Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets

Let $(X,\mathcal{T})$ be a topological space, let $\mathcal{B}$ be a basis for $\mathcal{T}$ (Basis and subbasis for a topology, and the topology generated by a family of sets) and let $A\subseteq X$. Interior and closure are as in Interior, closure, boundary, exterior, derived set and isolated point in a topological space.

• $A$ is dense in $X$ if $\bar{A}=X$.
• $A$ is codense in $X$ if $X\setminus A$ is dense.
• $A$ is nowhere dense in $X$ if $\mathrm{int}(\bar{A})=\mathrm{\varnothing }$.

Three equivalent forms of density, and the one used in practice. The following are equivalent:

1. $\bar{A}=X$;
2. $U\cap A\mathrm{\ne }\mathrm{\varnothing }$ for every nonempty open $U\subseteq X$;
3. $B\cap A\mathrm{\ne }\mathrm{\varnothing }$ for every nonempty $B\in \mathcal{B}$.

Proof. (1) $\Rightarrow$ (2): if $U$ is open and nonempty, pick $x\in U$; then $x\in \bar{A}$, so $U\cap A\mathrm{\ne }\mathrm{\varnothing }$ by clause (c) of A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$; the closure is the smallest closed superset and equals $A$ together with its derived set. (2) $\Rightarrow$ (3): a nonempty member of $\mathcal{B}$ is a nonempty open set. (3) $\Rightarrow$ (1): let $x\in X$; every $B\in \mathcal{B}$ with $x\in B$ is nonempty and so meets $A$, hence $x\in \bar{A}$ by clause (d) of A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$; the closure is the smallest closed superset and equals $A$ together with its derived set. Form 3 is what makes density checkable: for the Sorgenfrey line it is a statement about half-open intervals, and for a metric space a statement about balls.

Codensity is emptiness of the interior. $A$ is codense if and only if $\mathrm{int}(A)=\mathrm{\varnothing }$, because $X\setminus \mathrm{int}(A)=\overline{X\setminus A}$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space), so $\overline{X\setminus A}=X$ holds exactly when $\mathrm{int}(A)=\mathrm{\varnothing }$.

Nowhere dense implies codense, and the converse fails. If $\mathrm{int}(\bar{A})=\mathrm{\varnothing }$ then $\mathrm{int}(A)\subseteq \mathrm{int}(\bar{A})=\mathrm{\varnothing }$ by monotonicity of the interior, so $A$ is codense. The two notions can differ only on sets whose closure is larger than themselves, and there they sometimes do: a dense set with empty interior, such as the rationals inside the real line, is codense and is not nowhere dense, its closure being everything. They may also agree on such a set: $(0,1)$ inside $\mathbb{R}$ has closure $[0,1]$ and is neither codense nor nowhere dense. A closed set is nowhere dense precisely when it is codense, since then $\bar{A}=A$.

• Density is a property of the pair, not of the set. A subset dense in $X$ need not be dense in a space with a finer topology. In a nonempty indiscrete space every nonempty subset is dense, while $\mathrm{\varnothing }$ is not; in the empty space $\mathrm{\varnothing }$ is dense as well. Where a density claim is made below the topology is always named.

• The empty set. $\mathrm{\varnothing }$ is nowhere dense and codense in every space, and it is dense only in $X=\mathrm{\varnothing }$. $X$ itself is dense in $X$ and is nowhere dense only when $X=\mathrm{\varnothing }$.

• What is deliberately not defined here. Separability, meaning the existence of an at most countable dense subset, is a countability axiom not developed at this point in the reading order; it is defined later in Separability: the existence of an at most countable dense subset ↗. Where a space on the companion page has an at most countable dense subset, that is what is said in full.