Welcome to our community

Be a part of something great, join today!

Open set proof

  • Thread starter
  • Banned
  • #1

Poirot

Banned
Feb 15, 2012
250
Let X,Y be top. spaces and let p(x,y)=x be the projection mapping from X x Y with the product topology. Let W be a open set of X x Y. Prove that p(W) is an open set of X.

My attempt: for all (x,y) in W there exist open sets U in X and V in Y with x in U, y in V and U x V is a subset of W. Thus....?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Let X,Y be top. spaces and let p(x,y)=x be the projection mapping from X x Y with the product topology. Let W be a open set of X x Y. Prove that p(W) is an open set of X.

My attempt: for all (x,y) in W there exist open sets U in X and V in Y with x in U, y in V and U x V is a subset of W. Thus $\color{blue}{(x,y)\subset U\times V \subset W}$ and so $\color{blue}{x = p(x,y)\subset p(U\times V) \subset p(W)}$ ...
. .
 
  • Thread starter
  • Banned
  • #3

Poirot

Banned
Feb 15, 2012
250
Is some alternative definition of open sets being used here? I would have thought that I have to show it is a union or intersection of open sets.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Is some alternative definition of open sets being used here? I would have thought that I have to show it is a union or intersection of open sets.
I left out the fact that $p(U\times V) = U$. The structure of the proof is to show that given $x\in p(W)$ there exists an open set $U$ with $x\in U \subset p(W).$ Thus every point of $p(W)$ has an open neighbourhood in $p(W)$, from which it follows that $p(W)$ is open.