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

A topological space is separable if it has a countable dense subset. A subset $D$ is dense when its closure is the whole topological space, equivalently when every nonempty open set intersects $D$. Countable here permits a finite set, so finite topological spaces and the empty topological space fit the same definition.

A topological space is Lindelöf if every open cover has a countable subcover. We also prove the requested equivalence for metrizable spaces here, after the two source definition leaves.

Choose a compatible metric $d$. First suppose that $D$ is a countable dense subset. The balls $B(d_0,q)$ with centers $d_0\in D$ and positive rational radii $q$ form a countable topological basis. Indeed, if $x\in U$ and $B(x,ϵ)\subset U$, choose $d_0\in D$ with $d(x,d_0)<ϵ/3$ and a rational $q$ between $d(x,d_0)$ and $ϵ-d(x,d_0)$. The triangle inequality gives $x\in B(d_0,q)\subset U$.

For any open cover, retain the topological basis elements which are contained in some member of the cover, and choose one covering member for each such topological basis element. There are only countably many choices, and the resulting subfamily covers the topological space, since each point has a topological basis neighborhood inside a covering member. Thus a separable metrizable space is Lindelöf.

Conversely, suppose the metric space is Lindelöf. For each positive integer $n$, the balls $B(x,1/n)$ cover it, so select a countable subcover with centers in a countable set $D_n$. The union $D={\bigcup }_{n\ge 1}{D}_n$ is countable and dense: if $B(x,ϵ)$ is a nonempty open ball, choose $n$ with $1/n<ϵ$. One of the selected radius-$1/n$ balls contains $x$, so its center lies in $D\cap B(x,ϵ)$. ThereforeThe metric hypothesis is used in producing uniform small-ball covers and a countable ball topological basis; the equivalence need not hold for general topological spaces.

Choose countable dense sets $D_X\subseteq X$ and $D_Y\subseteq Y$. Their Cartesian product is countable. Every nonempty basic open set of the product topology has the form $U\times V$ with $U,V$ nonempty open in their respective topological spaces. Density gives $d_X\in D_X\cap U$ and $d_Y\in D_Y\cap V$, hence $(d_X,d_Y)\in (D_X\times D_Y)\cap (U\times V)$.

Every nonempty open set in the product contains such a basic open set. Thus $D_X\times D_Y$ is dense in $X\times Y$, proving topological separability of the product. If one factor is empty the product is empty and the conclusion holds directly.

Since $D$ is a discrete subset in the subspace topology, for each $d\in D$ choose an open set $U_d\subseteq Z$ with $U_d\cap D=\{d\}$. Closedness of $D$ makes $Z\setminus D$ open. The familyis therefore an open cover of $Z$. A countable subcover exists by the Lindelöf property. Each point $d$ can be covered only by its own $U_d$ among this family: the complement misses all of $D$, and every other $U_e$ meets $D$ only at $e$. Hence the countable subcover must include $U_d$ for every $d\in D$. This forcesBoth closedness and discreteness matter: they provide the open complement and the single-point traces, respectively.

The lower limit topology on the real line is the lower limit topology. Every nonempty basic interval $[a,b)$ contains a rational number, so the rational numbers are a countable dense subset. By the preceding product argument, ${\mathbb{Q}}^2$ is dense in the square, proving its topological separability.

Consider the uncountable anti-diagonalIt is a discrete subset: the basic rectangle $[t,t+ϵ)\times [-t,-t+ϵ)$ intersects $D$ only at $(t,-t)$, because both coordinates have increased weakly and their sum can remain zero only if neither increases.

It is also closed. At a point $(u,v)$ off the anti-diagonal, $u+v\ne 0$. If this sum is positive, every sufficiently small lower-limit rectangle based at $(u,v)$ still has positive coordinate sum. If it is negative, choose $ϵ>0$ with $2ϵ<-(u+v)$; then $[u,u+ϵ)\times [v,v+ϵ)$ has negative coordinate sum throughout. These neighborhoods avoid $D$, so its complement is open.

If the square were Lindelöf, its closed discrete subset $D$ would be countable by part (c), contradicting its bijection with $\mathbb{R}$. Hence

For each point which belongs to at least one $U_t$, its membership-index set is a nonempty subset of $[0,1]$, so has an infimum in $[0,1]$. Assigning the value one to the other points therefore defines $f:X\to [0,1]$ everywhere.

