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

Question

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

Problem:

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

Proof:

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.

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.

Answer 2

I have written a much more elemental proof, without the use of limit points or the diagonal's lemma.

Proof:

Let's see that $A:= X\setminus\{x\in X : f(x) = g(x)\}= \{x\in X : f(x)\neq g(x)\}$ is an open set by prooving that $A$ is a neighbourhood of each of its points.

Let $a\in A$, i.e $f(a)\neq g(a)$. Then, since $Y$ is a Hausdorff space, there are two open subsets of $Y$, $U$ and $V$ such that $f(a)\in U$, $g(a)\in V$ and $U\cap V = \emptyset$.

Now, we consider the set $B:=f^{-1}(U)\cap g^{-1}(V)$ wich is an open subset of $X$ since $f$ and $g$ are continous functions.

Also, $B$ is fully contained in $A$ because if there are some $x\in f^{-1}(U)\cap g^{-1}(V)$ such that $f(x)=g(x)$ then $f(x)\in U$ and $g(x)\in V$ but this is nosense since $U\cap V =\emptyset$, so $B\subseteq A$. So we have proved that $A$ is a neighbourhood of $a$.

As $A$ is an open subset of $X$, their complementary $X\setminus A = \{x\in X : f(x) = g(x)\}$ is a closed subset of $X$.

QED

I hope this will be useful.