Counter Example in Topology regarding Product Topology

Hello everyone,

I'm in need of some assistance with the following quesiton:

Let $\displaystyle \left\{ X_{\alpha}\right\} _{\alpha\in I} $ be collection of topological spaces and let $\displaystyle \left\{ A_{\alpha}\right\} _{\alpha\in I} $ be collection of subsets. Prove or bring a counter example showing whether:

$\displaystyle \prod_{\alpha\in I}int\left(A_{\alpha}\right)=int\left(\prod_{\alph a\in I}A_{\alpha}\right) $

in a) the product topology, b) the box topology.

I'm certain that it is incorrect in the product topology but I have yet to find a counter-example, as for the box topology I'm not sure.

I would greatly appreciate if someone could help me out here!

Thanks a lot.

Re: Counter Example in Topology regarding Product Topology

I've managed to resolve the question.

Regarding the Box Topology the claim is correct directly from the definition of the open sets in that topology. Since every interior is an open set in the according topological space the product of the interiors is a basis element of the box topology and thus open itself, meaning it is the same as its interior.

Regarding the product topology the claim is incorrect. I proved this using a small llema stating that the interior of the product is an empty set unless apart for a finite number of indexes it holds that $\displaystyle A_{\alpha}=X_{\alpha} $.

Thus is suffices to take as a counter example $\displaystyle \left(0,1\right)^{\mathbb{N}}\subset\mathbb{R^N}}}$ since

$\displaystyle \prod_{i\in\mathbb{N}}int\left(0,1\right)=\prod_{i \in\mathbb{N}}\left(0,1\right)=\left(0,1\right)^{\ \mathbb{N}}$

is clearly not empty.