For $0<r\le 1$,An index $t<r$ gives $f(x)\le t<r$. Conversely if the infimum is below $r$, some membership index must be below $r$. Thus this strict sublevel set is open.

For $0\le r<1$ we claimIf $x\notin \overline{U_t}$ and $t>r$, then $x$ cannot belong to any $U_s$ with $s\le t$, since $U_s\subseteq U_t$ for $s<t$. Hence $f(x)\ge t>r$. Conversely, if $f(x)>r$, density of $D$ supplies $r<t<u<f(x)$ with $t,u\in D$. Since $x\notin U_u$ and $\overline{U_t}\subseteq U_u$, one has $x\notin \overline{U_t}$. This proves the identity, and its right-hand side is open.

The remaining sublevel and superlevel sets, for thresholds outside these ranges, are either empty or all of $X$. Preimages of open rays are consequently open, and these rays generate the usual real topology. Therefore $f$ is continuous. The closure inclusion is what supplies openness of the strict superlevel sets; mere nesting would not suffice.

Urysohn's lemma states that, in a normal space, disjoint closed subsets $A,B$ can be separated by a continuous function $u:X\to [0,1]$ with $u=0$ on $A$ and $u=1$ on $B$.

First use the shrinking consequence of the separation property of a normal space: if a closed set $F$ lies in an open set $G$, there is an open $W$ with $F\subseteq W\subseteq \overline{W}\subseteq G$. To prove it, separate the disjoint closed sets $F$ and $X\setminus G$ by disjoint open neighborhoods $W,V$. Then $\overline{W}\subseteq X\setminus V\subseteq G$.

Let $D$ be the dyadic rationals in $[0,1]$. Set $U_1=X\setminus B$, and shrink around $A$ to choose $U_0$ with $A\subseteq U_0\subseteq \overline{U_0}\subseteq U_1$. At each subsequent dyadic stage, insert between each adjacent pair $s<t$ an open set $U_{(s+t)/2}$ such thatThe shrinking lemma applies because the first set is closed and contained in the last open set. Induction gives $\overline{U_s}\subseteq U_t$ for all dyadic $s<t$. Part (a) now constructs a continuous $u$. Each point of $A$ lies in $U_0$, so has value zero. Each point of $B$ lies in none of the $U_t$, so has value one. This proves the lemma.

For the embedding, let $I=C(X,[0,1])$, a set of continuous functions, and define the evaluation mapEvery coordinate is continuous, so $e$ is continuous in the product topology. The Hausdorff assumption makes singletons closed. For distinct points, Urysohn's lemma applied to their singletons gives a coordinate with different values, so $e$ is injective.

To check the inverse topology, take $x\in O$ with $O$ open. Apply the lemma to $\{x\}$ and $X\setminus O$, choosing a coordinate $u$ with $u(x)=1$ and $u=0$ outside $O$. ThenThis is the preimage of an open coordinate cylinder under $e$. Thus every original neighborhood is detected by coordinate neighborhoods, and $e^{-1}:e[X]\to X$ is continuous. The evaluation map is a topological embedding.

Let $\mathcal{B}$ be a topological basis consisting of clopen sets. For $U\in \mathcal{B}$ its characteristic function ${\chi }_U:X\to \{0,1\}$ is continuous, since the preimages of both singleton values are open. DefineThe coordinate criterion for product topology makes this map continuous. If $x\ne y$, the $T_0$ property gives an open set containing one and not the other. A topological basis element contained in that open set still distinguishes them, so their corresponding characteristic coordinates differ. Therefore $e$ is injective.

Finally, for each topological basis element $U$,The right-hand side is the preimage of an open cylinder in the product of discrete two-point topological spaces. These cylinders recover a topological basis for the topology of $X$, proving continuity of the inverse onto the image. Hence

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

A proper filter $\mathcal{F}$ on $X$ is a nonempty family of subsets, closed under finite intersection and passage to supersets, with $\varnothing \notin \mathcal{F}$. An ultrafilter is a maximal proper filter, equivalently one which contains, for each $A\subseteq X$, exactly one of $A$ and $X\setminus A$. The pushforward filter isInverse images preserve intersections and complements. Thus it is a proper filter, and the either-a-set-or-its-complement criterion shows it is an ultrafilter whenever $\mathcal{F}$ is one.

