Topological Subspace is Topological Space/Proof 1
Theorem
Let $\struct {X, \tau}$ be a topological space.
Let $H \subseteq X$ be a non-empty subset of $X$.
Let $\tau_H = \set {U \cap H: U \in \tau}$ be the subspace topology on $H$.
Then the topological subspace $\struct {H, \tau_H}$ is a topological space.
Proof
We verify the open set axioms for $\tau_H$ to be a topology on $H$.
Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets
Let $\AA \subseteq \tau_H$.
It is to be shown that:
- $\ds \bigcup \AA \in \tau_H$
Define:
- $\ds \AA' = \set {V \in \tau: V \cap H \subseteq \bigcup \AA} \subseteq \tau$
Let:
- $\ds U = \bigcup \AA'$
By the definition of a topology, we have $U \in \tau$.
Then, from Intersection Distributes over Union and by Union is Smallest Superset: Family of Sets:
- $\ds U \cap H = \bigcup_{V \mathop \in \AA'} \paren {V \cap H} \subseteq \bigcup \AA$
By the definition of $\tau_H$ and by Set is Subset of Union: General Result, we have:
- $\ds \forall S \in \AA: \exists V \in \tau: S = V \cap H \subseteq \bigcup \AA$
That is:
- $\forall S \in \AA: \exists V \in \AA': S = V \cap H$
By Set is Subset of Union: General Result, we have:
- $\ds \forall V \in \AA': V \subseteq U$
Since intersection preserves subsets, it follows that:
- $\forall S \in \AA: S \subseteq U \cap H$
By Union is Smallest Superset: General Result, we conclude that:
- $\ds \bigcup \AA \subseteq U \cap H$
Hence, by definition of set equality:
- $\ds \bigcup \AA = U \cap H \in \tau_H$
$\Box$
Open Set Axiom $\paren {\text O 2 }$: Pairwise Intersection of Open Sets
Let $A, B \in \tau_H$.
Let $U, V \in \tau$ be such that $A = U \cap H$ and $B = V \cap H$.
By the definition of a topology, we have $U \cap V \in \tau$.
From Set Intersection is Self-Distributive:
- $A \cap B = \paren {U \cap V} \cap H \in \tau_H$
$\Box$
Open Set Axiom $\paren {\text O 3 }$: Underlying Set is Element of Topology
By the definition of a topology, we have $X \in \tau$.
By Intersection with Subset is Subset, it follows that $H = X \cap H \in \tau_H$.
$\Box$
All the open set axioms are fulfilled, and the result follows.
$\blacksquare$