Topological Subspace is Topological Space

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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 1

We verify the open set axioms for $\tau_H$ to be a topology on $H$.


$(\text O 1)$: 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, by the distributivity of intersection 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:

$\ds \bigcup \AA = U \cap H \in \tau_H$

$\Box$


$(\text O 2)$: 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 Intersection Distributes over Intersection:

$A \cap B = \paren {U \cap V} \cap H \in \tau_H$

$\Box$


$(\text O 3)$: Set Itself

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.

$\blacksquare$


Proof 2

Follows directly from Subspace Topology is Initial Topology with respect to Inclusion Mapping.

$\blacksquare$


Also see


Sources