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Let $f,g\colon X \to Y$ be continuous maps. Let $Y$ be Hausdorff. Is the set $$A := \{x\in X \, : \, f(x)=g(x) \}$$ necessarily closed ?

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3 Answers 3

up vote 8 down vote accepted

Yes. Suppose that $f(x)\ne g(x)$. Since $Y$ is Hausdorff, there are disjoint open sets $U$ and $V$ such that $f(x)\in U$ and $g(x)\in V$. Let $W=f^{-1}[U]\cap g^{-1}[V]$; then $W$ is an open neighborhood of $x$. (Why?) What can you say about $f(p)$ and $g(p)$ if $p\in W$?

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That for $p\in W$ we have $f(p)\ne g(p)$, and so $x\in W \subseteq A^c$ (where $A^c$ is the complement of $A$). And finally, that $A^c$ is open, hence $A$ is closed. @Brian: Thanks. –  Teddy Sep 20 '12 at 9:41
    
@Teddy: You’ve got it. –  Brian M. Scott Sep 20 '12 at 9:44

The set $A$ is the inverse image under $h:X\to Y^2: x\mapsto(f(x),g(y))$ of the diagonal $\{(y,y)\mid y\in Y\}$ of $Y^2$, which is closed since $Y$ is Hausdorff.

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For the result about the diagonal, see here: $X$ is Hausdorff if and only if the diagonal of $X\times X$ is closed –  Martin Sleziak Sep 20 '12 at 10:02

If you are familiar with nets you can argue as follows:

Let $(x_d)_{d\in D}$ be a net such that each $x_d$ belongs to $A$ and $x_d\to x$. We want to show that $x\in A$. (Closedness is equivalent to closedness under limits of nets.)

By the continuity we get $f(x_d)\to f(x)$ and $g(x_d)\to g(x)$. We have $f(x_d)=g(x_d)$, so the same net converges to both $f(x)$ and $g(x)$.

Since $Y$ is Hausdorff this implies $f(x)=g(x)$. (In a Hausdorff space, limits of nets are unique.) The equality $f(x)=g(x)$ means that $x\in A$.

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