Throughout mathematics, we encounter subsets of [topological spaces](/page/Topology) that we need to study as spaces in their own right. The unit circle $S^1 \subset \mathbb{R}^2$, the closed unit ball $\overline{B}(0, 1) \subset \mathbb{R}^n$, the Cantor set $\mathcal{C} \subset [0, 1]$, and the solution set $\{x \in \mathbb{R}^n : f(x) = 0\}$ of a [continuous](/page/Continuity) equation are all defined as subsets of ambient spaces whose topology is already understood. The fundamental question is: **how should we topologise a subset so that the resulting space faithfully reflects the topological structure it inherits from the ambient space?**

The question is less innocent than it appears. A subset $A \subset X$ can carry many different topologies, and the wrong choice leads to pathologies. If we declare every subset of $A$ to be open (the discrete topology), we destroy continuity — a function that is continuous into $X$ will generally fail to be continuous into $A$. If we declare only $\varnothing$ and $A$ to be open (the indiscrete topology), we lose the ability to distinguish points. What we need is a topology on $A$ that is compatible with the topology of $X$ in a precise sense: the natural inclusion map $\iota: A \hookrightarrow X$ should be continuous, and any map into $A$ should be continuous exactly when the corresponding map into $X$ is.

[example: The Need for a Canonical Choice] Consider the closed interval $A = [0, 1]$ as a subset of $\mathbb{R}$ with the standard topology. The set $[0, 1/2)$ is not open in $\mathbb{R}$ — it fails the "open ball" condition at the point $0$, since every open ball $B(0, \varepsilon)$ contains negative numbers outside $[0, 1]$. Yet within the interval $[0, 1]$, the set $[0, 1/2)$ should be considered "open" in any reasonable sense: a continuous function $f: Y \to [0, 1]$ that maps a point $y_0$ into $[0, 1/2)$ should map a neighbourhood of $y_0$ into $[0, 1/2)$ as well. The resolution is that $[0, 1/2) = (-1/2, 1/2) \cap [0, 1]$ is the intersection of an open set in $\mathbb{R}$ with $A$. This observation — that the "open" subsets of $A$ should be exactly the intersections of open subsets of $X$ with $A$ — is the key insight that leads to the subspace topology. [/example]

This idea generalises far beyond Euclidean spaces. Whenever we study a submanifold of a manifold, a closed subgroup of a topological group, or the solution set of a system of equations, we are implicitly using the subspace topology. Its importance lies not in its complexity — the definition is one line — but in the subtleties that arise when one studies how topological properties interact with the passage to subsets.

## Definition

[definition: Subspace Topology] Let $(X, \tau)$ be a [topological space](/page/Topology) and let $A \subset X$ be a subset. The **subspace topology** (also called the **relative topology** or **induced topology**) on $A$ is the collection \begin{align*} \tau_A := \{ U \cap A : U \in \tau \}. \end{align*} The pair $(A, \tau_A)$ is called a **subspace** of $(X, \tau)$, and the [inclusion map](/page/Function) $\iota: A \hookrightarrow X$ defined by $\iota(a) = a$ is called the **canonical inclusion** (or **canonical embedding**). [/definition]

We should verify that $\tau_A$ is indeed a topology on $A$. The empty set $\varnothing = \varnothing \cap A$ is in $\tau_A$, and $A = X \cap A$ is in $\tau_A$. If $\{U_\alpha \cap A\}_{\alpha \in I}$ is a family of sets in $\tau_A$, then $\bigcup_{\alpha \in I} (U_\alpha \cap A) = (\bigcup_{\alpha \in I} U_\alpha) \cap A$, which is in $\tau_A$ since $\bigcup U_\alpha \in \tau$. Similarly, if $U_1 \cap A$ and $U_2 \cap A$ are in $\tau_A$, then $(U_1 \cap A) \cap (U_2 \cap A) = (U_1 \cap U_2) \cap A \in \tau_A$.

A set $V \subset A$ is called **open in $A$** (or **relatively open**) if $V \in \tau_A$, and **closed in $A$** (or **relatively closed**) if $A \setminus V$ is open in $A$. The closed sets of $\tau_A$ have an equally simple description: they are exactly the sets of the form $F \cap A$ where $F$ is [closed](/page/Closed%20Set) in $X$.

