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): $\varnothing = \varnothing \cap S$ and $S = X \cap S$ are traces. (T2): if $\mathcal{S}' \subseteq \mathcal{T}_S$, choose for each member a set of $\mathcal{T}$ tracing to it — no choice principle is needed, since $U' := \bigcup \{\, U \in \mathcal{T} : U \cap S \subseteq W \,\}$ is a canonical such set for $W \in \mathcal{T}_S$, being open by (T2) in $X$ and satisfying $U' \cap S = W$ — 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 $f(\overline{A}) \subseteq \overline{f(A)}$, 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

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

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 $f(\overline{A}) \subseteq \overline{f(A)}$.

Restriction of a continuous map. If $f : X \to Y$ is continuous and $S \subseteq X$, then $f|_S : S \to Y$ is continuous, since $(f|_S)^{-1}[V] = f^{-1}[V] \cap S$ 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 $f(\overline{A}) \subseteq \overline{f(A)}$).

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.