2.5Subspace Topology

Updated 27 May 2026Chapter (PDF)

[2.5.1]Definition#

Let $(X, \Tau_X)$ be a topological space, and let $Y \subseteq X$ . Then the collection

\Tau_Y = \{ Y \cap U : U \in \Tau_X \}

is the subspace topology on $Y$ .

Proof.

Note that $\emptyset, X \in \Tau_X$ , and these sets in $\Tau_X$ correspond directly to $\emptyset, Y$ in $\Tau_Y$ . Now suppose $\{Y_\lambda\}_{\lambda \in \Lambda}$ is a collection of sets in $\Tau_Y$ , where each $Y_\lambda = Y \cap U_\lambda$ . Then

\bigcup_{\lambda \in \Lambda} Y_\lambda = \bigcup_{\lambda \in \Lambda} Y \cap U_\lambda = Y \cap \underbrace{\left(\bigcup_{\lambda \in \Lambda} U_\lambda \right)}_{\in \Tau_X},

and so $\Tau_Y$ is closed under arbitrary union. An entirely symmetric argument follows to show $\Tau_Y$ is closed under finite intersection—thus $\Tau_Y$ is a topology.

[2.5.2]Theorem#

Suppose $(X, \Tau_X)$ is a topological space with basis $\mathcal{B}$ . Then, for any subspace $(Y, \Tau_Y)$ , the collection

\mathcal{C} = \{ B \cap Y : B \in \mathcal{B} \}

forms a basis for $\Tau_Y$ .

Proof.

Let $y \in Y$ . Then note some $O \in \Tau_X$ is such that $y \in O \subseteq X$ . But then $O \cap Y$ is a basis element, so $y \in (O \cap Y) \subseteq O \subseteq X$ . Thus $\mathcal{C}$ covers $Y$ and locally refines every element, making it a basis.

Note that if $X$ is a topological space and $Y \subseteq X$ is endowed with the subspace topology, then any open subset in $Y$ is simply the intersection of some open set in $X$ with $Y$ .

[2.5.3]Corollary(Subspace Topology is a Subset of Parent Topology for Open Subspaces)#

Suppose $(Y, \Tau_Y)$ is an open subspace of $(X, \Tau_X)$ . Then $\Tau_Y \subseteq \Tau_X$ . Equivalently, all open sets in $Y$ are open in $X$ .

Proof.

Suppose $O \in \Tau_Y$ . Then $O$ is an arbitrary union of basis elements of $\Tau_Y$ . Let $\mathcal{B}$ be a basis for $\Tau_X$ . Then we may write $O$ as the arbitrary union of basis elements, specifically with the basis induced by $\mathcal{B}$ . Then

O = \bigcup_{\lambda \in \Lambda} Y \cap U_\lambda = \underbrace{Y}_{\in \Tau_X} \cap \underbrace{\left(\bigcup_{\lambda \in \Lambda} U_\lambda \right)}_{\in \Tau_X} \in \Tau_X,

where $\{U_\lambda\}_{\lambda \in \Lambda}$ is a set of open sets in $X$ . Then the arbitrary union of each $U_\lambda$ is open in $X$ , and since $Y$ is open as well, the intersection of this arbitrary union with $Y$ is open as well. Thus $O$ is open in $X$ , meaning $\Tau_Y \subseteq \Tau_X$ .

[2.5.4]Theorem(Product Topology and Subspace Topology Coincide)#

Suppose $X, Y$ are topological spaces with subspaces $U \subseteq X$ and $V \subseteq Y$ , endowed with the subspace topology. Then the subspace topology on $U \times V \subseteq X \times Y$ is exactly the product topology.

Proof.

Let $\Tau_X, \Tau_Y, \Tau_U, \Tau_V$ be topologies endowed on the respective sets. Let $\Tau_S$ be the subspace topology for $U \times V \subseteq X \times Y$ , and let $\Tau_{U \times V}$ be the product topology on $U \times V$ . Now define these collections of sets as bases for the topologies:

\mathcal{B}_{S} = \{ (O_X \times O_Y) \cap (U \times V) : O_X \in \Tau_X, O_Y \in \Tau_Y \},

\mathcal{B}_{U \times V} = \{ (O_U \times O_V) : O_U \in \Tau_U, O_V \in \Tau_V \}.

