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 $B$. [example: Relative vs. Absolute Closure] Let $X = \mathbb{R}$, $A = (0, 2)$ with the subspace topology, and $B = (0, 1)$. **Closure in $\mathbb{R}$:** $\overline{B} = [0, 1]$. **Closure in $A$:** $\overline{B}^A = [0, 1] \cap (0, 2) = (0, 1]$. The point $0$ belongs to $\overline{B}$ but not to $\overline{B}^A$, because $0 \notin A$. The relative closure "cannot see" limit points that lie outside $A$. In the subspace $(0, 2)$, the set $(0, 1]$ is closed — its complement in $A$ is $(1, 2)$, which is open in the subspace topology since $(1, 2)$ is open in $\mathbb{R}$. Conversely, the point $1$ lies in both $\overline{B}$ and $\overline{B}^A$, since $1 \in A$. This illustrates the general principle: $\overline{B}^A \subset \overline{B}$, with equality if and only if $\overline{B} \subset A$ (i.e., all limit points of $B$ in $X$ already belong to $A$). [/example] ### Relative Interior The situation for interiors is more striking: the interior of $B$ in $A$ can be strictly larger than the interior of $B$ in $X$. [quotetheorem:1035] The strict inclusion is not a pathology — it reflects the fact that the subspace has "fewer dimensions" to worry about when testing whether a point is interior. A point that fails to be interior in $X$ (because open sets in $X$ extend in directions not available in $A$) may become interior in $A$ once those directions are removed. [example: Interior Depends on the Ambient Space] Let $X = \mathbb{R}^2$ and $A = \mathbb{R} \times \{0\}$ (the $x$-axis), equipped with the subspace topology. Consider the set $B = [0, 1] \times \{0\} \subset A$. **Interior in $X = \mathbb{R}^2$:** $\operatorname{int}_X(B) = \varnothing$, since no open ball $B_{\mathbb{R}^2}((x, 0), r)$ is contained in the $x$-axis. **Interior in $A$:** Under the subspace topology, $A$ is homeomorphic to $\mathbb{R}$ (via the projection $(x, 0) \mapsto x$), and $B$ corresponds to $[0, 1]$. The interior of $[0, 1]$ in $\mathbb{R}$ is $(0, 1)$, so $\operatorname{int}_A(B) = (0, 1) \times \{0\}$, which is non-empty. This phenomenon is ubiquitous in applications: submanifolds, solution curves, level sets — all have empty interior in the ambient space but possess rich internal structure when equipped with the subspace topology. [/example] ### Relative Boundary The boundary also depends on the ambient space. The boundary of $B$ in $A$ is $\partial_A B = \overline{B}^A \setminus \operatorname{int}_A(B)$, and this can differ substantially from $\partial_X B$. A point can lie on the boundary of $B$ in $X$ but be interior to $B$ in $A$ (or vice versa), or it can be a boundary point in $X$ but not even belong to $A$. [example: Relative Boundary] Let $X = \mathbb{R}$, $A = [0, 1]$, and $B = [0, 1/2]$. **Boundary in $\mathbb{R}$:** $\partial_X B = \{0, 1/2\}$. **Boundary in $A$:** $\overline{B}^A = [0, 1/2]$ (since $\overline{B} = [0, 1/2]$ and $[0, 1/2] \subset A$). The interior $\operatorname{int}_A(B) = [0, 1/2)$, since $[0, 1/2) = (-1, 1/2) \cap A$ is open in $A$. Thus $\partial_A B = [0, 1/2] \setminus [0, 1/2) = \{1/2\}$. The point $0$ is a boundary point of $B$ in $\mathbb{R}$ but an interior point of $B$ in $A$: within $[0, 1]$, the set $[0, 1/2]$ contains a relative neighbourhood of $0$, namely $[0, 1/4) = (-1, 1/4) \cap A$. [/example] ## Continuity and the Restriction of Maps The subspace topology interacts cleanly with [continuous maps](/page/Continuity). The basic principle is that restricting the domain or codomain of a continuous map preserves continuity — but enlarging the codomain from a subspace to the ambient space can create or destroy continuity depending on the direction. ### Restriction to Subspaces [quotetheorem:1036] Part (1) follows because if $V \subset Y$ is open, then $(f|_A)^{-1}(V) = f^{-1}(V) \cap A$, which is open in $A$ since $f^{-1}(V)$ is open in $X$. Part (2) follows from the universal property: the composition $\iota_B \circ f|_A^B = f|_A: A \to Y$ is continuous by Part (1), so $f|_A^B$ is continuous by the universal property of the subspace topology on $B$. The converse question — when does