I am really stuck on this exercise in my course notes.
Let $X$ and $Y$ be compact Hausdorff spaces and $f, g : X \to Y$ be continuous functions. Show that:
There is an $x \in X$ with $f(x) = g(x)$ if and only if for each open cover $C$ of $Y$, there is $x \in X$ and a $U \in C$ with $f(x), g(x) \in U$.
Proving the forwards direction is very easy but the reverse is not so easy.
I have tried to prove the contrapositive. If $f(x) \neq g(x)$ for all $x \in X$, then there exists an open cover $C$ of $Y$ such that for all $x \in X$ and $U \in C$, $f(x) \notin U$ or $g(x) \notin U$.
If $f(x) \neq g(x)$ then we could use that $Y$ is Hausdorff so for each $x\in X$ there exists open $U_{x}, V_{x}\subset X$ with $U_{x}\cap V_{x}=\emptyset$ and $f(x)\in U_{x}$, $g(x)\in V_x$. This would give an open cover of $f(X)$ and another open cover of $g(X)$.
After this I have no ideas. Any ideas of where I could go from here? (or where would be a better place to start?)
It seems the following.
Define a map $h:X\to Y\times Y$ by putting $h(x)=(f(x),g(x))$ for each point $x\in X$. Since both maps $f$ and $g$ are continuous then the map $h$ is continuous too. Since $X$ is a compact space, $h(X)$ is a closed subspace of a Hausdorff space $Y\times Y$. Moreover, since the set $h(X)$ does not intersect the diagonal $\Delta=\{(y,y)\in Y\times Y: y\in Y \}$, then each point $y$ of the space $Y$ has an open neighborhood $U_y$ such that $U_y\times U_y\cap h(X)=\varnothing$. Put $\mathcal C=\{U_y:y\in Y\}$. Then $\mathcal C$ is an open cover of the space $Y$ and there is no element $U\in\mathcal C$ and a point $x\in X$ such that both points $f(x)$ and $g(x)$ belong to $U$.