Convergence of a filter $\mathcal{F}\to x$ means that every neighborhood of $x$ belongs to $\mathcal{F}$. A mapping is continuous exactly when $\mathcal{F}\to x$ implies $f_{\ast }\mathcal{F}\to f(x)$ for all proper filters and points. The forward implication follows from inverse images of neighborhoods. For the converse apply the criterion to the neighborhood proper filter at $x$: each inverse image of a neighborhood of $f(x)$ must be a neighborhood of $x$, which is continuity at $x$.

A topological space is compact exactly when every ultrafilter converges to at least one point; equivalently every proper filter has a cluster point. For completeness, in a compact space the closed sets $\overline{A}$, $A\in \mathcal{U}$, have the finite intersection property, hence a common point $x$. Every neighborhood $N$ of $x$ belongs to the ultrafilter. Indeed, if $N$ did not, choose an open $V$ with $x\in V\subseteq N$. Then $V$ would not belong either, so $X\setminus V$ would belong. That set is closed and excludes $x$, contradicting the choice of $x$ in every closure. Conversely, if an open cover has no finite subcover, its closed complements generate a proper filter. An ultrafilter extension cannot converge anywhere, because each point has a covering neighborhood whose complement belongs to that ultrafilter. This proves the compactness criterion using the permitted extension principle.

Now suppose $b\in \overline{f[A]}$. The familyis a filter base of nonempty sets: closure guarantees each intersection is nonempty, and finite intersections correspond to intersections of neighborhoods. Extend its generated proper filter to an ultrafilter $\mathcal{H}$. It contains $A$ and every $f^{-1}[V]$, so $f_{\ast }\mathcal{H}\to b$.

Conversely, if such an ultrafilter exists, each neighborhood $V$ of $b$ has $f^{-1}[V]\in \mathcal{H}$. Its intersection with $A\in \mathcal{H}$ is nonempty. Thus $V$ meets $f[A]$, provingThis part includes the unnumbered proper filter definitions and criteria preceding the printed subparts.

For every ultrafilter $\mathcal{H}$ on $X$, its image is an ultrafilter on the compact space $Y$, so it has a limit. That limit is unique because $Y$ is Hausdorff: distinct limits would have disjoint neighborhoods both belonging to the same proper filter. Therefore $\Lambda$ is well-defined. Since $f[X]$ is dense, every $b\in Y$ lies in $\overline{f[X]}$. Part (a), with $A=X$, supplies an ultrafilter mapping to $b$, proving surjectivity.

Suppose disjoint subsets have disjoint image closures. If $\mathcal{H}\ne \mathcal{K}$, choose $A\in \mathcal{H}$ not in $\mathcal{K}$; the ultrafilter criterion gives $X\setminus A\in \mathcal{K}$. Part (a) givesThe two closures are disjoint by assumption, so the limits differ. Thus $\Lambda$ is injective.

Conversely, suppose $\Lambda$ is injective and disjoint $A,B\subseteq X$ had a common point $b$ in their image closures. Part (a) would supply ultrafilters $\mathcal{H},\mathcal{K}$ containing $A,B$ respectively, both with image limit $b$. They must be distinct, since a proper filter cannot contain the disjoint sets $A$ and $B$ simultaneously. This contradicts injectivity. Hence

Use the Stone-Čech extension property: a discrete topological space $X$ embeds densely into a compact Hausdorff space $\beta X$, and every continuous map from $X$ to a compact Hausdorff space extends continuously to $\beta X$. Apply it to the characteristic function ${\chi }_A:X\to \{0,1\}$ of an arbitrary subset $A$; discreteness makes it continuous. Its extension ${\tilde{\chi }}_A$ has clopen fibers, andIndeed that fiber is closed and contains $A$, while its openness and density of $X$ show that every point in the fiber is approached by points of $A$.

If $A$ and $B$ are disjoint, the extension of ${\chi }_A$ is one on $\overline{A}$ and zero on $\overline{B}$, so these closures are disjoint. The inclusion $X\to \beta X$ therefore satisfies the criterion from part (b), and its ultrafilter-limit map is both surjective and injective. ThusThis is a bijection of underlying sets; identifying the compatible topology on the ultrafilter topological space would give the familiar stronger homeomorphism statement, but is not needed for the requested conclusion.