# Projections in product spaces are not closed maps, false counter argument

I know that the projections are not closed maps in presence of the product topology. However, later in a chapter, they show that $\prod_\alpha F_\alpha$ is closed in $\prod_\alpha X_\alpha$ if and only if $\forall \alpha: F_\alpha$ is closed in $X_\alpha$.

But does this not contradict the fact that projections are not closed maps, since $P_\beta(\prod_\alpha F_\alpha) = F_\beta$.

Could anyone point out the flaw in my reasoning?

-

A map $f$ is closed if $f(E)$ is closed for every closed set $E$. It may be that $f(E)$ is always closed for closed sets $E$ of the form $E = \prod_\alpha F_\alpha$, but this does not imply that $f$ is closed, since there are closed sets $E$ that are not of the form $\prod_\alpha F_\alpha$.

-
But some closed set is just a subset of $\prod_\alpha X_\alpha$, is it not always of that form then? –  sxd Nov 27 '12 at 22:35
@DimitriSurinx: Not all closed sets are of the form $\prod_\alpha F_\alpha$. For instance, suppose you are working in the space $\mathbb{R}\times\mathbb{R}$. Then the line $\{(x,x) : x\in \mathbb{R}\}$ is closed in $\mathbb{R}\times\mathbb{R}$, but it is not of the form $E\times F$ where $E$ and $F$ are closed subsets of $\mathbb{R}$. –  froggie Nov 27 '12 at 22:37
Oh, right, how stupid, while staring at the generalised carthesian product definition I complety forgot that it actually in a sense combines every element from all the sets with one another -.- –  sxd Nov 27 '12 at 22:41

The set of positive $x$ and $y$ with $xy \geq 1$ is closed in the plane but its projection into either of the axes is clearly not closed.

-
I know this fact, clearly you did not read the question. –  sxd Nov 27 '12 at 22:48
Actually I did and I displayed a closed set in a product which isn't a product of closed sets and which does not project onto a closed set. I would have thought that this answers your question. It certainly was my intention to so do. –  jbc Nov 27 '12 at 22:56
I asked where the flaw was in my reasoning, since I know that proof you just presented. Thanks for the efford by the way, I did not want to sound rude or anything. –  sxd Nov 27 '12 at 23:43