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: $$A \in \vartheta$$ so $$H = A \cap H \in \vartheta_H$$. Similarly, $$\varnothing \in \vartheta$$ so $$\varnothing = \varnothing \cap H \in \vartheta_H$$.


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

As $$\vartheta$$ is a topology on $$A$$, $$\bigcup_I U_i \in \vartheta$$.

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

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

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

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


 * 3: 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.