removed.

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- May 1st 2010, 01:30 AMjohndonnull
removed.

- May 1st 2010, 11:13 AMnimon
I'm not sure what you mean by a 'net', but I can show you $\displaystyle A \Leftrightarrow C$ to get you started.

First, if $\displaystyle X$ is Hausdorff then, given any $\displaystyle (x,y)\in (X\times X) \backslash \Delta$ we can find disjoint, open neighbourhoods $\displaystyle U_{x},U_{y}\subseteq X$ of $\displaystyle x$ and $\displaystyle y$ respectively. Then $\displaystyle (x,y)\in U_{x}\times U_{y} \subseteq (X\times X) \backslash \Delta $ (the last inclusion is due to the disjointness of the neighbourhoods: if $\displaystyle (z,z) \in U_{x}\times U_{y}$ then $\displaystyle z\in U_{x} \cap U_{y}$ which is wrong by assumption). It follows that $\displaystyle (X\times X)\backslash \Delta$ is a union of open sets in $\displaystyle X \times X$, so must be open, and therefore $\displaystyle \Delta$ is closed.

The converse is similar. If $\displaystyle \Delta$ is closed then $\displaystyle (X\times X)\backslash \Delta$ is open, and so for any $\displaystyle (x,y)\in X\times X$ we can find an open set $\displaystyle V\subseteq (X\times X) \backslash \Delta$ such that $\displaystyle (x,y)\in V$, and it follows that $\displaystyle (x,y)\in U_{x}\times U_{y} \subseteq V$ for some $\displaystyle U_{x},U_{y}$ open in $\displaystyle X$. Since $\displaystyle U_{x}\times U_{y} \subseteq (X\times X) \backslash \Delta$, they must be disjoint, and so $\displaystyle X$ is Hausdorff.

Hope this helps. - May 2nd 2010, 08:23 PMDrexel28
I gave you a lot of help in your last thread, what do you think about this one?

Quote:

(C) The diagonal delta= {(x,x):x∈X }⊂ X x X is closed in X × X

**nimon**already did this.

Quote:

(ii) Let X,Y be topological spaces, A ⊂ X, and Ᾱ = X. Let f: X →Y ,

g: X →Y be continuous functions, and let f(a)= g(a) ∀ a∈A . Show that if

Y is a Hausdorff space, then f(x)= g(x) ∀ x∈X . [HINT: Use (B) or (C).]

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