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Let $A,B$ be topological spaces with $T_A,T_B$ their respective topologies and $\mathcal{B}_A,\mathcal{B}_B$ their respective bases.
How do I show that

$\{F\times G|F\in \mathcal{B}_A, G\in \mathcal{B}_B\}$ is a basis of $A\times B$?

What I thought:
Let $U\times V$ be open in $A\times B$, then $U=\cup_{i\in I}F_i$ and $V=\cup_{j\in J}G_j$.
I suppose that $U\times V=\cup_{i\in I}F_i\times\cup_{j\in J}G_j$. Is this correct? And how can I write this as a union of $F\times G$?

1 Answers1

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Start by proving that a set $U\subseteq A\times B$ that is open in the product topology can be written as a union of elements of the collection $\mathcal V:=\{F\times G\mid F\in\mathcal B_A,G\in\mathcal B_B\}$.

Let $\langle x,y\rangle\in U$.

On base of the definition of "product topology" we conclude that some $V_x\in T_A$ and some $W_y\in T_B$ exist with:$$\langle x,y\rangle\in V_x\times W_y\subseteq U$$

$\mathcal B_A$ is a base of $T_A$ so $x\in F_x\subseteq V_x$ for some $F_x\in\mathcal B_A$.

$\mathcal B_B$ is a base of $T_B$ so $y\in G_y\subseteq W_y$ for some $G_y\in\mathcal B_B$.

Then we have:$$\langle x,y\rangle\in F_x\times G_y\subseteq V_x\times W_y\subseteq U$$

We can do this for every $\langle x,y\rangle\in U$ enabling us to write $U$ as a union of elements of the collection:$$U=\bigcup_{\langle x,y\rangle\in U}F_x\times G_y$$

A second thing that has to be shown is that the intersection of two elements $F_1\times G_1$ and $F_2\times G_2$ of $\mathcal V$ can be written as a union of elements of $\mathcal V$. But it is evident that this is the intersection of two sets that are open in the product topology, hence the intersection is open in the product topology also. Above we allready proved that such a set can be written as union of elements of $\mathcal V$, so we are ready.

drhab
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    The last part is not needed. You just need to show that the product of bases is a base (for finitely many spaces), we already know here that all open times open sets are a base for the product topology. To then show that we can thin out this standard base to the base given in the exercise is what you have done in part one. – Henno Brandsma Mar 05 '17 at 15:38
  • @HennoBrandsma I don't understand the words "not needed" in your comment. If the last part is left out then in my view the proof is incomplete. I agree that another way to complete it is proving that the product of a finite number of bases is a base. – drhab Mar 07 '17 at 09:58
  • The question is not to show it is a base for some topology, but a base for the product topology. This is only part 1 – Henno Brandsma Mar 07 '17 at 11:57
  • @HennoBrandsma I think I understand now. If $\mathcal V\subseteq\tau$ and elements of $\tau$ can be written as union of elements of $\mathcal V$ then of course this is also true for finite intersections of elements of $\mathcal V$, since they are automatically also elements of $\tau$. Thank you. – drhab Mar 07 '17 at 12:13