# How do the closed subsets in the product topology look like

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?

thanks

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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.
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
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 at 10:43