Proving that, if two continuous functions $f$ and $g$ from $X$ to $Y$ are equivalent on $F \subset X$, and Y is Hausdorff, F must be closed. Let $X$ and $Y$ be topological spaces, and let $f$, $g\colon X \rightarrow Y$ be two continuous functions. Let $F = \{ x \in X \colon f(x) = g(x) \}$. Prove that, if $Y$ is Hausdorff, then $F$ is closed.
My attempt:
Assume $F$ is open. Let $U \subset Y$ be the image of $f$ on $F$, which is equivalent to the image of $g$ on $F$. Let $a$ and $b$ be two points in $U$. Since $Y$ is Hausdorff, there exist neighborhoods $N_a$, $N_b \subset U$ such that $N_a \cap N_b = \emptyset $.  
I have a hunch there is a contradiction to be drawn from the pre images of these two neighborhoods, but haven't been able to find it.
 A: Show that $G=\{x\in X:f(x)\ne g(x)\}$ is open in $X$ by showing that if $x\in G$ then there exists an open $A$ of $X$ with $x\in A\subset G,$ as follows: 
For $x\in G,$ let $U,V$ be open disjoint subsets of $Y$ with $f(x)\in U$ and $g(x)\in V.$ Let $A=(f^{-1}U)\cap (g^{-1}V).$  
If $y\in A$ we have $f(y)\in U$ and $g(y)\in V,$ but $ U\cap V=\phi$ so $f(y)\ne g(y).$
A: Consider the map $f \times g: X \times X \to Y \times Y$, defined in the most natural way. Note that $Y  \times Y$ is also Hausdorff from which we can conclude that the diagonal, $D$, is closed, and hence $(f \times g)^{-1}(D)$ is closed as well.
A: HINT: Let $h(x)=f(x)-g(x)$. Then $h^{-1}(\{0\})$ is closed (why?).
A: If $X\setminus F$ is empty (that is, $F=X$), then there is nothing to prove.
Otherwise, for each $w\in X\setminus F$ (that is, $f(w)\neq g(w)$), choose two disjoint open sets $U_w\subseteq Y$ and $V_w\subseteq Y$ such that $f(w)\in U_w$ and $g(w)\in V_w$. This is possible because $Y$ is a Hausdorff space. Since $f$ and $g$ are continuous, the set $f^{-1}(U_w)\cap g^{-1}(V_w)$ is an open subset of $X$.
Now I claim that $$X\setminus F=\bigcup_{w\in X\setminus F}f^{-1}(U_w)\cap g^{-1}(V_w),$$ which will show that $X\setminus F$ is open (and, of course, that $F$ is closed).
If $x\in X\setminus F$, then clearly $x\in f^{-1}(U_x)\cap g^{-1}(V_x)$ by construction. This proves the $\subseteq$ direction. On the other hand, if $x\in f^{-1}(U_w)\cap g^{-1}(V_w) $ for some $w\in X\setminus F$, then $f(x)\in U_w$ and $g(x)\in V_w$. But $U_w$ and $V_w$ are disjoint by construction, so that $f(x)\neq g(x)$. This reveals that $x\in X\setminus F$, proving the $\supseteq$ direction.
A: Here is a second, alternative answer.
Let $x\in X$ be in the closure of $F$. Then, there exists a net $\langle x_{\alpha}\rangle$ in $F$ converging to $x$. In particular, one has that $f(x_{\alpha})=g(x_{\alpha})$ for each index $\alpha$.
Since both $f$ and $g$ are continuous, the nets $\langle f(x_{\alpha})\rangle$ and $\langle g(x_{\alpha})\rangle$ in $Y$ converge to $f(x)$ and $g(x)$, respectively.
But the two nets $\langle f(x_{\alpha})\rangle$ and $\langle g(x_{\alpha})\rangle$ are, in fact, one and the same. Since $Y$ is a Hausdorff space, every net in it can converge to at most one point. This implies that $f(x)=g(x)$. That is, $x\in F$.
This proves that the closure of $F$ is a subset of $F$, so that $F$ is closed.
