Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2009 / A / 3

A filter base on a set $X$ is a nonempty family $\mathcal{F}$ of nonempty subsets such that for every $F,G\in \mathcal{F}$ there is $H\in \mathcal{F}$ with $H\subseteq F\cap G$. Repeated application gives a member contained in any specified finite intersection. In particular every finite intersection is nonempty, so the family has the finite intersection property. Its upward closure $\{A\subseteq X:\exists F\in \mathcal{F},F\subseteq A\}$ is the corresponding proper filter.

For a connected intersection of a closed filter base, the filter-base property implies the finite intersection property. Compactness, in its closed-set formulation, therefore gives $K=\bigcap \mathcal{F}\ne \varnothing$. It is closed in $X$, hence compact.

Suppose additionally that each base member is connected but $K$ is disconnected. Write $K=A\cup B$ with $A,B$ disjoint nonempty closed subsets of $K$. They are compact, hence closed in the Hausdorff space $X$. Normality of a compact Hausdorff space gives disjoint open sets $U,V$ containing $A,B$, respectively. The compact set $W=X\setminus (U\cup V)$ misses $K$. For every $w\in W$ some $F_w\in \mathcal{F}$ misses $w$, so the open sets $X\setminus F_w$ cover $W$. A finite subcover gives $F_1\cap \cdots \cap F_m\subseteq U\cup V$. Choose $F_0\in \mathcal{F}$ inside this finite intersection. If $W$ is empty, choose any base member instead.

Since $F_0$ contains $K$, it meets both $U$ and $V$; but it lies in their disjoint union. These two relatively open nonempty pieces disconnect $F_0$, a contradiction. Thus the intersection is nonempty and connected. Compactness is used to pass from an infinite exclusion condition to a single base member.

A Hausdorff compactification of $X$ is an embedding $h:X\to Y$ into a compact Hausdorff space with dense image. The complement $Y\setminus h[X]$ is the remainder; a one-point or two-point compactification has a remainder of that cardinality.

For one point, take the unit circle andThe angle runs over $(-\pi ,\pi )$, so this is a homeomorphism onto the circle with $(-1,0)$ removed. That image is dense, and the circle is compact Hausdorff. For two points, take $Y=[-\pi /2,\pi /2]$ and $h_2(x)=\arctan x$. This is a homeomorphism onto its dense open interval, leaving exactly the two endpoints. Therefore both required compactifications exist; in the first the positive and negative ends meet, whereas in the second they remain distinct.

Write $T_n=h(\{x\in \mathbb{R}:|x|>n\})$, making explicit that these are subsets of $Y$. If $p\notin h[\mathbb{R}]$ and $U$ is an open neighbourhood of $p$, then $h([-n,n])$ is compact and hence closed in $Y$. The open set $U\setminus h([-n,n])$ still contains $p$ and meets the dense image of $h$. Such an intersection point belongs to $T_n$, so $p\in \overline{T_n}$ for every $n$.

Conversely, if $p=h(x)$ choose an integer $n>|x|$. Since $h$ is an embedding, there is an open neighbourhood $U$ of $p$ in $Y$ whose intersection with $h[\mathbb{R}]$ is $h((-n,n))$. It misses $T_n$, so $p\notin \overline{T_n}$. HenceFor the two ends of a real-line compactification, let $A_n^{+}=\overline{h((n,\infty ))}$ and $A_n^{-}=\overline{h((-\infty ,-n))}$. Each is nonempty compact and connected, being the closure of a connected image. The two nested families therefore have nonempty connected intersections $C^{+}$ and $C^{-}$. Since closure commutes with finite unions, $\overline{T_n}=A_n^{+}\cup A_n^{-}$. MoreoverFor if a point fails both intersections, it fails one positive member and one negative member, hence both members at the larger index by nesting.

If the remainder were a finite three-point set, its Hausdorff subspace topology would be discrete. Each nonempty connected subset $C^{\pm }$ would then be a singleton, so their union would have at most two points. This contradiction proves there is no three-point Hausdorff compactification of the real line. The argument does not assert that every possible remainder is finite.