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This question already has an answer here:

Let X be a topological space. Prove that X is a T2-space if and only if $A = \{ (x,x) : x \in X \}$ is closed in $X \times X$.

I've been able to prove first part of the result but not the converse.

For the first part,

We have X is T2. hence X x X is also T2. Hence X x X is T1.

Then for any z belonging to A and y not belonging A, there exist an open set U in X x X such that y belongs to U but z doesn't belong to U.

Hence for every z belonging to A and for a fixed y not belonging to A , there exists an open set V(y) in X x X such that "V(y) intersection A" is empty.

hence (X x X) - A = Union of all V(y)'s such that y doesn't belong to A.

since arbitrary union of open sets is open, hence (X x X) - A is also open in X x X , Implying that A is closed in X x X. ( Is this proof valid ? )

Now how to prove the converse part ?

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marked as duplicate by Najib Idrissi, Community Apr 20 '15 at 9:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Martin Sleziak Apr 20 '15 at 10:33
  • $\begingroup$ Ok thanks. I was facing this difficulty of putting text in proper format. $\endgroup$ – Error 404 Apr 23 '15 at 17:12
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your prove is not correct at the part "hence (X x X) - A = Union of all V(y)'s such that y doesn't belong to A.". This question can be solved directly by using the fact that every points have neighbourhood $U_1 \times U_2$ with$U_1,U_2$ is open subset. try to use thit :)

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  • $\begingroup$ Okay. But those neighbourhoods can also intersect with A. Then how do we make sure that it does not happen ? $\endgroup$ – Error 404 Apr 20 '15 at 9:09

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