57

The original answer is wrong, as you can see in the comments to it and to the question. Like the commenters said, you don't need the open set to be in the basis.

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You can’t show it, because it isn’t necessarily true: there is no requirement that $\varnothing$ belong to a base for a topology. Your argument is incorrect, I’m afraid: if $B_1\cap B_2=\varnothing$, the requirement that you mention is vacuously satisfied: since there is no $x\in B_1\cap B_2$, no open set $B_3$ of any kind is needed. – Brian M. Scott Nov 16 '12 at 6:18
That's true, and I think that's part of why I'm a little uncomfortable about my answer: I used the fact I can separate two points before I actually had the topology on the set (that is, we haven't showed yet that the empty set is in the topology). It now seems wrong to me to use that before we ``accept" the topology on the set. – eliya gwetta Nov 16 '12 at 6:20
No, none of that is the problem. You’re not using Hausdorffness anywhere, and we’re given a topology on the set, namely, the topology induced by the metric. The problem is that you’ve seriously misunderstood what is meant by something being satisfied vacuously, badly enough that your argument doesn’t actually make sense. – Brian M. Scott Nov 16 '12 at 6:23
Totally unrelated, but quotes like ``word'' are for LaTeX source, whereas quotes like "word" work fine on Math.SE. – Austin Mohr Nov 16 '12 at 6:27
+1 for the honest revision. – Brian M. Scott Nov 16 '12 at 6:28

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