Subspace Topology

Statement

Let $X$ be a topological space with topology $\tau_X$, and let $S \subseteq X$. The subspace topology (or induced topology) on $S$ is:

This is the coarsest topology on $S$ making the inclusion $\iota : S \hookrightarrow X$, $\iota(s) = s$, continuous.

Continuity criterion: A function $f : Z \to S$ is continuous (with $S$ carrying the subspace topology) if and only if $\iota \circ f : Z \to X$ is continuous.

Equivalently, $\iota = \text{Subtype.val}$ is continuous, and any $f$ is continuous into $S$ iff it is continuous when viewed as a map into $X$.

Visualization

$\mathbb{Q}$ as a subspace of $\mathbb{R}$:

ℝ:  ────(────────────────────)────▶
         a                   b
         open interval (a,b) in ℝ

ℚ:  ─────●──●──●──●──●──●───────▶
          ↑                    (rational points)
    (a,b) ∩ ℚ is open in ℚ
    (a basic open set of the subspace topology)

Open set in $\mathbb{R}$	Corresponding open in $\mathbb{Q}$	Contains irrationals?
$(-1, 1)$	$(-1,1) \cap \mathbb{Q}$	No
$(0, \sqrt{2})$	$(0, \sqrt{2}) \cap \mathbb{Q}$	No
$\mathbb{R}$	$\mathbb{Q}$	No

Note that $(0, \sqrt{2}) \cap \mathbb{Q}$ is open in $\mathbb{Q}$ even though $\sqrt{2} \notin \mathbb{Q}$: the set $\{q \in \mathbb{Q} : 0 < q < \sqrt{2}\} = \{q \in \mathbb{Q} : q^2 < 2, q > 0\}$ is also equal to $(0, \sqrt{2}) \cap \mathbb{Q}$ and is open in the subspace topology.

Continuity example: The function $f : \mathbb{Q} \to \mathbb{Q}$, $f(q) = q^2$ is continuous in the subspace topology because $g(q) = q^2$ is continuous as a map $\mathbb{R} \to \mathbb{R}$ (a polynomial), and $f = g|_{\mathbb{Q}}$.

Proof Sketch

1. $\tau_S$ is a topology: Empty set: $\emptyset = \emptyset \cap S$. Whole space: $S = X \cap S$. Unions: $\bigcup (U_\alpha \cap S) = (\bigcup U_\alpha) \cap S$. Finite intersections: $(U \cap S) \cap (V \cap S) = (U \cap V) \cap S$.

2. Inclusion is continuous: For $U \in \tau_X$, $\iota^{-1}(U) = U \cap S \in \tau_S$ by definition.

3. Coarsest such topology: Any topology $\tau'$ on $S$ making $\iota$ continuous must contain $\iota^{-1}(U) = U \cap S$ for all $U \in \tau_X$. So $\tau_S \subseteq \tau'$.

4. Universal property: Given $f : Z \to S$, note $\iota \circ f : Z \to X$. If $\iota \circ f$ is continuous and $V \in \tau_S$ with $V = U \cap S$, then $f^{-1}(V) = (\iota \circ f)^{-1}(U)$, which is open. Conversely, if $f$ is continuous, so is $\iota \circ f$ (composition of continuous maps).

Connections

• Heine–Borel TheoremHeine–Borel Theorem$$K \subseteq \mathbb{R}^n \;\text{compact} \iff K \;\text{closed and bounded}$$In ℝⁿ, a subset is compact if and only if it is closed and boundedRead more → — the subspace topology on a closed bounded set $K \subseteq \mathbb{R}^n$ makes $K$ compact; Heine–Borel characterises exactly which subspaces are compact.
• Bolzano–Weierstrass TheoremBolzano–Weierstrass Theorem$$\text{bounded } (x_n) \subseteq \mathbb{R}^n \implies \exists\, \text{convergent subsequence}$$Every bounded sequence in ℝⁿ has a convergent subsequenceRead more → — every bounded sequence in $\mathbb{R}^n$ takes values in a compact subspace (a closed ball), and the Bolzano–Weierstrass theorem is the statement that sequentially compact subspaces are compact.
• Urysohn's LemmaUrysohn's Lemma$$\exists\, f : X \to [0,1],\quad f|_s = 0,\; f|_t = 1$$In a normal space, disjoint closed sets can be separated by a continuous functionRead more → — subspaces of normal spaces need not be normal in general, but Urysohn's lemma constructs continuous functions separating closed sets, a tool used to show closed subspaces of normal spaces are normal.

import Mathlib.Topology.Constructions

/-- The inclusion of a subtype is continuous in the subspace topology. -/
theorem subtype_inclusion_continuous
    {X : Type*} [TopologicalSpace X] (p : X → Prop) :
    Continuous (Subtype.val (p := p)) :=
  continuous_subtype_val