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Show that if B is a basis for a topology on X , then the topology generated by B is equal the intersection of all topologies on X that contains 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? – Pete L. Clark Apr 11 '11 at 6:33
@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. – joriki Apr 11 '11 at 7:07
@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...) – Pete L. Clark Apr 11 '11 at 14:33

Perhaps, as joriki suggests, the question is to show that for a base $\mathcal{B}$ for a topology on $X$, the family $\tau_{\mathcal{B}}$ of arbitrary unions of elements of $\mathcal{B}$ -- including, I suppose, the empty union, which gives the empty set -- is the smallest topology containing $\mathcal{B}$.

If so, this is certainly true and can be shown as follows:

Step 1: Since any topology is closed under arbitrary unions, any topology $\tau$ containing $\mathcal{B}$ must contain $\tau_{\mathcal{B}}$, so it is enough to show that $\tau_{\mathcal{B}}$ is a topology.

Recall that what we are assuming about $\mathcal{B}$ is that

(B1) $\bigcup_{B \in \mathcal{B}} B = X$ and
(B2) For all $B_1, B_2 \in \mathcal{B}$, if $x \in B_1 \cap B_2$, then there exists $B_3 \in \mathcal{B}$ such that $x \in B_3 \subset B_1 \cap B_2$.

Step 2: Thus $\emptyset, X$ are unions of elements of $\mathcal{B}$: the former by taking the empty union, the latter by (B1).

Step 3: Being the set of all unions of a certain family of sets, $\tau_{\mathcal{B}}$ is certainly closed under arbitrary unions.

Step 4: So the matter of it is to show that $\tau_{\mathcal{B}}$ is closed under finite intersections. For this, it is enough to show that if $U_1,U_2 \in \tau_{\mathcal{B}}$, then so is $U_1 \cap U_2$. To show this we need to use condition (B2), which notice has not yet been used. This verification takes two or three lines. I urge the OP to try it herself and tell us whether she succeeded and if not what she tried.

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