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

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

• Seems to be a correct proof – RMWGNE96 May 11 '19 at 12:30
• 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. – DanielWainfleet May 12 '19 at 5:42