Topology: Subspaces

First, suppose (X, d) is a metric space. If Y is a subset of X, then one can restrict the metric to $d' = d|_{Y\times Y}:Y\times Y\to\mathbf{R}$ , i.e. for any $y,y'\in Y$ , we set d’(y, y’) := d(y, y’). It’s not hard to show that the resulting function is a metric on Y. The resulting pair (Y, d’) is called a metric subspace of (X, d).

The open subsets of Y are related to those of X as follows.

Proposition. A subset $V\subseteq Y$ is open if and only if $V = U\cap Y$ for some open $U\subseteq X$ .

Proof.

First note that if $a\in Y$ , then an open ball in Y is of the form:

$N_{d'}(a,\epsilon) = \{y\in Y: d'(y,a)<\epsilon\} = \{x\in Y:d(y,a)<\epsilon\} \cap Y = N_d(a,\epsilon) \cap Y.$

Suppose U is open in X and V = U ∩ Y. Each $a\in V$ is also in U, and so $N_d(a,\epsilon) \subseteq U$ for some ε>0. Hence $N_{d'}(a,\epsilon) = N_d(a,\epsilon)\cap Y \subseteq U\cap Y = V$ . So V is open in Y.
Conversely, let V be open in Y. For each a in V, we have $N_{d'}(a,\epsilon)\subseteq V$ for some ε>0 depending on a. Take the union U of all such $N_d(a,\epsilon)$ ; then U is open in X since it’s a union of open balls in X. Also

$U\cap Y =\cup_{a\in V}(N_d(a,\epsilon)\cap Y) = \cup_{a\in V} N_{d'}(a, \epsilon)=V.$ ♦

Exercise

If U is an open ball in X, is U ∩ Y necessarily an open ball in Y?

[ Answer (highlight to read): no, consider X=R and Y=R-{0}. Then (-1, +1) is an open ball N(0, 1) in X but (-1, +1) ∩ Y is the disjoint union of (-1, 0) and (0, +1) which is not an open ball in Y. ]

This inspires us to extend the definition of subspaces to general topological spaces.

Definition. Let (X, T) be a topological space and Y be a subset of X. The subspace topology on Y is defined by:

$T_Y =\{U\cap Y : U\in T\},$

i.e. $V\subseteq Y$ is open if and only if it’s of the form V = U ∩ Y for some open subset U of X.

Equivalently, the class of closed subsets of Y is given by D = Y – (U∩Y) = (X–U) ∩ Y = C ∩ Y for some closed subset C of X.

Examples

A subspace of (X, d) with the discrete metric is still discrete.
Pick the half-open interval $Y=[0, 1)\subset \mathbf{R}$ . Then $[0, 1/2) = (-1, 1/2)\cap Y$ is open in Y but not open in R.
Consider the subset $Y=(-\infty, -1]\cup\{0\}\cup[1, \infty)$ of the real line R. The singleton set {0} is an open subset of Y since {0} = N(0, 1/2). Furthermore, it is closed since the complement is a union of two open subsets.
Consider the subset Z of R under the usual metric. Then the resulting subspace is the discrete space even though the induced metric d(m, n) = |m–n| is not exactly the discrete metric.
Let $X=\mathbf{R}^2$ and Y be the set of points (x, y) satisfying $x^2 + y^2 = 1$ . Geometrically, Y is a circle. Here, we’ll think of it as a topological space with the subspace topology inherited from X. The space is denoted $S^1$ . More generally, for each positive integer n, the space $S^n$ is the subspace of $\mathbf{R}^{n+1}$ comprising of all points $(x_1, x_2, \ldots, x_{n+1})$ satisfying $\sum_{i=1}^{n+1} x_i^2 = 1.$
Consider X = N under the right order topology.
1. If Y = {1, 2, 3}, then the subspace topology gives { (empty set), {3}, {2, 3}, Y }.
2. If Y is the set of even numbers, then the bijection $X\to Y, n\mapsto 2n$ preserves the structure of topological spaces. We say that the two spaces are homeomorphic. We will say more about this in a later article.

Since we’re often dealing with multiple spaces and subspaces, when describing open/closed subsets, it’s essential to qualify with “(XX) is open/closed in (YY)” instead of merely saying “(XX) is open”. E.g. in example 2 above, Y = [0, 1) is a subspace of R and the subset [0, 1/2) is open in Y but not in R.

The following diagram illustrates some open subsets of Y in example 3.

