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I'd just like to see if the following proof is reasonable - and if not, what went wrong. Thx for taking a look!

$F \subset X= \prod_{\alpha}x_\alpha \ \alpha \in I$ is closed if and only if $F$ is the intersection of finite unions of $F_1 \times F_2 \times F_3 \times ... \times F_n$ where $F_i \subset x_i$ is closed $\forall i$.

Proof

(-1) Suppose each $F_i \subset x_i$ is closed $\forall i$.

(0) This means $C(F_1),\ C(F_2)$ are open.

(1) In turn saying $C(F_1) \times X_1 \ \ \& \ \ C(F_2) \times X_2$ are open in $X_1 \times X_2$.

(2) Thus, $C(x_1 \times F_2) = C(F_1) \times X_1 \ \cup \ C(F_2) \times X_2$ is open.

(3) It follows that $F_1 \times F_2$ is closed...

(4) And by induction $F_1 \times F_2 \times ... \times F_n$ is closed! Any intersection of unions of these, becoming F, will be closed.

For the other direction we've gotta first suppose that $F \subset X$ is closed.

In light of the comments below, I've erased what was in this spot. I'm working on finishing this direction, or rewriting the above to show both directions.

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  • $\begingroup$ It is not correct to say that a closed set $F$ may be expressed as a finite product of closed sets. E.g. take a closed ball in $\mathbb R$. It cannot be expressed like that. $\endgroup$ – drhab Mar 29 '15 at 18:36
  • $\begingroup$ But if F is a subset of a product, can't I say F is itself a product of sets? $\endgroup$ – Joshua Bunce Mar 29 '15 at 18:39
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    $\begingroup$ No. The set $\{\langle x,y\rangle\mid x^2+y^2\leq1\}$ is a subset of product $\mathbb R\times\mathbb R$, but it is not a product of sets itself. $\endgroup$ – drhab Mar 29 '15 at 18:42
  • $\begingroup$ Of course! Thank you. $\endgroup$ – Joshua Bunce Mar 29 '15 at 18:47
  • $\begingroup$ How about the finite union of products of closed sets? $\endgroup$ – Joshua Bunce Mar 29 '15 at 19:18

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