📂Topology

Subspace Topology, Relative Topology

Definition ¹

Let’s assume that a topological space $(X,\mathscr{T})$ and a subset $A \subset X$ are given. Then, the following set

$$\mathscr{T}_{A} =\left\{ A\cap U\ :\ U\in \mathscr{T} \right\}$$

is a topology on $A$. In this case, $\mathscr{T}_{A}$ is referred to as the Subspace Topology or Relative Topology. Moreover, the topological space $(A, \mathscr{T}_{A})$ is called the Subspace of $(X,\mathscr{T})$.

Theorem

• [0]: For a topological space $(X, \mathscr{T}$) and a subset $A \subset X$,

$$\mathscr{T}_{A} = \left\{ A\cap U\ :\ U \in \mathscr{T}\right\}$$

becomes a topology on $A$.

Given a topological space $X$ and a subspace $A$, the equivalence condition for a set in subspace $A$ to be open or closed is as follows:

• [a1]: The necessary and sufficient condition for $V\subset A$ to be an open set in $A$ is the existence of an open set $U$ in $X$ that satisfies $V= A\cap U$.
• [b1]: The necessary and sufficient condition for $F\subset A$ to be a closed set in $A$ is the existence of a closed set $E$ in $X$ that satisfies $F=A\cap E$.

Being an open (closed) set in a subspace does not guarantee the same property in the entire space. If a subspace is an open (closed) set with respect to the entire space, then this property is preserved in the entire space. Let’s consider a topological space $(X,\mathscr{T})$, a subspace $(A,\mathscr{T}_{A})$, and a subset $B\subset A\subset X$:

• [a2]: If $B$ is an open set in the subspace $A$ and $A$ is an open set in $X$, then $B$ is an open set in $X$.
• [b2]: If $B$ is a closed set in the subspace $A$ and $A$ is a closed set in $X$, then $B$ is a closed set in $X$.
• [3]: Let’s say $\mathscr{B}$ is a basis of topological space $(X,\mathscr{T})$. Then,

$$\mathscr{B}_{A} =\left\{ A\cap B\ :\ B\in \mathscr{B} \right\}$$

is a basis for the subspace $(A,\mathscr{T}_{A})$.

Explanation

To avoid confusion, let’s clarify several notations. If $(X,\mathscr{T})$ is the entire space,

$$\mathscr{T}_{A}=\left\{A\cap U\ :\ U \in \mathscr{T} \right\}$$

is the topology of the subset $A$. Hence, it forms the subspace $(A,\mathscr{T}_{A})$. $\mathscr{B}$ is a basis of the entire set $X$. $\mathscr{B}_{A}$ is a collection of intersections between each element of the entire set’s basis and $A$. This becomes the basis of the subset $A$, as per the theorem.

$$\mathscr{T}_{\mathscr{B}_{A}}=\left\{U_{A}\subset A\ :\ \forall\ x \in U_{A},\ \exists\ (A\cap B) \in \mathscr{B}_{A}\ \ \text{s.t.}\ x\in (A\cap B) \subset U_{A}\right\}$$

Moreover, the core is that $\mathscr{T}_{\mathscr{B}_{A}}$ is equivalent to $\mathscr{T}_{A}$. The content may feel complex. To summarize, it goes as follows:

1. Intersections between elements of the entire space’s basis and $A$ form the basis of $A$.
2. The topology generated by this basis is $\mathscr{T}_{\mathscr{B}_{A}}$.
3. The topology $\mathscr{T}_{\mathscr{B}_{A}}$ generated in 2 consists of sets that are intersections between open sets of $X$ and $A$, which is $\mathscr{T}_{A}$.

Proof

[0]

• $(T1)$: Since $A \cap \varnothing =\varnothing$ and $A \cap X=A$, both the empty set and the entire set belong to $\mathscr{T}_{A}$.
• $(T2)$: Let’s assume $V_\alpha \in \mathscr{T}_{A}( \alpha \in \Lambda)$. By the definition of $\mathscr{T}_{A}$, for each $V_\alpha$, there exists $U_\alpha$ satisfying $V_\alpha = A \cap U_\alpha$. By the definition of topology, $U=\cup_{\alpha \in \Lambda} U_\alpha \in \mathscr{T}$. Thus,

$$\bigcup_{\alpha \in \Lambda} V_\alpha = \bigcup_{\alpha \in \Lambda} (A \cap U_\alpha ) =A\cap (\cup_{\alpha \in \Lambda} U_\alpha ) =A\cap U \in \mathscr{T}_{A}$$

and hence $\bigcup _{\alpha \in \Lambda} V_\alpha \in \mathscr{T}_{A}$ is true.

• $(T3)$: Let’s assume $V_{1},\ \cdots\ ,V_{n} \in \mathscr{T}_{A}$. Similarly, for each $V_{i}$, there exists $U_{i}$ satisfying $V_{i} =A \cap U_{i}$. And since $U=\cap _{i} U_{i} \in \mathscr{T}$,

$$\bigcap _{i=1}^n V_{i} = \bigcap_{i=1}^n (A\cap U_{i}) = A\cap \left( \bigcap_{i=1}^n U_{i} \right) =A\cap U \in \mathscr{T}_{A}$$

is true. Therefore, $\bigcap_{i=1}^n V_{i} \in \mathscr{T}_{A}$ is true.

