Evaulation of a Topology Proof

Question

I have been using Munkres' Topology in my Topology class and we have been going over homework exercises. One of my classmates submitted a pretty sizable proof to a question, but mine was only a few lines. I am now uncertain whether or not I made some mistake in my proof. The question and my solution are included below.

Chapter 31 Exercise 5: Let $f, g : X \to Y$ be continuous: assume that $Y$ is Hausdorff. Show that $\{x | f(x) = g(x)\}$ is closed in $X$.

Proof: Note that $S = X - \{x | f(x) = g(x)\} = \{x | f(x) \neq g(x)\}$. Clearly $S$ is open in $X$ as the arbitary union of elements in $X$. Hence $\{x | f(x) = g(x)\}$ is closed in $X$.

Did I make a major mistake in this proof?

Answer 1

You just sort of stated that you were done, without providing much justification.

My answer had an error in it. The condition $\{x:f(x)=g(x)\}$ is closed is in fact equivalent to $Y$ being Hausdorff. See @HennoBrandsma's response.

Answer 2

Unions of arbitrary elements are not always open, so that sentence doesn't work. What does it mean to be an open set? I think you'll get more mileage out of considering the boundary of a set.

Answer 3

A slick solution: $\Delta_Y=\{(y,y): y \in Y\}$ is closed in $Y^2$ iff $Y$ is Hausdorff. And so if $Y$ is Hausdorff $\{x: f(x)=g(x)\}=(f \nabla g)^{-1}[\Delta_Y]$ is closed in $X$, where $f\nabla g: X \to Y^2$ is the map that sends $x$ to $(f(x),g(x)) \in Y^2$ and which is continuous when $f$ and $g$ are.

If we weaken $Y$ to $T_1$ we don't have this conclusion: Let $X=\Bbb Z$ in the cofinite topology and let $Y=X$, which is $T_1$ but not Hausdorff. Then for the continuous $f(n)=n$ and $g(n)=|n|$ (the preimage of a finite set is finite, so continuity follows) we have that $\{x: f(x)=g(x)\}= \Bbb Z^+_0$ is not closed in $X$, as it's infinite.