Is a product nowhere dense?

Question

I'm reading Haworth1977, Baire spaces.

On page 7 the authors say

...the finite product of subsets is nowhere dense iff at least one of the subsets is itself nowhere dense. However, this is not necessarily true for infinite product

and they give an example of countable product that is nowhere dense but every component is somewhere dense. Later they state the following proposition

For each $\alpha \in A$ let $N_\alpha$ be a subset of the space $X_\alpha$. Then, $\prod_{\alpha\in A}N_\alpha$ is nowhere dense in $\prod_{\alpha\in A}X_\alpha$ iff for some $\beta\in A$, $N_\beta$ is nowhere dense in $X_\beta$ or $\text{cl}N_\alpha\ne X_\alpha$ for infinitely many $\alpha\in A$.

I must miss something because to me the proposition says exactly the contrary of the previous comment.

If someone has the book Kuratowski Topologie I (1958, 4th ed., I assume written in French) the proposition above corresponds to some proposition on page 154 of the given edition of Kuratowski, and it would be interesting to have the proposition of that book, French is also fine. (Haworth1977 refers to Kuratowski)

Answer 1

Let $A$ be a set. For all $\alpha\in A$ let $X_\alpha$ be a topological space and let $N_\alpha$ be a subset of $X_\alpha$.

If $A$ is finite then

$\prod_{\alpha\in A} N_\alpha$ is nowhere dense in $\prod_{\alpha\in A} X_\alpha$ if and only if there exists $\beta\in A$ such that $N_\beta$ is nowhere dense in $X_\beta$.

If $A$ is infinite (countable or larger) then

$\prod_{\alpha\in A} N_\alpha$ is nowhere dense in $\prod_{\alpha\in A} X_\alpha$ if and only if there exists $\beta\in A$ such that $N_\beta$ is nowhere dense in $X_\beta$ OR there exists infinitely many $\beta\in A$ such that the closure of $N_\beta$ is not $X_\beta$ (i.e. $N_\beta$ is not dense in $X_\beta$).

Note that if $A$ is finite then the condition "there exists infinitely many $\beta\in A$ such that..." can't be satisfied, so the second statement and the first statement are equivalent for finite $A$.

The main idea in this condition is that the product topology on a product space contains a "co-finite" cylinder set (it contains a product of all but finitely many of the $X_\alpha$) and so if the closure of $N_\beta$ is not $X_\beta$ for infinitely many $\beta$ then the closure of $\prod_{\alpha\in A} N_\alpha$ can't contain an open set.