# If $B_a$ is a basis for the topology on $X_a$, how to show that $\{\prod b_a\mid b_a\in B_a\}$ is a basis for the box topology on $\prod X_a$?

If not, what is a counterexample?

I'm trying to prove Theorem 19.2 from Munkres's Topology.

Suppose the topology on each $X_a$ is given by a basis $B_a$. The collection $B$ of all sets of the form $\prod b_a$, where $b_a \in B_a$ for all $a$, will serve as a basis for the box topology on $\prod X_a$.

At first, I thought it was enough just to show that $B$ satisfies the basis definition. Do I also need to show that the topology generated by $B$ equals the box topology?

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You will need to show that for every element $x$ in the product and every box topology neighborhood $U$ of $x$, there is an element of $B$ such that $x\in B\subseteq U$.

Edit: Once you've done that, you need only confirm that every element of $B$ is open in the box topology on $X$--which should be a fairly simple task, as elements of a basis $B_a$ of $X_a$ are necessarily open in $X_a$. Then by Lemma 13.2 of Munkres, $B$ will be a basis for the box topology on $X$, as desired.

Note that this last detail is actually essential to include! (I neglected to mention it because it should be clear that the elements of $B$ are open in the box topology on $X$, but in hindsight, I shouldn't have omitted that detail. Apologies for any confusion!) For example, if we take $B'$ to be the set of all singletons $\{x\}$ (for $x\in X$), the set $B'$ rather trivially satisfies the condition I mentioned in the first place, but while it is a basis, it is a basis for the discrete topology--strictly a finer topology (in general) than the box topology.

All the condition I mentioned at first really shows is that if $B$ is a basis for some topology, then it generates a topology finer than the box topology on $X$ (possibly strictly finer). Knowing that the elements of $B$ are open in the box topology allows us to conclude that (1) $B$ is a basis for a topology on $X$, and that (2) the generated topology is coarser than (so equivalent to, since also finer than) the box topology on $X$.

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Thanks. I was confused by the wording of the question. Isn't it asking if the topology generated by the given basis equals the box topology? –  Steven Li May 17 '12 at 17:34
@sli: Yes, provided that you recognize that this includes showing that $B$ is a base at all. It’s asking you to show that $B$ is a base for a topology $\tau$ on the product and that this $\tau$ is the box topology. –  Brian M. Scott May 17 '12 at 20:31
Apologies, @sli. I left out one fairly critical detail--which fortunately is almost trivial to see. See my edit for the addition. –  Cameron Buie May 18 '12 at 1:38
Thanks for the help! –  Steven Li May 22 '12 at 2:43
@sli: No problem at all! Did you get through it, then? –  Cameron Buie May 22 '12 at 4:39