Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2008 / A / 3

A Hausdorff compactification of $X$ is a pair $(h,Y)$ in which $Y$ is compact Hausdorff and $h:X\to Y$ is a topological embedding with dense image.

For a locally compact Hausdorff noncompact topological space, form $X^{\ast }=X\cup \{\infty \}$ with one new point. Keep all open subsets of $X$, and take the neighborhoods of infinity to contain setsCompact subsets of a Hausdorff space are closed. Finite unions of compact sets are compact, so finite intersections of these basic neighborhoods are again of the same form; their intersections with ordinary open sets are open in $X$. This specifies a topology restricting to the original one on $X$.

It is compact: from an open cover choose a member containing infinity. Its complement is contained in some compact $K$, and finitely many more members cover that compact remainder. It is Hausdorff: two points of $X$ are separated as before; to separate $x\in X$ from infinity, choose an open neighborhood $U$ of $x$ lying inside a compact neighborhood $K$. The open set $U$ and the infinity neighborhood $X^{\ast }\setminus K$ are disjoint.

The inclusion of $X$ is an embedding. Every basic neighborhood of infinity meets $X$, because otherwise its compact complement $K$ would equal all of $X$, contrary to noncompactness. Thus $X$ is dense in $X^{\ast }$. This proves that $X^{\ast }$ is the one-point compactification.

First prove the suggested density fact. If $Z$ is dense in $Y$, $V$ is open and $y\in V$, then every open neighborhood $W$ of $y$ has $W\cap V$ nonempty open. It meets $Z$, so $W$ meets $V\cap Z$. HenceIdentify $X$ with its dense embedded image in the hausdorff compactification. For $x\in X$, local compactness supplies an open neighborhood $U$ in $X$ and a compact set $K\subseteq X$ with $x\in U\subseteq K$. Choose an open set $V$ in $Y$ with $V\cap X=U$. The inclusion into $Y$ is continuous, so $K$ is compact in $Y$ and therefore closed, since $Y$ is Hausdorff. The density fact now yieldsThus every point of the embedded $X$ has a $Y$-open neighborhood contained in $X$. Consequently