Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

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 $\prod_\alpha F_\alpha$.

share|improve this answer
    
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.

share|improve this answer
    
I know this fact, clearly you did not read the question. –  sxd Nov 27 '12 at 22:48
2  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.