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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.2012-11-16
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    Every open set must be a union of basis elements, so doesn't that imply that the empty set must be in the basis?2012-11-16
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    $\varnothing$ 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 $\varnothing=\bigcup\varnothing$.2012-11-16
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    Oh, okay, that makes sense. Thank you very much!2012-11-16

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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.2012-11-16
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    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.2012-11-16
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    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.2012-11-16
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    Totally unrelated, but quotes like ``word'' are for LaTeX source, whereas quotes like "word" work fine on Math.SE.2012-11-16
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    You're right again that Hausdorffness isn't really used (I changed my comment around the same time you answered). You're also right about me using vacuously true wrong. To be honest, I think I'm not entirely sure what it really means. I've seen a similar argument in a real analysis book where open sets were defined to be sets where for each $x$ in the set, there's a $\delta >0$ such that $(x-\delta,x+\delta)$ is in the set. Then the book says that the open set satisfies the definition vacuously because there are no elements in $\emptyset$.2012-11-16
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    I ran out of space in the previous comment. Then the author proceeds by saying that for any $x$ in the empty set and any $\delta >0$, $(x-\delta,x+\delta) \subset \emptyset$. I see the bigger problem: I said that because the hypothesis is false, then the conclusion is true. And of course, that's nonsense. Sorry for the misleading answer and for taking the discussion completely off track. But I learned a valuable lesson out of it, so I guess it's not that bad. P.S. I see how my answer is a little different from what that book said. Sorry again for turning this thread into a huge mess.2012-11-16