[remark: The Characterising Universal Property] The subspace topology is the **coarsest** topology on $A$ that makes the inclusion $\iota: A \hookrightarrow X$ continuous. It is characterised by the following universal property: for any topological space $Y$ and any function $f: Y \to A$, the map $f$ is continuous (with respect to $\tau_A$) if and only if the composition $\iota \circ f: Y \to X$ is continuous. This means that continuity of a map into a subspace $A$ is equivalent to continuity of the same map viewed as a map into the ambient space $X$. [/remark]

The universal property is the deeper reason why the subspace topology is the "right" choice. Any other topology on $A$ making $\iota$ continuous would either be too fine (declaring too many sets open, potentially making some maps into $A$ discontinuous even though the corresponding maps into $X$ are continuous) or would not make $\iota$ continuous at all.

[example: Open and Closed Sets in Subspaces] The distinction between "open in $A$" and "open in $X$" is the single most important source of confusion when working with subspaces. Consider $X = \mathbb{R}$ with the standard topology. **1.** Let $A = [0, 2]$. The set $[0, 1) = (-1, 1) \cap [0, 2]$ is open in $A$ but not open in $\mathbb{R}$. The set $[1, 2] = [1, 3] \cap [0, 2]$ is closed in $A$ (as the intersection of a closed set in $\mathbb{R}$ with $A$), and also closed in $\mathbb{R}$. **2.** Let $A = \mathbb{Z} \subset \mathbb{R}$. Every singleton $\{n\} = (n - 1/2, n + 1/2) \cap \mathbb{Z}$ is open in $A$, since $(n - 1/2, n + 1/2)$ is open in $\mathbb{R}$. Hence every subset of $\mathbb{Z}$ is a union of singletons, hence open. The subspace topology on $\mathbb{Z}$ is the **discrete topology**. More generally, any subset of $\mathbb{R}$ with no accumulation points inherits the discrete topology. **3.** Let $A = [0, 1] \cup [2, 3]$. The set $[0, 1] = (-1/2, 3/2) \cap A$ is open in $A$, and its complement $[2, 3]$ is also open in $A$. Thus $[0, 1]$ is a **clopen** (simultaneously open and closed) subset of $A$. This means $A$ is [disconnected](/page/Connectedness): it splits as the disjoint union of two non-empty open sets. [/example]

### Bases for the Subspace Topology

If the topology $\tau$ on $X$ is generated by a basis $\mathcal{B}$, then the subspace topology on $A$ is generated by the basis $\mathcal{B}_A = \{B \cap A : B \in \mathcal{B}\}$. This is immediate: every open set $U \in \tau$ is a union of basis elements $U = \bigcup_\alpha B_\alpha$, so $U \cap A = \bigcup_\alpha (B_\alpha \cap A)$ is a union of elements of $\mathcal{B}_A$.

For metric spaces, this has a concrete interpretation: if $(X, d)$ is a [metric space](/page/Metric%20Space) and $A \subset X$, then the open balls in the subspace metric $d|_{A \times A}$ are exactly $B_X(x, r) \cap A$ for $x \in A$ and $r > 0$. Hence the metric topology on $(A, d|_{A \times A})$ coincides with the subspace topology inherited from $(X, d)$. This compatibility ensures that the abstract definition agrees with the metric-space notion of "restricting the metric."

## Closure, Interior, and Boundary in Subspaces

One of the most delicate aspects of working with subspaces is that operations like [closure](/page/Closure), [interior](/page/Interior), and [boundary](/page/Boundary) depend on the ambient space. The closure of a set $B$ in the subspace $A$ can differ from its closure in $X$, and the interior of $B$ in $A$ can be non-empty even when its interior in $X$ is empty. Failing to track which space the operation is taken in is the source of many errors.

### Relative Closure

The fundamental formula for relative closure is clean: the closure of $B$ in the subspace $A$ is the trace of its closure in $X$.

[quotetheorem:1011]

The formula follows from the characterisation of closed sets in the subspace: a subset $C \subset A$ is closed in $A$ if and only if $C = F \cap A$ for some closed $F \subset X$. The closure $\overline{B}^A$ is the smallest such set containing $B$, and one verifies that $\overline{B} \cap A$ is closed in $A$ (since $\overline{B}$ is closed in $X$), contains $B$, and is contained in every closed subset of $A$ that contains $B$.