# $X$ is Hausdorff iff $X×X$ is Hausdorff, where $X$ is a topological space?

One side is true i.e. if $$X$$ is Hausdorff then $$X\times X$$ is Hausdorff. But is it true that if $$X\times X$$ is Hausdorff then $$X$$ is also Hausdorff?

My proof: since $$X\times X$$ is Hausdorff, take two points $$(x_1,y)$$ and $$(x_2,y)$$ hence there are to basis elements say $$U_1\times V_1$$ and $$U_2\times V_2$$ containing them respectively and are disjoint. Hence $$U_1$$ and $$U_2$$ are required open sets in $$X$$ which are disjoint and contain $$x_1,x_2$$ respectively.

Since $$x_1$$, $$x_2$$ are arbitrary hence $$X$$ is Hausdorff.

Any subspace of a Hausdorff space is Hausdorff, and $$X$$ can obviously be embedded in $$X\times X$$, so if $$X\times X$$ is Hausdorff, so is $$X$$. Even more general, if there exists a non-empty topological space $$Y$$ such that $$X\times Y$$ is Hausdorff, then $$X$$ is Hausdorff. So we even find for two non-empty topological spaces $$X$$ and $$Y$$ that $$X\times Y$$ is Hausdorff if, and only if, both $$X$$ and $$Y$$ are Hausdorff.
• "Even more general, if there exists a non-empty topological space $Y$ such that $X×Y$ is Hausdorff, then $X$ is Hausdorff" . I think $Y$ also need to be Hausdorff for statement to be true . – user707473 Sep 22 at 12:15
• Well $Y$ needs to be Hausdorff for $X\times Y$ to be Hausdorff, so that would be an unnecessary constraint. – SmileyCraft Sep 22 at 12:15
Assuming that $$X\times X$$ is Hausdorff, take $$x,x^\ast\in X$$. You want to prove that there are open sets $$A,A^\ast\subset X$$ such that $$x\in A$$, $$x^\ast\in A^\ast$$, and $$A\cap A^\ast=\emptyset$$. Consider $$(x,x),(x^\ast,x)\in X\times X$$. There are open sets $$B,B^\ast\subset X\times X$$ such that $$(x,x)\in B$$, $$(x,x^\ast)\in B^\ast$$, and $$B\cap B^\ast=\emptyset$$. You can write $$B$$ as an union $$\bigcup_{\lambda\in\Lambda}B_\lambda\times B_\lambda'$$, where each $$B_\lambda$$ and each $$B_\lambda'$$ is an open subset of $$X$$. So, fix $$\lambda\in\Lambda$$ such that $$(x,x)\in B_\lambda\times B_\lambda^\ast\subset B$$. By the same argument, $$(x,x^\ast)$$ belongs to some product $$C\times C^\ast$$, where $$C$$ and $$C^\ast$$ are open subsets of $$X$$ and $$C\times C^\ast\subset B^\ast$$. Since $$B\cap B^\ast=\emptyset$$, $$(B_\lambda\times B_\lambda^\ast)\cap(C\times C^\ast)=\emptyset$$. But then$$\bigl(B_\lambda\times(B_\lambda^\ast\cap C^\ast)\bigr)\cap\bigl(C\times(B_\lambda^\ast\cap C^\ast)\bigr)=\emptyset$$and, since $$B_\lambda^\ast\cap C^\ast\neq\emptyset$$, you can deduce from this that $$B_\lambda\cap C=\emptyset$$.
If $$X \times X$$ is Hausdorff, and $$x \neq x'$$ in $$X$$ we can separate $$(x,x)$$ and $$(x',x)$$ by disjoint open sets in $$X \times X$$, so there are basic open sets $$U \times V$$ containing $$(x,x)$$ and $$U' \times V'$$ containing $$(x',x)$$ such that $$(U \times V) \cap (U' \cap V')= \emptyset$$. But then $$U$$ is an open neighbourhood of $$x$$ and $$U'$$ is an open neighbourhood of $$x'$$ such that $$U \cap U' = \emptyset$$ (if $$p$$ would be in the intersection, $$(p,x)$$ would be in $$(U \times V) \cap (U' \cap V')$$, contradiction.). So $$X$$ is Hausdorff.
Alternatively, apply a theorem that $$X$$ is an open image of $$X \times X$$ and as such also Hausdorff. Or use that fact that $$X$$ embeds into $$X \times X$$ (the diagonal e.g.) and subspaces are Hausdorff.