# Polytope as the intersection of closed half-spaces

I am stuck at a problem which looks very simple but which I still cannot prove. Let $P$ be a $d$-polytope. Say that $P$ can be represented as the intersection of a given finite set of closed halfspaces $\{H_i\}_{i \in I}$, i.e. $P = \bigcap\limits_{i \in I} H_i$. Let $F$ be a facet of $P$, ie. $F$ is a $(d-1)$-face of $P$. Conjecture: Then, there is a $k \in I$ such that $F$ is contained in the hyperplane given by the half-space, i.e. $F \subset \partial H_k$.

Is this true? How can I prove this?

Let me know if any clarification is needed. Thanks in advance!

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For infinite $I$ this is clearly wrong: Consider the square $C = \{(x,y)\,:\, 0 \leq x,y\leq 1\}$ and write it as $C = \{x \geq 0\} \cap \{x \leq 1\} \cap \{y \leq 1\} \cap \bigcap_{n=1}^{\infty} \{y \geq -\frac{1}{n}\}$. The face $\{(x,0)\,:\,0\leq x \leq 1\}$ isn't contained in the boundary of any half-space. You can easily generalize this to higher dimensions. –  t.b. Nov 2 '11 at 9:32
Oh, indeed. I've edited my question according to your remark. Thank you! –  Tom Jonathan Nov 2 '11 at 9:37