Let $O_S \in \Tau_S$ . Then note $O_S$ is the arbitrary union of $\{B_{S_\lambda}\}_{\lambda \in \Lambda}$ , and so

O_S = \bigcup_{\lambda \in \Lambda} (O_{X_\lambda} \times O_{Y_\lambda}) \cap (U \times V) = \bigcup_{\lambda \in \Lambda} \underbrace{ \underbrace{(O_{X_\lambda} \cap U)}_{\in \Tau_U} \times \underbrace{(O_{Y_\lambda} \cap V)}_{\in \Tau_V}}_{\in \Tau_{U \times V}} \in \Tau_{U \times V}.

Thus $\Tau_S \subseteq \Tau_{U \times V}$ . Conversely, suppose $O_{U \times V} \in \Tau_{U \times V}$ . Then

O_{U \times V} = \bigcup_{\lambda \in \Lambda} (O_{U_\lambda} \times O_{V_\lambda}) = \left( \bigcup_{\lambda \in \Lambda} (O_{X_\lambda} \times O_{Y_\lambda}) \right) \cap (U \times V) \in \Tau_{S},

and so $\Tau_S \supseteq \Tau_{U \times V}$ , completing the proof.

The product topology is nice in this regard, but note the order topology is not. That is, if $(X, <)$ is an ordered set and $(Y, <)$ is a subspace with the same ordering, the order topology is not necessarily the inherited subspace topology. We provide an example where the subspace and order topology do and don't coincide.

[2.5.5]Example(Subspace Topology vs. Order Topology)#

For our first example, let $Y_1 = [0, 1]$ be a subset of $\bR$ . Consider its subspace and order topologies. The basis for the subspace topology takes the form

\mathcal{B}_S = \{(a, b) \cap Y_1 : a, b \in \bR\} = \begin{cases} (a, b) & a, b \in Y_1, \\ [0, b) & b \in Y_1, \\ (a, 1] & a \in Y_1, \\ Y \text{ or } \emptyset & \text{otherwise.} \end{cases}

Note this coincides exactly with the basis for the order topology (excluding $Y, \emptyset$ , but they are redundant when generating the topology). Thus the subspace and order topologies are equivalent.

Now let $Y_2 = [0, 1) \cup \{2\}$ . The basis for the subspace topology takes the form

\mathcal{B}_S = \{(a, b) \cap ([0, 1) \cup \{2\}): a, b \in \bR\} = \{[(a, b) \cap [0, 1)] \cup [(a, b) \cap \{2\}] : a, b \in \bR \}.

Namely, note that if we choose $(a, b) = (1.5, 2.5)$ , we have that $(a, b) \cap Y = \{2\} \in \mathcal{B}_S$ , and so it is a member of the subspace topology. But note that any element of the order topology that contains $2$ must either be $[0, 1) \cup \{2\}$ or $(a, 1) \cup \{2\}$ , where $0 < a < 1$ . In any case, $\{2\}$ is not in the order topology, meaning the order topology and subspace topology do not agree for $Y_2$ .

[2.5.6]Definition(Convex Set)#

Suppose $(X, <)$ is an ordered set and $(Y, <) \subseteq (X, <)$ . We say $Y$ is convex in $X$ if, for any $a < b \in Y$ , the open interval $(a, b)_X \subseteq Y$ .

For example, $[0, 1]$ is a convex subset of $\bR$ , but $[0, 1) \cup \{2\}$ is not (since $(1, 2)_\bR$ contains elements like $1.5$ ).

[2.5.7]Theorem(Order Topology and Subspace Topology Coincide for Convex Subsets)#

Suppose $(X, <)$ is an topological space endowed with the order topology. Let $(Y, <)$ be a subspace of $X$ , where $Y \subseteq X$ is convex. Then the subspace topology and order topology coincide.

Proof.

Without loss of generality, suppose $Y$ has a maximum and minimum. Then a basis element for the subspace topology on $Y$ takes the form

x \in (\mathcal{B}_S = \{(a, b) \cap Y : a, b \in X\}) = \begin{cases} (a, b) & a, b \in Y, \\ [\min(Y), b) & b \in Y, \\ (a, \max(Y)] & a \in Y, \\ Y \text{ or } \emptyset & \text{otherwise,} \end{cases}

which is exactly the order topology. Of course, note that the second and third forms of basis elements are removed if $Y$ does not have a minimum or maximum, respectively. Nevertheless, this coincides perfectly with the order topology on $Y$ , completing the proof.

2.5.1Exercises#

[2.5.8]Problem#

Suppose $Y$ is a subspace of $X$ and $A \subseteq Y$ . Show that the subspace topology $A$ inherits from $Y$ is the same as the subspace topology $A$ inherits from $X$ .

Proof.

