For continuous $f, g : Z \to Y$ with $Y$ Hausdorff the agreement set $\{ z \in Z : f(z) = g(z) \}$ is closed in $Z$

Let $Z$ be a topological space, let $Y$ be a Hausdorff space (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) and let $f, g : Z \to Y$ be continuous (Continuity of a map of topological spaces at a point and globally). Then the agreement set

$E(f,g) \;:=\; \{\, z \in Z : f(z) = g(z) \,\}$

is closed in $Z$.

No hypothesis is placed on $Z$: the separation hypothesis is on the codomain alone, and it is not decoration. Let $Y_0 = \{a,b\}$ with $a \ne b$ carry the indiscrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), which is not Hausdorff. Every function $Z \to Y_0$ is continuous, the only preimages to check being those of $\varnothing$ and $Y_0$, namely $\varnothing$ and $Z$. So for any subset $S \subseteq Z$ the constant map $f_0 \equiv a$ and the map $g_0$ taking the value $a$ on $S$ and $b$ off $S$ are continuous with $E(f_0,g_0) = S$, closed or not.

• Why the diagonal criterion is the right tool here. The condition "$f(z) = g(z)$" is a condition on the pair of values, so it becomes a membership condition once the two maps are packaged into one map into the square; the criterion then converts the separation hypothesis on $Y$ into the closedness of the set that condition names. Nothing is proved twice: the whole content is A space is Hausdorff if and only if its diagonal is closed in the square carrying the product topology together with the preimage identity of The diagonal $\Delta_X \subseteq X \times X$, the diagonal map $\delta_X$, and the pairing $\langle f, g \rangle$ of two maps.

• Both hypotheses are used, and only these. Continuity of $f$ and $g$ enters only through [L1], and the Hausdorff condition only through [L2]. In particular no countability, compactness or separation hypothesis on $Z$ appears anywhere in the argument.

• The complement is what the statement is often used for. $Z \setminus E(f,g)$ is open, so if $f$ and $g$ differ at a point they differ throughout some open neighbourhood of it. Equivalently, $E(f,g)$ contains the closure of every subset of $Z$ on which $f$ and $g$ agree, which is the form in which a statement about a dense set is obtained from this one.