Intersection of Subgroups is Subgroup

Theorem
The intersection of two subgroups of a group is itself a subgroup of that group:

$$\forall H_1, H_2 \le \left({G, \circ}\right): H_1 \cap H_2 \le G$$

It also follows that $$H_1 \cap H_2 \le H_1$$ and $$H_1 \cap H_2 \le H_2$$.

Generalized Result
The intersection of any number of subgroups of a group is itself a subgroup of that group:

$$\forall H_k \le \left({G, \circ}\right): H = \bigcap_k {H_k} \le G$$

Proof
Let $$H = H_1 \cap H_2$$ where $$H_1, H_2 \le \left({G, \circ}\right)$$. Then:

As $$H \subseteq H_1$$ and $$H \subseteq H_2$$, the other results follow directly.

Generalized Proof
Let $$a, b \in H_k$$.

$$a, b \in H \Longrightarrow \forall k: a, b \in H_k \Longrightarrow \forall k: a \circ b^{-1} \in H_k \Longrightarrow a \circ b^{-1} \in H$$