Let $\mathcal{T}_{ST}$ denote the subspace topology endowed on set $S$ as a subspace of $T$ . Let $\Tau_X$ be the topology on $X$ . We show $\mathcal{T}_{AY} \subseteq \mathcal{T}_{AX} \subseteq \mathcal{T}_{AY}$ .

First, suppose $O \in \mathcal{T}_{AY}$ . Then note there exist a family of open sets $\{O_{Y_\lambda}\}_{\lambda \in \Lambda} \subseteq \mathcal{T}_{YX}$ such that

O = \bigcup_{\lambda \in \Lambda} (O_{Y_\lambda} \cap A) = A \cap \left( \bigcup_{\lambda \in \Lambda} (O_{Y_\lambda}) \right) = A \cap \left( \bigcup_{\lambda \in \Lambda} (O_{X_\lambda} \cap Y) \right) = \overset{A}{\cancel{(A \cap Y)}} \cap \left( \bigcup_{\lambda \in \Lambda} O_{X_\lambda} \right) \in \Tau_{AX}.

Conversely, let $O \in \Tau_{AX}$ . Then note there exist a family of open sets $\{O_{Y_\lambda}\}_{\lambda \in \Lambda} \subseteq \mathcal{T}_{X}$ such that

O = \bigcup_{\lambda \in \Lambda} (O_{X_\lambda} \cap A) = A \cap \left( \bigcup_{\lambda \in \Lambda} (O_{X_\lambda}) \right) = (A \cap Y) \cap \left( \bigcup_{\lambda \in \Lambda} (O_{X_\lambda}) \right) = A \cap \left( \bigcup_{\lambda \in \Lambda} (O_{X_\lambda} \cap Y) \right) \in \Tau_{AY},

completing the proof.

[2.5.9]Corollary#

Suppose $f: X \to Y$ is a function of sets, and $\Lambda$ is an index set. Show that

f\left( \bigcup_{\lambda \in \Lambda} A_\lambda \right) = \bigcup_{\lambda \in \Lambda} f(A_\lambda).

Proof.

Suppose $y \in f\left( \bigcup_{\lambda \in \Lambda} A_\lambda \right)$ . Then there is some $\mu \in \Lambda$ such that $x \in A_\mu$ and $y = f(x)$ . Then, of course, $y \in f(A_\mu) \subseteq \bigcup_{\lambda \in \Lambda} f(A_\lambda)$ . Conversely, suppose $y \in \bigcup_{\lambda \in \Lambda} f(A_\lambda)$ . There is some $\mu \in \Lambda$ such that there is some $x \in A_\mu$ such that $y = f(x) \in f(A_\mu) \subseteq f\left( \bigcup_{\lambda \in \Lambda} A_\lambda \right)$ , completing the proof.

[2.5.10]Problem#

A map $f: X \to Y$ is said to be an open map if, for every open set $U \subseteq X$ , the image $f(U)$ is open in $Y$ . Show $\pi_X: X \times Y \to X$ and $\pi_Y: X \times Y \to Y$ are open maps.

Solution.

Let $U$ be an open subset of $X \times Y$ . Then note there exists an indexing set $\Lambda$ such that $U = \bigcup_{\lambda \in \Lambda} (O_{X_\lambda} \times O_{Y_\lambda})$ . Then

\pi_X(U) = \pi_X\left(\bigcup_{\lambda \in \Lambda} (O_{X_\lambda} \times O_{Y_\lambda})\right) = \bigcup_{\lambda \in \Lambda} \pi_X(O_{X_\lambda} \times O_{Y_\lambda}) = \bigcup_{\lambda \in \Lambda} O_{X_\lambda},

which is open in $X$ . An analogous procedure can be applied for $\pi_Y$ .

[2.5.11]Problem#

Show that

\mathcal{B} = \{(a, b) \times (c, d): a < b, c < d \text{ and } a, b, c, d \in \bQ\}

is a basis for the standard topology on $\bR^2$ .

Solution.

Let $O \subseteq \bR^2$ be open, and let $x \in O$ , where $x = (x_1, x_2)$ . Since $O$ is the arbitrary union of open boxes, there is some $a' < b', c' < d' \in \bR$ such that $x \in [(a', b') \times (c', d')] \subseteq O$ . This means $a' < x_1 < b'$ and $c' < x_2 < d'$ . By the density of the rationals, there exist $a, b, c, d \in \bQ$ such that

a' < a < x_1 < b < b',

c' < c < x_2 < d < d'.

Then $x \in [(a, b) \times (c, d)] \subseteq [(a', b') \times (c', d')] \subseteq O$ , meaning $\mathcal{B}$ covers $\bR^2$ and locally refines each open set, completing the proof.