Interior of cartesian product

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Is it true in general that for an arbitrary family of sets $(A_i)_{i\in I}, A_i\subseteq X_i$, where $(X_i,\tau_i)$ are topological spaces, we have that:

$$\mathrm{Int}\left (\prod_{i\in I} A_i\right)=\prod_{i\in I} \mathrm{Int}{A_i}$$? The left hand side is considered in the product topology $(X,\tau)$, where $X=\prod_{i\in I} X_i$.

It is known that for finite families of sets it is true.

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Not quite. Consider the space $\mathbb{R}^\infty$ with product topology. Let $A_i = I= (0,1)$. Then there is no basic open sets which are contained in $\prod_i A_i$. Therefore $\prod_i A_i$ has empty interior whereas $\prod \hbox{Int}A_i = \prod A_i$ is not empty.

Remember that in the product topology, basic opensets are those $\prod U_i$ where

  1. $U_i$ is open in $X_i$
  2. $U_i = X_i$ for all but finitely many $i$s.