Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2008 / A / 1

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