Topological Subspace is Topological Space/Proof 1

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

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

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

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

By Intersection with Subset is Subset, it follows that $H = X \cap H \in \tau_H$.

All the open set axioms are fulfilled, and the result follows.

Also see

 * Definition:Topological Subspace