continuity on subspaces imply continuity on the whole space? — is more subtle. ### The Pasting Lemma In many constructions, a function $f: X \to Y$ is defined "piecewise" by giving continuous formulas on subsets that cover $X$. The question is: when is the resulting function continuous? The answer depends on whether the covering sets are open or closed. [quotetheorem:1037] The open version is immediate: if $V \subset Y$ is open, then $f^{-1}(V) = \bigcup_\alpha (f|_{U_\alpha})^{-1}(V)$. Each $(f|_{U_\alpha})^{-1}(V)$ is open in $U_\alpha$, hence open in $X$ (since $U_\alpha$ is open in $X$ and an open subset of an open subspace is open in the ambient space). The union is therefore open in $X$. The closed version for two sets follows by considering preimages of closed sets: if $F \subset Y$ is closed, then $f^{-1}(F) = (f|_{C_1})^{-1}(F) \cup (f|_{C_2})^{-1}(F)$. Each term is closed in $C_i$, hence closed in $X$ (since $C_i$ is closed in $X$), so the union is closed in $X$. [example: Failure of Closed Pasting for Infinitely Many Pieces] The closed pasting lemma does **not** extend to infinite closed covers. Define $f: \mathbb{R} \to \mathbb{R}$ by $f(x) = 0$ for $x \le 0$ and $f(x) = 1$ for $x > 0$. Cover $\mathbb{R}$ by the closed sets $C_0 = (-\infty, 0]$ and $C_n = [1/n, \infty)$ for $n \in \mathbb{N}$. Each restriction $f|_{C_0}$ and $f|_{C_n}$ is continuous (being constant or constant on a closed interval), yet $f$ itself is discontinuous at $0$. The more fundamental failure arises even with a genuinely infinite closed cover: take $X = [0, 1]$ and cover it by the singletons $\{x\}$ for $x \in [0, 1]$. Every function $f: [0, 1] \to Y$ is continuous when restricted to any singleton (every function on a one-point space is continuous, since the only open sets in a one-point space are $\varnothing$ and the space itself). But not every function on $[0, 1]$ is continuous. The issue is that an arbitrary union of closed sets need not be closed, so the preimage argument breaks down. [/example] ## Hereditary Properties A natural question pervades the theory: which topological properties pass from a space to its subspaces? A topological property $\mathcal{P}$ is called **hereditary** if every subspace of a space with property $\mathcal{P}$ also has property $\mathcal{P}$, and **closed-hereditary** (respectively, **open-hereditary**) if every closed (respectively, open) subspace inherits $\mathcal{P}$. Understanding which properties are hereditary and which are not is essential for working with subspaces, because it determines which tools remain available when we pass to a subset. ### Properties Inherited by All Subspaces The [Hausdorff property](/page/Topology) (T2) is hereditary: if $X$ is Hausdorff and $A \subset X$, then distinct points $a, b \in A$ can be separated by open sets $U, V$ in $X$ (with $a \in U$, $b \in V$, and $U \cap V = \varnothing$), and the sets $U \cap A$ and $V \cap A$ separate $a$ and $b$ in $A$. The same argument works for the weaker separation axioms T0 and T1: points that are distinguished by open sets in $X$ remain distinguished in $A$. [Metrizability](/page/Metrizable%20Space) is hereditary: if $d$ is a metric on $X$ inducing $\tau$, then $d|_{A \times A}$ is a metric on $A$ inducing $\tau_A$, as verified earlier. [Second countability](/page/Countable%20Set) is hereditary: if $\mathcal{B}$ is a countable basis for $\tau$, then $\{B \cap A : B \in \mathcal{B}\}$ is a countable basis for $\tau_A$. First countability is hereditary by the same argument applied to local bases. ### Properties Inherited Only by Closed Subspaces [Compactness](/page/Compact%20Space) is **not** hereditary: the open interval $(0, 1)$ is a non-compact subspace of the compact space $[0, 1]$. However, compactness is **closed-hereditary**: a closed subset of a compact space is compact. The proof is standard: if $\{U_\alpha\}$ is an open cover of the closed set $C$, then $\{U_\alpha\} \cup \{X \setminus C\}$ is an open cover of $X$, which has a finite subcover, and removing $X \setminus C$ yields a finite subcover of $C$. [Completeness](/page/Complete%20Metric%20Space) (for