Exercises

Consider the subspace Y = {1/n : n positive integer} of R, under the usual metric. Is this space discrete? What about $Y^* := Y\cup\{0\}$ ?
Example 2 gives us the concept of clopen subsets, i.e. subsets of a topological space which are both open and closed. In any topological space X, $\emptyset$ and X are always clopen. Are there any other clopen subsets of Q, where Q inherits the subspace topology from R?

Answers

Y is discrete since each {1/n} = (1/n – ε, 1/n + ε) ∩ Y for some small ε>0. On the other hand Y* is not discrete since any open subset containing 0 must also contain some 1/n.
Yes, Q has infinitely (in fact, uncountably) many clopen subsets. If r is irrational, then (-∞, r) ∩ Q and (r, ∞) ∩ Q form a disjoint union of Q by open subsets.

Basic Properties of Subspaces

The following basic property is often taken for granted.

Theorem. If (X, T) is a topological space and $Z\subseteq Y\subseteq X$ are subsets, then we can form the subspace topology on Z in two ways:

by taking the subspace topology $T\mapsto T_Z$ from $Z\subseteq X$ ; and

by taking successive subspaces $T\mapsto T_Y$ which is a topology on Y, then $T_Y\mapsto (T_Y)_Z$ .

The two topologies are identical.

The proof is straightforward: in the first case, the class of all open subsets of Z is given by U ∩ Z for open subsets U of X; in the second case, the class is given by (U ∩ Y) ∩ Z = U ∩ Z for open subsets U of X. They’re identical. ♦

The following properties are also surprisingly useful in practice.

Theorem. Let Y be a subspace of (X, T). If Y is open in X, then any open subset of Y is an open subset of X. If Y is closed in X, then any closed subset of Y is a closed subset of X.

Proof.

For the first statement, an open subset of Y is of the form V = U ∩ Y for some open subset U of X. Since U and Y are both open in X, so is V = U ∩ Y. The same proof holds for the second statement by replacing ‘open’ with ‘closed’. ♦

Furthermore, for bases and subbases, we have:

Theorem. Let $(X, T)$ be a topological space with subspace $(Y, T_Y)$ .

If B is a basis for T, then $B_Y :=\{U\cap Y : U\in B\}$ is a basis for Y.

If S is a subbasis for T, then $S_Y:=\{U\cap Y: U\in S\}$ is a subbasis for Y.

Proof.

For the first statement, we first verify that $B_Y$ is indeed a basis of some topology over Y:

$\cup_{V\in B_Y} V = \cup_{U\in B} (U\cap Y) = \left(\cup_{U\in B} U\right)\cap Y = X\cap Y = Y.$
Any two elements of $B_Y$ are of the form $U_1\cap Y, U_2\cap Y$ for some basic open subsets $U_1, U_2\in B$ . Since B is a basis, $U_1 \cap U_2 = \cup_i V_i$ for some $V_i \in B$ . This gives $(U_1 \cap Y)\cap (U_2\cap Y) = (U_1\cap U_2)\cap Y = \left(\cup_i V_i\right)\cap Y=\cup_i (V_i\cap Y)$ which is a union of elements in $B_Y$ .

Finally, we need to show $B_Y$ generates the topology $T_Y$ .

Every element $U\cap Y \in B_Y$ (for some element U in B) is open in Y by definition. So $B_Y\subseteq T_Y$ .
Conversely, any element of $T_Y$ is of the form $U\cap Y$ for some open subset U of X. Since B is a basis for (X, T), $U=\cup_i V_i$ for some $V_i\in B$ . So $U\cap Y = \cup_i (V_i\cap Y)$ which is a union of elements in $B_Y$ .

This concludes the proof for the first statement.

For the second, suppose the intersections of finitely many elements of S form basis B. It suffices to show that the intersections of finitely many elements of $S_Y$ form basis $B_Y$ .

In one direction, note that $S\subseteq B\implies S_Y \subseteq B_Y$ .
Conversely, any element of $B_Y$ is of the form U ∩ Y for some $U\in B$ . So $U=V_1\cap V_2 \cap \ldots \cap V_n$ for finitely many $V_i\in S$ . But this gives $U\cap Y = \cap_i (V_i\cap Y)$ for finitely many $V_i\cap Y\in S_Y$ .

And the proof is complete. ♦

Summary.

Any subset of a topological space X inherits a topology from it. The inheritance is consistent across inclusion chains of topological spaces.

With spaces and subspaces, one should be more careful when talking about “open sets”, i.e. mention what it’s open in.

If Y is open (resp. closed) in X and Z is open (resp. closed) in Y, then Z is open (resp. closed) in X.

If Y is a subspace of X, then a basis (resp. subbasis) of X restrict to give a basis (resp. subbasis) of Y.