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Given a topological space $X$, I want to find a criteria such that for all continuous maps $f,g:X\rightarrow X$, the set $A=\{x|f(x)=g(x)\}$ is closed - Property (1).

Consider the map $H:X\rightarrow X\times X$ such that $H(x)=(f(x),g(x))$. $H$ is continuous and $A=H^{-1}(\{(x,x)\})$, so a good criteria would be the set $\{(x,x)|x\in X\}$ is closed, or the set $K=\{(x,y)|x\neq y\}$ is open.

If $X$ is Hausdorff, for all $(x,y)\in K$ there exists $U_x, U_y$ such that $x\in U_x, y\in U_y$ and $U_x\times U_y\subset K$. Taking the union over $K$ we get $K$ is open.

So if the space is Hausdorff, $(1)$ is satisfied. How can we weaken the sufficient condition?

anonymous67
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  • Define $h(x) = f(x)-g(x)$, since both $f$ and $g$ are continuous, so is $h$, then $A=h^{-1}({0})$. So it is only necessary that ${0}$ is closed in $X$. – Nigel Overmars Apr 17 '16 at 10:52
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    That's my preferred condition too but since Mr.T is talking of general spaces we're probably not guaranteed an algebraic '-' structure as with the vector spaces. – Squid Apr 17 '16 at 10:56
  • For spaces with such a (continuous) group structure, having the unit element closed is equivalent to being Hausdorff (as this argument shows). – Henno Brandsma Apr 17 '16 at 10:59

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The set $\{(x,x): x \in X \}$ is closed in $X \times X$ iff $X$ is Hausdorff. This is classical and well-known. So Hausdorff spaces are the natural class of spaces here, I think.

Henno Brandsma
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