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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....?

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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....?

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- Feb 7, 2012

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. .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 $\color{blue}{x = p(x,y)\subset p(U\times V) \subset p(W)}$ ...

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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.