2
$\begingroup$

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!

Thanks ahead of time.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.