metric spaces) is closed-hereditary but not hereditary: a closed subspace of a complete metric space is complete (a Cauchy sequence in the subspace is Cauchy in the ambient space, hence convergent; the limit lies in the subspace because the subspace is closed). An open subspace of a complete metric space need not be complete — for instance, $(0, 1)$ with the metric inherited from $\mathbb{R}$ is not complete (the sequence $1/n$ is Cauchy but does not converge in $(0, 1)$). The [Lindelof property](/page/Topology) (every open cover has a countable subcover) is closed-hereditary but not hereditary. The **Sorgenfrey line** $\mathbb{R}_\ell$ (the real line with the lower limit topology, generated by half-open intervals $[a, b)$) is Lindelof, but the subspace $\mathbb{Q}$ of rationals is Lindelof in any case (being second-countable). A more informative example: the Sorgenfrey plane $\mathbb{R}_\ell \times \mathbb{R}_\ell$ is not Lindelof, even though $\mathbb{R}_\ell$ is. ### Properties Not Inherited by Subspaces [Connectedness](/page/Connectedness) is not hereditary, and fails dramatically. The space $\mathbb{R}$ is connected, but the subspace $\mathbb{Q}$ is totally disconnected: for any two rationals $p < q$, the irrational number $\alpha \in (p, q)$ provides a separation $\mathbb{Q} = (\mathbb{Q} \cap (-\infty, \alpha)) \cup (\mathbb{Q} \cap (\alpha, \infty))$. Connectedness is not even closed-hereditary. Consider $X = \mathbb{R}$ and the closed subspace $A = \{0\} \cup \{1\}$. The ambient space $\mathbb{R}$ is connected, $A$ is closed, yet $A$ is disconnected — each singleton is clopen in the subspace topology (since $\{0\} = (-1/2, 1/2) \cap A$ is open, and $\{1\} = (1/2, 3/2) \cap A$ is open). More generally, any finite subset of $\mathbb{R}$ with at least two points is closed and disconnected. This reflects the fundamental difference between compactness (which is closed-hereditary) and connectedness: removing part of a connected space can sever the "links" that hold it together, even if the remaining piece is closed. Compactness and connectedness both fail to be hereditary, but for different reasons. Compactness fails because subsets can "lose" limit points (open sets are "missing" boundary points). Connectedness fails because the "glue" holding pieces together may not survive restriction. [example: A Compact Hausdorff Space Whose Subspace is Not Normal] The failure of heredity can be quite dramatic even for strong properties. Normality (the T4 axiom: disjoint closed sets can be separated by open sets) is not hereditary, even for compact Hausdorff spaces. The **Tychonoff plank** is the product $T = [0, \omega_1] \times [0, \omega]$, where $[0, \omega_1]$ and $[0, \omega]$ are ordinal spaces with the order topology, and $\omega_1$ is the first uncountable ordinal, $\omega$ the first infinite ordinal. Both factors are compact and Hausdorff, so $T$ is compact and Hausdorff, hence normal (compact Hausdorff spaces are normal). The subspace $T^* = T \setminus \{(\omega_1, \omega)\}$ — obtained by removing the single "corner point" — is locally compact and Hausdorff, but **not** normal. The two disjoint closed subsets $\{\omega_1\} \times [0, \omega)$ and $[0, \omega_1) \times \{\omega\}$ of $T^*$ cannot be separated by open sets. This example shows that removing a single point from a normal space can destroy normality. It underscores why normality is a "fragile" property compared to the Hausdorff condition or regularity, both of which are hereditary. [/example] ## Embeddings and Subspace Recognition In practice, subspaces arise not only by taking subsets but also by recognising that a given topological space is "isomorphic" to a subspace of another. The correct notion here is the **topological embedding**, which generalises the canonical inclusion $\iota: A \hookrightarrow X$. The problem that embeddings solve is the following: given a continuous injection $f: Y \to X$, under what conditions can we regard $Y$ as a subspace of $X$ (via $f$)? A continuous injection is not sufficient — the map $f$ might distort the topology of $Y$, mapping open sets to non-open