Topological Subspace is Topological Space/Proof 1

Theorem
Let $T = \left({A, \vartheta}\right)$ be a topological space.

Let $\varnothing \subset H \subseteq A$ be a non-null subset of $T$.

Let $\vartheta_H = \left\{{U \cap H: U \in \vartheta}\right\}$.

Then $T_H = \left({H, \vartheta_H}\right)$ is a topological space.

Proof

 * $(1): \quad A \in \vartheta$ so $H = A \cap H \in \vartheta_H$. Similarly, $\varnothing \in \vartheta$ so $\varnothing = \varnothing \cap H \in \vartheta_H$.


 * $(2): \quad$ Let $\displaystyle \bigcup_I U_i$ be the union of arbitrarily many sets $U_i \in \vartheta$.

As $\vartheta$ is a topology on $A$, it follows that $\displaystyle \bigcup_I U_i \in \vartheta$.

Then $\displaystyle \left({\bigcup_I U_i}\right) \cap H \in \vartheta_H$.

From Intersection Distributes over Union we have that $\displaystyle \left({\bigcup_I U_i}\right) \cap H = \bigcup_I \left({U_i \cap H}\right)$.

But $U_i \cap H \in \vartheta_H$.

So $\displaystyle \left({\bigcup_I U_i}\right) \cap H$ is the union of arbitrarily many sets of $\vartheta_H$.


 * $(3): \quad$ Let $U = U_1 \cap U_2$ where both $U_1 \in \vartheta$ and $U_2 \in \vartheta$.

Then $U \cap H = U_1 \cap U_2 \cap H = \left({U_1 \cap H}\right) \cap \left({U_2 \cap H}\right) \in \vartheta_H$.

Hence $\vartheta_H$ is a topology on $H$ and $T_H = \left({H, \vartheta_H}\right)$ is a topological space.