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?

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    $\begingroup$ But $\{0\}\times[0,1]$ is homeomorphic to the closed unit disk... $\endgroup$ – Daniel Robert-Nicoud Sep 28 '16 at 19:26
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    $\begingroup$ A compact convex set in $\mathbb{R}^n$ is homeomorphic to the closed unit ball if and only if it has nonempty interior. $\endgroup$ – Daniel Fischer Sep 28 '16 at 19:28
  • $\begingroup$ The generalization you mention is actually stated and proved for Euclidean spaces of any dimension. $\endgroup$ – Lee Mosher Sep 28 '16 at 19:29
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    $\begingroup$ Right. Generally, a nonempty compact convex subset of $\mathbb{R}^n$ is homeomorphic to the closed unit ball in the affine subspace it spans. $\endgroup$ – Daniel Fischer Sep 28 '16 at 19:39
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    $\begingroup$ @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. $\endgroup$ – Kaj Hansen Sep 28 '16 at 19:46

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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