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Show that if $B$ is a basis for a topology on $X$ , then the topology generated by $B$ is equal to the intersection of all topologies on $X$ that contain $B$.

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    I'm a little confused: for any family $\mathcal{F}$ of subsets of a set $X$, the topology generated by $\mathcal{F}$ is **by definition** the intersection of all topologies on $X$ that contain $\mathcal{F}$. So what is there to show?2011-04-11
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    @Pete: Isn't there also another definition of the topology generated by $B$ as the set of arbitrary unions of elements of $B$? If that is so, the question would amount to showing the equivalence of these two definitions.2011-04-11
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    @joriki: Thanks for you comment: it is indeed a plausible interpretation of the question. (I actually had a similar thought after I made my comment, but rather than address it I chose to go to sleep...)2011-04-11

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