# Topological Subspace is Topological Space

## Theorem

Let $\left({X, \tau}\right)$ be a topological space.

Let $H \subseteq X$ be a non-empty subset of $X$.

Let $\tau_H = \left\{{U \cap H: U \in \tau}\right\}$ be the subspace topology on $H$.

Then the topological subspace $\left({H, \tau_H}\right)$ is a topological space.

## Proof 1

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

### $({O1})$: Union of Open Sets

Let $\mathcal A \subseteq \tau_H$.

It is to be shown that:

$\displaystyle \bigcup \mathcal A \in \tau_H$

Define:

$\displaystyle \mathcal A' = \left\{{V \in \tau: V \cap H \subseteq \bigcup \mathcal A}\right\} \subseteq \tau$

Let:

$\displaystyle U = \bigcup \mathcal A'$

By the definition of a topology, we have $U \in \tau$.

$\displaystyle U \cap H = \bigcup_{V \mathop \in \mathcal A'} \left({V \cap H}\right) \subseteq \bigcup \mathcal A$

By the definition of $\tau_H$ and by Set is Subset of Union: General Result, we have:

$\displaystyle \forall S \in \mathcal A: \exists V \in \tau: S = V \cap H \subseteq \bigcup \mathcal A$

That is:

$\forall S \in \mathcal A: \exists V \in \mathcal A': S = V \cap H$

By Set is Subset of Union: General Result, we have:

$\displaystyle \forall V \in \mathcal A': V \subseteq U$

Since intersection preserves subsets, it follows that:

$\forall S \in \mathcal A: S \subseteq U \cap H$

By Union is Smallest Superset: General Result, we conclude that:

$\displaystyle \bigcup \mathcal A \subseteq U \cap H$

Hence, by definition of set equality:

$\displaystyle \bigcup \mathcal A = U \cap H \in \tau_H$

$\Box$

### $({O2})$: 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 $U \cap V \in \tau$.

$A \cap B = \left({U \cap V}\right) \cap H \in \tau_H$

$\Box$

### $({O3})$: 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.

$\blacksquare$