Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 4 / 2G / Solution

The closure $\overline{Z}$ is the intersection of all closed subsets containing $Z$; equivalently, every neighbourhood of a point of $\overline{Z}$ meets $Z$. The subspace is dense when $\overline{Z}=X$. A space is Hausdorff when distinct points have disjoint neighbourhoods.

For Hausdorff $Y$, the diagonal ${\Delta }_Y$ is closed in $Y\times Y$. Thusis closed. Since it contains the dense set $Z$, it equals $X$.

The conclusion fails without Hausdorffness. Let $X=Y=\{0,1\}$ have the Sierpiński space topology $\{\varnothing ,\{1\},X\}$ and let $Z=\{1\}$, which is dense. The identity map and the constant map with value $1$ are continuous and agree on $Z$, but differ at $0$.