# For a polytope, why is the intersection of faces a face?

First I fix definitions:

A polytope in $$V$$ is a subset that is bounded and is the intersection of half-spaces of $$V$$. A half-space is defined with respect to an affine form, and is that part of $$V$$ where the affine form is nonnegative.

Let $$P$$ be a polytope, $$P = \cap_{n=1}^{m}H_n+$$, for half-spaces $$H_i+$$.

A face is defined as the intersection of a support hyperplane with $$P$$. Where a support hyperplane is one such that the polytope is contained in either the nonnegative part of the affine form that vanishes on it, or the nonpositive part of that affine form.

Let $$F_1 = P\cap H_1$$, and $$F_2 = P\cap H_2$$. Then $$F_1 \cap F_2 = P\cap H_1 \cap H_2$$.

For $$F_1 \cap F_2$$ to be a face, there must be a support hyperplane $$H_3$$ such that $$(H_1 \cap H_2) \subseteq H_3$$. Intuitively it seems like this should be the case, since the intersection of $$H_1 and H_2$$ is a set of lower dimension on the boundary of the polytope, so it seems plausible that one can find a support hyperplane containing it.

How can I see this more clearly/rigorously?