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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.

H_1317
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    $f(x)=g(x)\Longleftrightarrow f(x)-g(x)=0$. – Rushabh Mehta Apr 07 '19 at 02:47
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    no that wouldnt be correct here as Y is a general topological space and the operation of subtraction is not even defined. @DonThousand – H_1317 Apr 07 '19 at 02:51
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    This proof looks good to me. – Moya Apr 07 '19 at 03:02
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    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. – Henno Brandsma Apr 07 '19 at 06:03
  • 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 – H_1317 Apr 07 '19 at 06:42
  • I think the parenthetical just confuses things. The structure of the proof is: We want to show $\overline{C} \subseteq C$ (which is equivalent to $C$ being closed) so let $x \in \overline{C}$. Assume $x \notin C$... contradiction, so $x \in C$ and the inclusion has been shown. QED. $x \in \overline{C}$ is the "outer assumption" as it were, to start the inclusion proof, so that stays valid after the contradiction, which only invalidates the last assumption you made, i.e. $x \notin C$. – Henno Brandsma Apr 07 '19 at 10:16
  • The parenthetical is probably more confusing but both should follow from the contradiction — at this point im more just curious.... Since either C is closed or it is not closed, we then obtain a contradiction in the case C is not closed. Hence C must be closed. So the statement in the proof where we USE the fact that C is not closed must be invalid. This statement is our initial assumption and is “ x1 is not in C where x1 is a limit point of C. “ there are 3 ways to invalidate the statement: 1.) x1 is really in C or 2.) x1 is not a limit point or 3.) both 1 and 2. @HennoBrandsma thanks – H_1317 Apr 07 '19 at 21:03

2 Answers2

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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.

YuiTo Cheng
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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.

TheProfetOfBourbaki
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