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I know that the open subsets in the product topology of $X=X_1\times X_2\times...\times X_3$, where $X_1,X_2,...,X_n$ topological spaces, are the union of subsets of X: $U_1\times U_2\times ...\times U_n$, where $U_1,U_2,...,U_n$ are open subsets of $X_1,X_2,...,X_n$ respectively.

I have a question, how the closed subsets look like? can I say that they are the union of the subsets of X: $F_1\times F_2 \times...\times F_n$, where $F_1,F_2,...,F_n$ are closed subsets of $X_1, X_2,..., X_n$ respectively? In fact, what they are?


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up vote 7 down vote accepted

Sets of the form $F_1\times\dots\times F_n$, with $F_j$ closed for $1\leq j\leq n$, are closed since their complement is open: it is indeed $$\bigcup_{j=1}^n\left(\prod_{i=1}^{j-1}X_i\times (X_i\setminus F_i)\times \prod_{i=j+1}^nX_i\right).$$ But not all the closed sets have this form: for example, in the plane $\Bbb R^2$, the unit disk $\{(x_1,x_2)\in\Bbb R^2, x_1^2+x_2^2\leq 1\}$ cannot be written as a cartesian product.

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when you said $F_j$ are you talking about every $F_j$? – user42912 Oct 9 '12 at 13:54
Yes, and I should say "the" $F_j$. – Davide Giraudo Oct 9 '12 at 13:59
Two arbitrary disjoint squares in the plane does not seem to be an intersection of products of closed sets to me. You would at least need to take finite unions of products of closed sets before you take intersections. Is the result true then? If so can you provide a proof? It doesn't seem clear to me. – Seth Feb 25 '14 at 1:11
Ok, I checked it and it's true that closed sets are intersections of finite unions and products of closed sets. – Seth Feb 28 '14 at 23:44
Dear Davide, you write "But closed sets are arbitrary intersection of sets of the form $F_1\times\dots\times F_n$". That is definitely not true, as illustrated by your very example of a closed disc in the plane. Moreover any intersection of sets of the form $F_1\times\dots\times F_n$ with $F_i$ closed has the exact same form: you get nothing new by taking intersections. – Georges Elencwajg Mar 9 '15 at 10:43

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