# Subbasis for Cartesian Product Topology

Let $\aleph(\mathcal{A})$ be arbitrary (i.e., let $\mathcal{A}$ be an indexing set of arbitrary cardinality). Then, in the space $\prod \{Y_{\alpha}|\alpha \in \mathcal{A}\}$, for each $\alpha \in \mathcal{A}$, let $\displaystyle \Sigma_{\alpha}$ be a subbasis for the topology $\mathcal{T}_{\alpha}$ of $Y_{\alpha}$.

I need to prove that the family $\{ \langle V_{\beta}\rangle \,|\,\text{all}\,v_{\beta}\in \Sigma_{\beta}; \, \text{all}\, \beta \in \mathcal{A} \}$ is also a subbasis for the cartesian product topology in $\displaystyle \prod_{\alpha}Y_{\alpha}$

Please note that this is NOT homework! I have my final exam on Wednesday and I am trying to read through Chapter IV of Dugundji's Topology to help me prepare, but I still am having a lot of trouble even understanding how this cartesian product topology works.

This problem appears in the text as part of a given theorem, and then for the proof, it says "This is immediate from I, 9.5, and is left for the reader".

I. 9.5 states that

"Let $\{Y_{\alpha}|\alpha \in \mathcal{A}\}$ be a family of nonempty sets; for each $\alpha \in \mathcal{A}$, let $A_{\alpha}, B_{\alpha}$ be subsets of $Y_{\alpha}$. Then, $$(1) \, \displaystyle \prod_{\alpha}A_{\alpha} \cap \prod_{\alpha}B_{\alpha}$$ $$(2) \displaystyle \prod_{\alpha}A_{\alpha} \cup \prod_{\alpha}B_{\alpha} \subset \prod_{\alpha}\left(A_{\alpha} \cup B_{\alpha} \right)"$$

I don't really see how this result is "immediate" from this - and even if it were, why would it be left to the reader? If something is left to the reader that implies that it requires a little bit of work and mechanics and manipulation, right?

On that front, I believe the way to show that this family $\{\langle V_{\beta} \rangle\}$ is a subbasis for the product topology, call it $\mathcal{T}_{p}$ is to show that the topology generated by $\{\langle V_{\beta} \rangle \}$, call it just $\mathcal{T}$ is the same topology as $\mathcal{T}_{p}$. However, I am not sure how to go about doing this. Especially since we need to be a little bit careful, because isn't it true that a set of the form $\displaystyle \prod U_{\alpha}$ where each $U_{\alpha}$ is a proper open subset of $Y_{\alpha}$ is never an open set in the cartesian product $\displaystyle \prod Y_{\alpha}$? (This is what I meant when I said I was having trouble understanding how the cartesian topology works.)

Anyway, to start, I'd say, suppose $\langle V_{\beta} \rangle = p_{\beta}^{-1}(V_{\beta})$ is a subbasis for $\mathcal{T}$. Then, taking finite intersections of these $p_{\beta}^{-1}(V_{\beta})$, we obtain basic elements for the topology $\mathcal{T}$. Now, from a different result related to I. 9.5, we have that $\displaystyle \cap_{\beta} \langle V_{\beta} \rangle = \prod_{\beta} V_{\beta}$, and so our basic elements look like $\displaystyle \cup_{\beta}\prod_{\beta}V_{\beta}$.

Then, by I. 9.5, we have that $\displaystyle \cup_{\beta}\prod_{\beta}V_{\beta} \subset \prod_{\beta}\left( \cup_{\beta}V_{\beta} \right)$. But then, how do I get that this gives me the open sets in $\mathcal{T}_{p}$?

Then, going in the other direction is also a problem for me. How do I show that $\mathcal{T}_{p} \subset \mathcal{T}$ here?

There is a question already posted on here where the OP asked for a proof of this result using the approach of proving that the product topology is the smallest topology containing $\{ V_{\beta} \}$, but I'm not sure I really understand the language the answer uses regarding maximal elements, or even why proving that it is the smallest such topology is helpful (for reference, the question/answer is here).

What I would like is for someone to provide me with a worked out detailed proof building on what I've already said here (supposing it's right, of course) explaining all the why's. Like I said, this is not homework - I was just trying to work through it on my own to prepare for my exam, and I got stuck. I have a tendency of doing that - getting to a point and then getting stuck for hours in a quagmire of details and losing loads of valuable study time because of it. Therefore, any help you could give me would be greatly appreciated!

So the product topology on $\Pi_{\alpha \in \mathcal A}Y_{\alpha}$ is defined as the topology generated by subbase $\cup_{\alpha \in \mathcal A}\mathcal C_{\alpha}$ (i.e. the coasest topology for which all the projections $\{\pi_{\alpha}\}$ are continuous), where $$\mathcal C_{\alpha} = \{\pi_{\alpha}^{-1}(A): \, A \in \mathcal T_{\alpha}\}$$
Now consider $$\mathcal D_{\alpha} = \{\pi_{\alpha}^{-1}(A): \, A \in \Sigma_{\alpha}\}$$
Since $\Sigma_{\alpha}$ is a subbase of $\mathcal T_{\alpha}$, $\forall O \in \mathcal T_{\alpha}$, it could be expressed as arbitrary union of sets, and those sets are expressed by finite intersection of elements in $\Sigma_{\alpha}$
Also we know that $f^{-1}(\cap_i A_i)=\cap_i f^{-1}(A_i)$ and $f^{-1}(\cup_i A_i)=\cup_i f^{-1}(A_i)$
Thus any element in $\mathcal C_{\alpha}$ could be expressed by arbitrary unions of sets, and those sets could be expressed as finite intersection of elements in $\mathcal D_{\alpha}$
Since $\cup_{\alpha \in \mathcal A}\mathcal C_{\alpha}$ is a subbase of the product topology, so is $\cup_{\alpha \in \mathcal A}\mathcal D_{\alpha}$. So we are done.