By satisfying three conditions of topology, $\mathscr{T}_{A}$ is a topology on $A$.

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[a1]

Refer to the proof in metric spaces. It’s trivial by the definition of $\mathscr{T}_{A}$.

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[b1]

$(\implies)$ If $F$ is a closed set in $A$, then $A-F$ is an open set in $A$. Hence, by [a1], there exists an open set $U$ in $X$ satisfying $A-F=A\cap U$. Since $U$ is an open set, $E=X-U$ is a closed set in $X$. Then,

$$A\cap E=A\cap (X-U)=A-(A\cap U)=A-(A-F)=F$$

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$(\Longleftarrow )$ If $E$ is a closed set in $X$, then $X-E$ is an open set in $X$. Thus, by [a1], $A \cap (X-E)$ is an open set in $A$. Since $F^c=A-(A\cap E)=A\cap(X-E)$, $F ^c$ is an open set in $A$. Therefore, $F$ is a closed set in $A$.

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[a2]

If $B$ is an open set in $A$, by [a1], there exists an open set $U$ in $X$ satisfying $B=A\cap U\ (U\in \mathscr{T})$. By assumption, $A$ is an open set in $X$. Therefore, $B$ is the intersection of open sets in $X$, thus an open set in $X$.

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[b2]

If $B$ is a closed set in

$A$, by [b1], there exists a closed set $E$ in $X$ satisfying $B=A\cap E$. By assumption, $A$ is a closed set in $X$, and $B$ is the intersection of closed sets, thus $B$ is also a closed set in $X$.

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[3]

Part 1. $\mathscr{B}_{A}$ is a basis for $A$.

[b1]: For any $x\in A$, since $A\subset X$, $x\in X$ is true. Since $\mathscr{B}$ is a basis of $X$, by definition, there exists $B$ satisfying $x \in B \in \mathscr{B}$. Hence, there exists $A\cap B \in \mathscr{B}_{A}$ satisfying $x\in (A\cap B ) \in \mathscr{B}_{A}$. [b2]: For any $A\cap B_{1}$, $A\cap B_{2}$, and $x\in \Big( (A\cap B_{1} ) \cap (A \cap B_{2}) \Big)$, $$(A\cap B_{1})\cap (A \cap B_{2})=A\cap B_{1}\cap B_{2}$$ so, $x\in (B_{1}\cap B_{2})$ is true. Since $\mathscr{B}$ is a basis of $X$, by definition, there exists $B_{3}$ satisfying $x\in B_{3} \subset ( B_{1}\cap B_{2})$. Therefore, $$x \in (A\cap B_{3})\subset \Big( A\cap (B_{1}\cap B_{2}) \Big)=(A\cap B_{1}) \cap (A\cap B_{2})$$ As it satisfies the two conditions for being a basis according to conditions for a basis, $\mathscr{B}_{A}$ is the basis for subset $A$.

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Part 2. $\mathscr{T}_{\mathscr{B}_{A}}=\mathscr{T}_{A}$ is true.

$(\subset)$ Since $\mathscr{B}$ is a basis of $(X,\mathscr{T})$, $\mathscr{T}_{\mathscr{B}}=\mathscr{T}$ and, therefore, $\mathscr{B}\subset \mathscr{T_{\mathscr{B}}}=\mathscr{T}$ is true. Hence, for every $B \in \mathscr{B}$, $B\in \mathscr{T}$ is true. By the definition of $\mathscr{T}_{A}$, $A\cap B \in \mathscr{T}_{A}$ is true. Thus, $$\mathscr{B}_{A} \subset \mathscr{T}_{A}$$ $\mathscr{T}_{\mathscr{B}_{A}}$ is the smallest topology containing $\mathscr{B}_{A}$, so $$\mathscr{T}_{\mathscr{B}_{A}} \subset \mathscr{T}_{A}$$

$(\supset )$ Assume $V \in \mathscr{T}_{A}$. By [a1], there exists $U\in \mathscr{T}$ satisfying $V=A\cap U$. For any point $x\in V \subset A$ of $V$, $x \in U$ is true. Since $\mathscr{B}$ is the basis generating $\mathscr{T}$, there exists $B\in \mathscr{B}$ satisfying $x\in B \subset U$. Therefore, $A \cap B \in \mathscr{B}_{A}$ satisfies $$x\in (A\cap B) \subset (A\cap U) =V$$ This meets the condition for $V$ to belong to the topology $\mathscr{T}_{\mathscr{B}_{A}}$ generated by $\mathscr{B}_{A}$, so $V \in \mathscr{T}_{\mathscr{B}_{A}}$ is true. Therefore, $$\mathscr{T}_{\mathscr{B}_{A}} \supset \mathscr{T}_{A}$$

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