Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $S\subseteq X$. The subspace topology (also relative topology) on $S$ is

${\mathcal{T}}_S:=\{U\cap S:U\in \mathcal{T}\},$

the family of traces on $S$ of the open sets of $X$. The pair $(S,{\mathcal{T}}_S)$ is a subspace of $X$. A subset of $S$ that lies in ${\mathcal{T}}_S$ is said to be open in $S$, and relatively open where the ambient space needs emphasis.

${\mathcal{T}}_S$ is a topology, and this is discharged here. (T1): $\mathrm{\varnothing }=\mathrm{\varnothing }\cap S$ and $S=X\cap S$ are traces. (T2): if ${\mathcal{S}}^{\mathrm{\prime }}\subseteq {\mathcal{T}}_S$, choose for each member a set of $\mathcal{T}$ tracing to it — no choice principle is needed, since $U^{\mathrm{\prime }}:=\bigcup \{U\in \mathcal{T}:U\cap S\subseteq W\}$ is a canonical such set for $W\in {\mathcal{T}}_S$, being open by (T2) in $X$ and satisfying $U^{\mathrm{\prime }}\cap S=W$ — and then ${\bigcup }_i(U_i\cap S)=({\bigcup }_i{U}_i)\cap S\in {\mathcal{T}}_S$ by (T2) in $X$. (T3): $(U\cap S)\cap (V\cap S)=(U\cap V)\cap S\in {\mathcal{T}}_S$ by (T3) in $X$.

Closed sets of a subspace are the traces of the closed sets. A set $C\subseteq S$ is closed in $S$ if and only if $C=F\cap S$ for some closed $F\subseteq X$. Indeed $S\setminus (U\cap S)=(X\setminus U)\cap S$ and $S\setminus (F\cap S)=(X\setminus F)\cap S$, so complementation inside $S$ matches complementation inside $X$ under tracing.

Bases and subbases trace as well. If $\mathcal{B}$ is a basis for $\mathcal{T}$ (Basis and subbasis for a topology, and the topology generated by a family of sets) then ${\mathcal{B}}_S:=\{B\cap S:B\in \mathcal{B}\}$ is a basis for ${\mathcal{T}}_S$: its members are open in $S$, and for $W=U\cap S$ open in $S$ and $x\in W$ there is $B\in \mathcal{B}$ with $x\in B\subseteq U$, whence $x\in B\cap S\subseteq W$. The same computation with a subbasis $\mathcal{S}$ shows that $\{S_0\cap S:S_0\in \mathcal{S}\}$ is a subbasis for ${\mathcal{T}}_S$, since tracing commutes with finite intersections and with unions.

The inclusion is continuous. The inclusion map $\iota :S\to X$, $\iota (s)=s$, satisfies ${\iota }^{-1}[U]=U\cap S$ for every $U\subseteq X$, so preimages of open sets are open and $\iota$ is continuous (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f(\bar{A})\subseteq \overline{f(A)}$, clause (b)). Moreover ${\mathcal{T}}_S$ is the coarsest topology on $S$ making $\iota$ continuous: any topology on $S$ for which $\iota$ is continuous must contain every ${\iota }^{-1}[U]=U\cap S$, hence contain ${\mathcal{T}}_S$.

Characteristic property of a map into a subspace. Let $(Z,{\mathcal{T}}_Z)$ be a topological space and let $g:Z\to S$ be a function. Then

$g\text{ is continuous as a map }Z\to (S,{\mathcal{T}}_S)⟺\iota \circ g\text{ is continuous as a map }Z\to (X,\mathcal{T})\mathrm{.}$

Proof. For $U\in \mathcal{T}$ one has $(\iota \circ g{)}^{-1}[U]=g^{-1}[{\iota }^{-1}[U]]=g^{-1}[U\cap S]$. If $g$ is continuous then each $g^{-1}[U\cap S]$ is open, so $\iota \circ g$ is continuous; conversely if $\iota \circ g$ is continuous then for any $W=U\cap S$ open in $S$ the set $g^{-1}[W]=(\iota \circ g{)}^{-1}[U]$ is open, so $g$ is continuous. Both directions use only clause (b) of For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f(\bar{A})\subseteq \overline{f(A)}$.

Restriction of a continuous map. If $f:X\to Y$ is continuous and $S\subseteq X$, then $f{\mathrm{\mid }}_S:S\to Y$ is continuous, since $(f{\mathrm{\mid }}_S{)}^{-1}[V]=f^{-1}[V]\cap S$ is open in $S$ for every open $V\subseteq Y$ (Continuity of a map of topological spaces at a point and globally, For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f(\bar{A})\subseteq \overline{f(A)}$).

When relative and ambient agree. If $S$ is open in $X$ then a subset of $S$ is open in $S$ if and only if it is open in $X$: a trace $U\cap S$ is then an intersection of two open sets of $X$, and conversely an open subset of $X$ contained in $S$ is its own trace. The same statement with "closed" throughout holds when $S$ is closed in $X$. Both are used in the pasting lemma of the next item, and both fail without the hypothesis: $S$ itself is always open and closed in $S$, and need be neither in $X$.

• The subspace topology is what makes a subset a space. Before it, a statement such as "the restriction of $f$ to $C$ is continuous" has no meaning, because $C$ carries no topology. Every restriction below is taken with respect to the subspace topology and with no other convention available.

• Openness and closedness are not absolute. $[0,1)$ is open in $[0,2)$ and is neither open nor closed in $\mathbb{R}$; the interval $(0,1)$ is closed in itself. A sentence of the form "$A$ is open" is incomplete unless the space is named, and this library names it whenever more than one is in play.

• Transitivity. If $S\subseteq T\subseteq X$ then the subspace topology $S$ inherits from $(T,{\mathcal{T}}_T)$ is the subspace topology it inherits from $X$, since $(U\cap T)\cap S=U\cap S$ for $U\in \mathcal{T}$. So no ambiguity arises from the route by which a subset is reached.