### On Bare Points

S. J. Bernau
1969 Journal of the Australian Mathematical Society
This note shows that the set of bare points of a compact convex subset of a normed linear space is, in general, a proper subset of its set of exposed points. We begin with some notation and standard definitions. All linear spaces are assumed real. Let M be a subset of a real normed linear space E; cl M and conv M denote the closure and convex hull of M respectively. If M is convex an extreme point of M is a point of M which is not interior to any line segment contained in M; and an exposed
more » ... and an exposed point of M is a point p of M such that for some closed supporting hyperplane H, H n M = {/>}. We write ext M and exp M for the sets of extreme and exposed points of M. If U is the unit ball of E and 5 the unit sphere then U is rotund if S = ext U and smooth if at each point of S there is only one supporting hyperplane of U [3, VII § 2]. We will say that the norm in E is rotund or smooth if the unit ball has the corresponding property. In  G. H. Orland defined a bare point of a compact convex plane set A to be a point p of A such that there exists a closed disc containing A and having p on its circumference. Orland proves that A = cl conv B where B is the set of bare points of A. In  S. K. Berberian points out that Orland's definition and proof extend at once to inner product spaces and give there a proof of the Krein-Mil'man Theorem which does not require the use of Zorn's Lemma. (It is clear that bare points are extreme points in this context.) The idea of a bare point of a bounded convex set is readily defined in any normed space. It is a point of the set on the boundary sphere of some closed ball containing the set. The notion is essentially a metric one. Changing to an equivalent norm will almost certainly change sets of bare points. In particular, if the unit ball is not rotund there will be bare points of the unit ball which are not extreme points. In [4, Theorem 2.1] Klee showed that, if A is a compact convex set in a normed linear space, then A = cl conv exp A. Klee's proof of this result consists first, of replacing the original norm by an equivalent one 25 use, available at https://www.cambridge.org/core/terms. https://doi.