sets. What we need is that $f$ is a homeomorphism onto its image. [definition: Topological Embedding] Let $X$ and $Y$ be topological spaces. A **topological embedding** (or simply **embedding**) is a continuous injection $f: Y \to X$ such that the corestriction \begin{align*} f: Y \to f(Y) \end{align*} is a [homeomorphism](/page/Topology), where $f(Y)$ carries the subspace topology inherited from $X$. Equivalently, $f$ is an embedding if and only if $f$ is injective and a map $U \subset Y$ is open in $Y$ if and only if $f(U)$ is open in $f(Y)$. [/definition] Every canonical inclusion $\iota: A \hookrightarrow X$ is an embedding (this is immediate from the definition of the subspace topology). The converse question — when is a continuous injection an embedding? — is more interesting. [quotetheorem:1038] This follows from the Closed Map Lemma: a continuous map from a compact space to a Hausdorff space is a closed map, and a continuous closed injection is an embedding (since it maps closed sets to closed sets, it maps open complements to relatively open sets, so the corestriction is a homeomorphism). [example: A Continuous Injection That Is Not an Embedding] The compact-to-Hausdorff hypothesis is essential. Consider the map \begin{align*} f: [0, 2\pi) &\to S^1 \subset \mathbb{R}^2 \\ t &\mapsto (\cos t, \sin t), \end{align*} where $[0, 2\pi)$ carries the subspace topology from $\mathbb{R}$ and $S^1$ carries the subspace topology from $\mathbb{R}^2$. This map is a continuous bijection, but it is **not** an embedding: the set $[0, 1)$ is open in $[0, 2\pi)$, but $f([0, 1))$ is not open in $S^1$. The arc $f([0, 1))$ contains the point $(1, 0)$ but does not contain a full open arc around it — the "missing" arc near $(1, 0)$ from the direction of $t = 2\pi$ prevents $f([0, 1))$ from being open. The failure occurs because $[0, 2\pi)$ is not compact: the sequence $t_n = 2\pi - 1/n$ converges in $S^1$ (to $(1, 0)$) but the preimage sequence does not converge in $[0, 2\pi)$ to the preimage of the limit. The inverse function $f^{-1}$ is discontinuous at $(1, 0)$. [/example] ### Recognising Subspaces via Embeddings The notion of embedding allows us to identify when an abstract topological space "is" a subspace of a larger space, even when it is not literally presented as a subset. [example: The Cantor Set as a Subspace of $[0, 1]$] The [Cantor set](/page/Product%20Topology) $\mathcal{C}$ is constructed iteratively by removing middle thirds from $[0, 1]$. As a subset of $[0, 1]$, it inherits the subspace topology. But the Cantor set is also homeomorphic to the product space $\{0, 2\}^{\mathbb{N}}$ (with the product topology), via the map \begin{align*} \phi: \{0, 2\}^{\mathbb{N}} &\to \mathcal{C} \\ (a_k)_{k=1}^\infty &\mapsto \sum_{k=1}^\infty \frac{a_k}{3^k}. \end{align*} Since $\{0, 2\}^{\mathbb{N}}$ is compact (by Tychonoff's theorem) and $\mathcal{C} \subset \mathbb{R}$ is Hausdorff, this continuous bijection is automatically an embedding. The product structure $\{0, 2\}^{\mathbb{N}}$ reveals properties of $\mathcal{C}$ — such as its total disconnectedness and self-similarity — that are obscured by the subset description. [/example] ## Subspaces and Products A natural question arises when two constructions — subspaces and products — interact: given subspaces $A \subset X$ and $B \subset Y$, does the product $A \times B$ with the [product topology](/page/Product%20Topology) agree with $A \times B$ viewed as a subspace of $X \times Y$? The answer is yes, and this compatibility is essential for many arguments. [quotetheorem:1039] The verification uses bases: a basis element for the product topology on $A \times B$ is $(U \cap A) \times (V \cap B)$ where $U \in \tau_X$ and $V \in \tau_Y$, and this equals $(U \times V) \cap (A \times B)$, which is a typical basis element for the subspace topology on $A \times B$ inherited from $X \times Y$. The two bases coincide, so the topologies agree. This result extends to arbitrary products: if $A_\alpha \subset X_\alpha$ for each $\alpha$ in an index set $I$, then the product topology on $\prod_{\alpha \in I} A_\alpha$ coincides with the subspace topology inherited from $\prod_{\alpha \in I} X_\alpha$. [example: The Torus as a Subspace and as a Product] The torus $\mathbb{T}^2 = S^1 \times S^1$ can be constructed in two ways: 1. As the product of two copies of the circle $S^1$, each carrying the subspace topology from $\mathbb{R}^2$. 