It would be very appreciated if someone could review my proof written below. Thanks!


Let $f, g : X \to Y$ be continuous; assume $Y$ is Hausdorff. Show that $\{x \mid f(x) = g(x)\}$ is closed in $X$.


Let $f, g: X \rightarrow Y$ be continuous where $Y$ is Hausdorff.

Let $C = \{x \mid f(x) = g(x)\}$

Suppose $C$ is not closed. Then let $x_1 \notin C$ where $x_1$ is a limit point of $C$. Then $f(x_1) \neq g(x_1)$ in $Y$. Then since $Y$ is Hausdorff we can find disjoint open sets $U$ and $V$ containing $f(x_1)$ and $g(x_1)$ respectively. Then $f^{-1}[U]$ and $g^{-1}[V]$ are open sets in $X$ since $f$ and $g$ are continuous and both sets contain the point $x_1$.

Then consider the set $A = f^{-1}[U] \cap g^{-1}[V]$. This set is open and must also contain the point $x_1$. Since $x_1$ is a limit point of $C$ the set $A$ must contain some point $z \in C$. Then $f(z) = g(z)$ and since $z \in A$ we have $f(z) \in U$ and $g(z) \in V$. But since $f(z) = g(z)$ we have $U \cap V \neq \emptyset$ as both sets contain $f(z)$. Hence we have obtained a contradiction since $U$ and $V$ were chosen to be disjoint and thus $x_1$ must be a point in $C$ (or equivalently $x_1$ is not a limit point and is not a member of C )

So $C$ must contain all its limit points and thus $C$ is closed.

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    $\begingroup$ $f(x)=g(x)\Longleftrightarrow f(x)-g(x)=0$. $\endgroup$ Apr 7, 2019 at 2:47
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    $\begingroup$ no that wouldnt be correct here as Y is a general topological space and the operation of subtraction is not even defined. @DonThousand $\endgroup$
    – H_1317
    Apr 7, 2019 at 2:51
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    $\begingroup$ This proof looks good to me. $\endgroup$
    – Moya
    Apr 7, 2019 at 3:02
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    $\begingroup$ Start by taking $x_1$ to be an adherence point (or limit point as you call it) of $C$, as you do. The contradiction follows from the assumption $x_1 \notin C$. So after the contradiction you know $x_1 \in C$, not that $x_1$ is not a limit point. Then use the last sentence right away and you're done. $\endgroup$ Apr 7, 2019 at 6:03
  • $\begingroup$ Thanks again @HennoBrandsma, you are correct. I added a parenthetical which i believe could also be a valid conclusion? Let me know if you agree. thanks $\endgroup$
    – H_1317
    Apr 7, 2019 at 6:42

1 Answer 1


Your proof is correct. Here is a possibly simpler proof exploiting the fact that the diagonal of Hausdorff spaces are closed:

Let $F(x)=(f(x),g(x))$. Clearly, $F$ is continuous. Then $\{x:f(x)=g(x)\}=F^{-1}(A)$ where $A$ is the diagonal of $Y$. Since $F$ is continuous, $\{x:f(x)=g(x)\}$ is closed in $X$, being the preimage of a closed set.


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