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A generalization of Brower's fixed point theorem says that any continuous map from compact, convex set in the plane $K$ to itself, $f:K \to K$, must have a fixed point.

It is easy to see that any set homeomorphic to the closed unit disk must have a Brower fixed point theorem. However, the above generalization includes a wider array of allowed sets than just those homeomorphic to the closed unit disk. For instance, it includes the set $\{0\} \times [0,1]$.

In some sense, that is not surprising as that just reduces to a lower dimensional case. But this makes me wonder, what additional sets are being added? Is it just lower dimensional cases? In other words, what are the conditions for a compact, convex set to be homeomorphic to the closed unit ball? Purely that it is 2 dimensional?

abnry
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    But ${0}\times[0,1]$ is homeomorphic to the closed unit disk... – Daniel Robert-Nicoud Sep 28 '16 at 19:26
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    A compact convex set in $\mathbb{R}^n$ is homeomorphic to the closed unit ball if and only if it has nonempty interior. – Daniel Fischer Sep 28 '16 at 19:28
  • The generalization you mention is actually stated and proved for Euclidean spaces of any dimension. – Lee Mosher Sep 28 '16 at 19:29
  • I'm confused about the homeomorphism... I can't see how there can be continuous bijective map. There might be a homotopy, I think. Can you clarify? – abnry Sep 28 '16 at 19:35
  • @DanielFischer, that would be equivalent to there being three non collinear points in the set, correct? – abnry Sep 28 '16 at 19:37
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    Right. Generally, a nonempty compact convex subset of $\mathbb{R}^n$ is homeomorphic to the closed unit ball in the affine subspace it spans. – Daniel Fischer Sep 28 '16 at 19:39
  • I don't think the closed unit disk can be homeomorphic to $[0,1]$. There would be a homeomorphism between an open interval and an open ball. But then there would be a homeomorphism between R1 and R2, which is impossible. – abnry Sep 28 '16 at 19:45
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    @abnry, another way to see it is that it is possible to remove a single point from $[0, 1]$ so that it becomes disconnected. On the other hand, this is impossible with the closed unit disk. – Kaj Hansen Sep 28 '16 at 19:46

1 Answers1

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Yes, indeed, every nonempty compact convex subset of $R^n$ is homeomorphic to the closed unit ball of suitable dimension.

See my answer here. In the answer I was using the assumption that the set in question is symmetric and has nonempty interior; in your case it need not be symmetric so you use an asymmetric norm $p(x)$ instead. The point is that every convex subset $C$ with nonempty interior in the $n$-dimensional affine space $A^n$ is absorbing with respect to the origin that belongs to the interior of $C$ and, hence, defines a norm on the corresponding vector space. If you have a nonempty convex subset $C\subset R^n$ with empty interior, then $C$ will have nonempty interior in its affine hull, i.e. the smallest affine subspace of $R^n$ containing $C$, see for instance, Rahul's answer here.

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Moishe Kohan
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