2. As a subspace of $\mathbb{R}^4$ via the embedding $(e^{i\theta}, e^{i\phi}) \mapsto (\cos\theta, \sin\theta, \cos\phi, \sin\phi)$. The theorem guarantees that these two constructions yield the same topology. The product topology on $S^1 \times S^1$ — formed from the subspace topologies on each $S^1$ — agrees with the subspace topology on the image of the embedding in $\mathbb{R}^4$. [/example] ## Dense Subspaces and Completions A subspace $A$ of a topological space $X$ is [dense](/page/Dense%20Subset) if $\overline{A} = X$: every point of $X$ is a limit point of $A$ (or belongs to $A$). Dense subspaces are important because they allow us to extend continuous functions and to characterise spaces by their "approximating" subspaces. The key difficulty with dense subspaces is the following: a continuous function on a dense subspace need not extend continuously to the whole space, even when the ambient space is very well-behaved. When such an extension does exist, it is typically unique, and the conditions for existence involve [completeness](/page/Complete%20Metric%20Space) or compactness. [quotetheorem:1040] The uniqueness follows because the set $\{x \in X : f(x) = g(x)\}$ is closed (as the preimage of the diagonal $\Delta_Y = \{(y,y) : y \in Y\}$ under the continuous map $x \mapsto (f(x), g(x))$, and the diagonal is closed in $Y \times Y$ when $Y$ is Hausdorff) and contains the dense set $A$, hence equals $X$. Uniqueness does not guarantee existence. The function $f: \mathbb{Q} \to \mathbb{R}$ defined by $f(x) = 1/(x^2 - 2)$ is continuous on $\mathbb{Q}$ (since $x^2 - 2 \neq 0$ for $x \in \mathbb{Q}$) but does not extend continuously to $\mathbb{R}$ (it blows up at $\pm\sqrt{2}$). [quotetheorem:964] This result is the foundation of the **completion** construction: given a metric space $(X, d)$, the completion $(\overline{X}, \overline{d})$ is a complete metric space containing $X$ as a dense subspace. The identity map $\operatorname{id}: X \to X$ extends to the identity of $X$ within $\overline{X}$, and the construction is unique up to isometry. The real numbers $\mathbb{R}$ arise as the completion of $\mathbb{Q}$; the space $L^p(U)$ arises as the completion of $C_c(U)$ under the $L^p$ norm. [example: Dense Subspaces Determine the Topology] Consider the spaces $\mathbb{Q} \subset \mathbb{R}$ and $\mathbb{Q} \subset \mathbb{R}_\ell$ (where $\mathbb{R}_\ell$ is the Sorgenfrey line). The rationals $\mathbb{Q}$ are dense in both spaces, but the subspace topologies on $\mathbb{Q}$ differ: in $\mathbb{R}$, the sets $\mathbb{Q} \cap (a, b)$ form a basis for the subspace topology on $\mathbb{Q}$, while in $\mathbb{R}_\ell$, the sets $\mathbb{Q} \cap [a, b)$ form a basis. These topologies are genuinely different: in the Sorgenfrey subspace topology, every set $\mathbb{Q} \cap [q, q + 1)$ for $q \in \mathbb{Q}$ is clopen, while in the standard subspace topology, no bounded clopen set exists. This shows that the ambient topology matters: the same underlying set $\mathbb{Q}$ inherits different topological structures depending on the ambient space, even though it is dense in both. [/example] ## Standard Techniques for Subspace Arguments This section collects the reasoning patterns that arise repeatedly when working with the subspace topology. Each technique is paired with a context in which it is most useful. ### The "Open in Open" and "Closed in Closed" Principle [quotetheorem:1041] For Part (1): $B = U \cap A$ for some $U$ open in $X$, and $A$ is open in $X$, so $B = U \cap A$ is an intersection of two open sets in $X$, hence open in $X$. Part (2) follows by the same argument with "open" replaced by "closed." This principle does **not** extend to mixed cases. A set that is open in a closed subspace, or closed in an open subspace, need not be either open or closed in $X$. For instance, $[0, 1)$ is open in the closed subspace $[0, 1]$ of $\mathbb{R}$, but $[0, 1)$ is neither open nor closed in $\mathbb{R}$. ### Extending Local Properties via Open Covers If a topological property holds on a neighbourhood of every point in a subspace $A$, then it often holds on all of $A$. The technique is: 1. For each $a \in A$, find an open set $U_a$ in $X$ containing $a$ on which the desired property holds. 