proooooof
the$$\beta ^{th}$$ projection map $$\pi _{\beta} : \Pi _{\alpha \in \Delta } \rightarrow X _{\beta}$$ is continuous and open but not be closed

Question

proooooof 
the$$\beta ^{th}$$ projection map $$\pi _{\beta} : \Pi _{\alpha \in \Delta } \rightarrow X _{\beta}$$ is continuous and open but not be closed

perl · Answer

hmm, is there a definition for projection map? i only know projection of a function , like projection f(x,y)= (0,y)

anonymous · Answer

nooooo

anonymous · Answer

projection map not like projection of a function

perl · Answer

sorry i dont know what that is. what class is this? or book

anonymous · Answer

this is  the Luther of the book  paul e. long

anonymous · Answer

oooooh here its  like 
$$\Pi _{\alpha \in \Delta } X _{\alpha} \rightarrow X _{\beta}$$

anonymous · Answer

Ah, got it

anonymous · Answer

ok let me see

anonymous · Answer

So, if U is open in X_beta, then
$$\pi_\beta^{-1}(U) = U \times \Pi_{\alpha \in \Delta\backslash\beta}$$, which is in the basis according to the definition of a product topology.

anonymous · Answer

So, that satisfies the definition for continuity. (That Latex was so hard!)

anonymous · Answer

how to proof if it open

anonymous · Answer

what about need not to be closed

anonymous · Answer

Oh, whoops. Let me work on those

anonymous · Answer

where the continuous proof and you know that projection map is onto but not 1-1

anonymous · Answer

Ok, so, just take an arbitrary element in the basis of the product topology (so, a union of inverse projections of open sets from the X's). Then apply some projection map to it. If the projection map goes to one of the X's that contributed an inverse projection to that basis element, then the map goes back to the original open set, and it's open. Otherwise, the map goes back to an entire X, which is also open. Either way, the projection maps basis elements to open sets, so the entire map is open.

anonymous · Answer

It need not be closed because you could go through the same argument for a basis of closed sets, but then when you map back to an entire X, that's an open set, so the argument doesn't hold.

anonymous · Answer

but what will happens  if  closure $$  of \left( X _{\beta} \right)$$

anonymous · Answer

What?

anonymous · Answer

πβ:Πα∈Δ→ closure (Xβ)

anonymous · Answer

I'm not sure what you mean...you can only take the closure of subset, not the entire topology.

anonymous · Answer

but it is in product spaces

anonymous · Answer

Oh, I guess I see what you mean. But the projection still maps to the set itself, not its closure.

anonymous · Answer

but what if it be in closure dose the the projection map will find some elements to map

anonymous · Answer

Sorry, not sure what you're saying?