I'm trying to show that the intersection of two faces of a closed set is a face. Our definition of a face is that it is the intersection of the set is is a face of and a hyperplane.
So to prove this I've already shown: Say $F_1, F_2$ faces of a closed set $K$, then there are hyperplanes $$H_1 = \{x\in\mathbb{R}^d | <x,y_1>=\alpha_1\}$$ $$H_2 = \{x\in\mathbb{R}^d | <x,y_2>=\alpha_2\}$$ such that $F_1=H_1\cap K$ and $F_2=H_2\cap K$. So $F_1\cap F_2=H_1\cap K\cap H_2\cap K = K\cap H_1\cap H_2$. So I've proved that $F_1\cap F_2$ is a face if $H_1\cap H_2$ is a hyperplane.
But thats where it gets tricky because $H_1\cap H_2 = \{x\in\mathbb{R}^d | <x,y_1>=\alpha_1 \text{ and }<x,y_2>=\alpha_2 \}$.
So now my question is, can I say this set is equal to the set $\{x\in\mathbb{R}^d | <x,y_1+y_2>=\alpha_1+\alpha_2 \}$? Because then I am all done? (Even though it feels like this is not correct and I would get more elements than only the elements that are in $H_1\cap H_2$.)
How else could i approach this?