2. The collection $\{U_a \cap A\}_{a \in A}$ is an open cover of $A$ in the subspace topology. 3. Apply the Pasting Lemma (for continuity) or another "local-to-global" principle. This pattern is used, for instance, to verify that a map defined on a submanifold is continuous: check continuity in each coordinate chart (which amounts to checking continuity on open subsets of the subspace), then paste using the open pasting lemma. ### Compact Subspaces and Extraction Arguments A common pattern combines the subspace topology with compactness to extract global conclusions from local data: 1. Start with a compact subspace $K \subset X$. 2. Cover $K$ with open sets (open in $X$, or open in $K$) having a desired property. 3. Extract a finite subcover from the compactness of $K$. 4. Use finiteness to pass from local estimates to global bounds. This pattern underlies many fundamental results: the proof that a continuous function on a compact space attains its bounds, the proof that compact subsets of Hausdorff spaces are closed, and the Tube Lemma for finite products. [example: Compactness and the Distance to a Closed Set] Let $(X, d)$ be a metric space, $K \subset X$ compact, and $F \subset X$ closed, with $K \cap F = \varnothing$. We show that $\operatorname{dist}(K, F) := \inf\{d(k, f) : k \in K, f \in F\} > 0$. For each $k \in K$, since $k \notin F$ and $F$ is closed, there exists $\varepsilon_k > 0$ with $B(k, \varepsilon_k) \cap F = \varnothing$, which gives $\operatorname{dist}(k, F) \ge \varepsilon_k$. The open cover $\{B(k, \varepsilon_k / 2)\}_{k \in K}$ of the compact space $K$ has a finite subcover $\{B(k_i, \varepsilon_{k_i}/2)\}_{i=1}^N$. Set $\delta = \min_{1 \le i \le N} \varepsilon_{k_i}/2 > 0$ (a minimum of finitely many positive numbers). For any $k \in K$, there exists $k_i$ with $d(k, k_i) < \varepsilon_{k_i}/2$. For any $f \in F$, we have $d(k_i, f) \ge \varepsilon_{k_i}$ (since $B(k_i, \varepsilon_{k_i}) \cap F = \varnothing$). By the triangle inequality: \begin{align*} d(k, f) \ge d(k_i, f) - d(k, k_i) > \varepsilon_{k_i} - \varepsilon_{k_i}/2 = \varepsilon_{k_i}/2 \ge \delta. \end{align*} Hence $\operatorname{dist}(K, F) \ge \delta > 0$. The compactness of $K$ is essential: without it, the infimum of the distances $\varepsilon_k$ over all $k$ could be zero (as happens with $K = \{1/n : n \in \mathbb{N}\}$ and $F = \{0\}$ in $\mathbb{R}$, where $\operatorname{dist}(K, F) = 0$ even though $K \cap F = \varnothing$). [/example] ### Checking a Property in the Subspace vs. the Ambient Space A recurring task is to verify that a subset $A$, equipped with the subspace topology, has some topological property (compactness, connectedness, etc.). The standard approach depends on the property: **Compactness.** It is often easier to work in $X$: show that $A$ is closed and contained in a compact subset of $X$. This avoids needing to handle covers of $A$ directly. **Connectedness.** It is often easier to work in $A$: show that $A$ cannot be partitioned into two non-empty relatively open sets, or exhibit a continuous surjection from a connected space onto $A$. **Hausdorff, metrizable, second-countable.** These are automatic (hereditary), so there is nothing to check — they follow from the ambient space. **Completeness.** For a metric subspace $A \subset X$, check that $A$ is closed in $X$ (if $X$ is complete), or construct Cauchy sequences in $A$ and verify their limits lie in $A$. ## References 1. Munkres, J. R., *[Topology](/page/Topology)* (2000). 2. Willard, S., *General Topology* (1970). 3. Engelking, R., *General Topology* (1989). 4. Kelley, J. L., *General Topology* (1955). 5. Bourbaki, N., *General Topology, Chapters 1-4* (1995).