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This is sort of similar to this question I just posted, so hopefully I did a little better on this one. Does this proof work?

"Let $f:X \to Y$ be a continuous function and assume that $Y$ is $T_2$. Prove that $\{(x_1,x_2) \in X \times X\,|\,f(x_1)=f(x_2)\}$ is a closed subset of $X \times X$."

Correct proofs below.

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Alex Petzke
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    I think you should explicitly mention that $O_i=f^{-1}(U_i)$ ($i=1,2$) and $O_1\times O_2 \subseteq (X\times X)\setminus A$. Otherwise it's alright. – Sayantan Jan 24 '13 at 03:36
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    A cleaner approach is to recall (or prove) that being Hausdorff is equivalent to the diagonal $\Delta = { (y, y) : y \in Y }$ being closed. – Qiaochu Yuan Jan 24 '13 at 04:01
  • As an alternative: if $(x_i, y_i)_{i \in I}$ is a net in $A$, converging to $(a,b)$ then $x_i \rightarrow a$, so $f(x_i) \rightarrow f(a)$ and also $y_i \rightarrow b$, and $f(x_i) = g(y_i) \rightarrow g(b)$, so by unicity of limits in the Hausdorff space $Y$, $f(a) = g(b)$ so $(a,b) \in A$ as well, so $A$ is closed. – Henno Brandsma Jan 24 '13 at 18:54
  • @Qiaochu: Can you check out my edit and let me know if my proof is valid and if it can be simplified any further? Thanks. – Alex Petzke Jan 24 '13 at 21:07
  • @Alex: yes, that's the proof I had in mind. – Qiaochu Yuan Jan 24 '13 at 22:14
  • @Qiaochu: Good, thanks. – Alex Petzke Jan 24 '13 at 23:14

1 Answers1

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Let $A:=\{(x_1,x_2) \in X \times X\,|\,f(x_1)=f(x_2)\}$. We will show that $(X \times X)\setminus A =\{(x_1,x_2) \in X \times X\,|\,f(x_1) \ne f(x_2)\}$ is open. Since $f(x_1) \ne f(x_2)$, there are disjoint open sets $U \ni f(x_1)$ and $V \ni f(x_2)$ by the fact that $Y$ is $T_2$. Since $f$ is continuous, there are open sets $O_1 \ni x_1$ and $O_2 \ni x_2$ in $X$, such that $f[O_1] \subset U$ and $f[O_2] \subset V$. Then $O_1 \times O_2$ is an open set containing $(x_1,x_2)$ in the product topology on $X \times X$. When $(x,y)$ is in $O_1 \times O_2$, then $f(x) \in U, f(y) \in V$, so $f(x) \neq f(y)$, as $U$ and $V$ are disjoint, which shows that $O_1 \times O_2 \subset (X \times X) \setminus A$, so that $(X \times X)\setminus A$ is the union of such open sets, and so is open itself. It follows that $A$ is closed.

Here's a proof using the idea in Qiaochu's comment:

Let $A:=\{(x_1,x_2)\in X \times X\,|\,f(x_1)=f(x_2)\}$ and consider the map $(f,f):X \times X \to Y \times Y$. Since $Y$ is $T_2$, $\Delta_Y$ is closed (I do understand the proof of this). Since if $(x_1,x_2) \in A$ then $f(x_1)=f(x_2)$, it follows that $(f,f)(A) \subseteq \Delta_Y$. Then $(f,f)^{-1}(\Delta_Y)=A$, and since $f$ is continuous then $(f,f)$ is continuous and it follows that $A$ is closed.

Henno Brandsma
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Alex Petzke
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