How to prove that "If $X$ is a Hausdorff space equivalent to $\varDelta \left(x\right),$ it is a closed set in $X\times X$"？
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Can you provide the definition of $\Delta(x)$? – kccu Jan 29 '20 at 14:36
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@kccu \varDelta $\left(x\right)$=(x,x), x\in X.(Is actually a diagonal mapping.) – user142088 Jan 29 '20 at 14:40
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Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. – saulspatz Jan 29 '20 at 14:44
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Suppose that $\Delta$ is closed and $X$ is not Hausdorff, there exists $x,y\in X$ such that for every open subset, $U,V$ $x\in U, y\in V$, $U\cap V$ is not empty. $(x,y)$ is not in $\Delta$. A basis of the topology of $X\times X$ is $U\times V$, $U, V$ open. For every, $x\in U, y\in V$ open, $U\cap V$ is not empty, let $z$ an element of $U\cap V$, $(z,z)\in \Delta$ implies that $U\times V\cap \Delta$ is not empty and $X\times X-\Delta$ is not open.
Suppose that $X$ is Hausdorff, let $(x,y), x\neq y$, there exists $x\in U, y\in V$ open such that $U\cap V$ is empty. $U\times V$ contains $(x,y)$ and $U\times V\cap \Delta$ is empty implies that $X\times X-\Delta$ is open.
Tsemo Aristide
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