3
$\begingroup$

I realize this question has been asked, But i just wrote up this proof and it's a little different so I was hoping somebody would check it.

Let $f,g: X \rightarrow Y$ be continuous and $Y$ be Hausdorff. Prove that $Z=\{x | f(x) = g(x)\}$ is a closed subset of $X$.

Proof: Let $x \in X-Z$

Then $f(x) \neq g(x)$, and since $Y$ is hausdorff this implies that $\exists U_1, U_2$ open in $Y$ s.t. $f(x) \in U_1$, $g(x) \in U_2$ and $U_1 \cap U_2 = \emptyset$.

Since $f$ is continuous and $U_1$ is open in $Y$, we have $f^{-1}(U_1)$ is open in $X$ and $x \in f^{-1}(U_1)$.

Since $g$ is continuous and $U_2$ is open in $Y$, we have $g^{-1}(U_2)$ is open in $X$ and $x \in g^{-1}(U_2)$.

Claim: $g^{-1}(U_2) \cap f^{-1}(U_1)$ is a neighborhood of $x$ contained in $X-Z$.

Suppose $z \in g^{-1}(U_2) \cap f^{-1}(U_1)$, then $f(z) \in U_1$, $g(z) \in U_2$. But $U_1 \cap U_2 = \emptyset$ and so $f(z) \neq g(z)$ and thus $z \in X-Z$.

Thus $X-Z$ is open and thus $Z$ is closed

$\endgroup$
  • 2
    $\begingroup$ Seems to be a correct proof $\endgroup$ – RMWGNE96 May 11 '19 at 12:30
  • $\begingroup$ Perfectly right. A useful corollary is that if $Y$ is Hausdorff and $f:X\to Y,\,g:X\to Y$ are continuous and agree on a dense subset of $X$ then $f=g.$ For example, in the proof of the Jones Lemma. $\endgroup$ – DanielWainfleet May 12 '19 at 5:42
1
$\begingroup$

The proof you have given is correct, but the identical proof has been given on this site too. E.g. this one is very close. Or this one. Or this. It's the most "obvious" proof.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.