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- If the affine span of is not all of , then extend the affine span to a supporting hyperplane. Else, is disjoint from , so apply the above theorem. (en)
- Let and be any pair of points, and let . Since is compact, it is contained in some ball centered on ; let the radius of this ball be . Let be the intersection of with a closed ball of radius around . Then is compact and nonempty because it contains . Since the distance function is continuous, there exist points and whose distance is the minimum over all pairs of points in . It remains to show that and in fact have the minimum distance over all pairs of points in . Suppose for contradiction that there exist points and such that . Then in particular, , and by the triangle inequality, . Therefore is contained in , which contradicts the fact that and had minimum distance over . (en)
- We first prove the second case.
WLOG, is compact. By the lemma, there exist points and of minimum distance to each other.
Since and are disjoint, we have . Now, construct two hyperplanes perpendicular to line segment , with across and across . We claim that neither nor enters the space between , and thus the perpendicular hyperplanes to satisfy the requirement of the theorem.
Algebraically, the hyperplanes are defined by the vector , and two constants , such that . Our claim is that and .
Suppose there is some such that , then let be the foot of perpendicular from to the line segment . Since is convex, is inside , and by planar geometry, is closer to than , contradiction. Similar argument applies to .
Now for the first case.
Approach both from the inside by and , such that each is closed and compact, and the unions are the relative interiors .
Now by the second case, for each pair there exists some unit vector and real number , such that .
Since the unit sphere is compact, we can take a convergent subsequence, so that . Let . We claim that , thus separating .
Assume not, then there exists some such that , then since , for large enough , we have , contradiction. (en)
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