# Sets homeomorphic to convex sets

I have recently encountered the "Brouwer fixed-point theorem". The theorem states (per Wikipedia) that any convex compact set has the fixed point property. This statement struck me as odd as the theorem requires convexity, which isn't a topological property (i.e preserved by homeomorphisms) while the fixed-point property is.

My question then: Is there a simple or well known topological property which is equivalent to "homeomorphic to a convex set"

In the case of $A \subseteq \mathbb{R}$ it's not hard to see that $A$ is homeomorphic to (and is) a convex subset of $\mathbb{R}$ if and only if it is connected. My intuition tells me that in the case of $\mathbb{R}^2$ this extends to simply connected sets, and in higher dimensions to sets where all homotopy groups are trivial, but I don't know if my intuition is right on this.

EDIT: thinking on the comments and answers so far it seems another necessary condition is that the set also be locally connected. However that is still not sufficient as any 'Y shaped' subset of $\mathbb{R}^2$ still isn't homeomorphic to a convex set. Would being a simply connected manifold, possibly with boundary, be sufficient in $\mathbb{R}^2$?

Please note when answering that I'm mostly familiar with point-set topology and only know very little about algebraic topology.

• Isn't "homeomorphic to a convex set" already a topological property? (Of course, the question still remains whether that topological property can be expressed in terms of a reasonable "internal" specification.) – Daniel Schepler Mar 23 '18 at 17:30
• If you take the topologist's sine curve and attach a path connecting the two path components, you get a simply-connected space (with trivial homotopy groups) that is not homeomorphic to anything convex. – MartianInvader Mar 23 '18 at 17:51
• Brouwer fixed-point theorem for compact convex sets follows from the (usual) Brouwer fixed-point theorem for the closed unit ball as any compact convex set with nonempty interior is homeomorphic to a closed unit ball. – positrón0802 Mar 24 '18 at 3:49

If $A\subset\mathbb{R}^2$ open and simply connected, then by Riemann mapping theorem it is homeomorphic to the unit disc.

Let $A\subset \mathbb{R}^n$ open and homeomorphic to a convex set $K\subset \mathbb{R}^n$. Then $K$ has to be open as well (invariance of domain) and by this argument it has to be homeomorphic to $\mathbb{R}^n$. Hence $A$ is a contractible manifold. However, for $n\ge 3$ there are contractible manifolds (such as the Whitehead manifold - see last link) which are not homeomorphic to $\mathbb{R}^n$.

• If I understand correctly there are no simple known sufficient conditions for the case $n\geq3$ even when the set is open? – Bar Alon Mar 26 '18 at 19:08
• Well, the argument shows that 'homeomorphic to a convex set' is not a homotopy invariant property, and thus cannot be checked by looking at simply connectedness (resp. higher homotopy or singular homology groups). – Jan Bohr Mar 27 '18 at 6:10
• Whether there is another sufficient condition that is useful to you I don't know. However the argument shows that for open sets it is equivalent to look for sufficient conditions to be homeomorphic to euclidean space. – Jan Bohr Mar 27 '18 at 6:12
• I just noticed that you were wondering why convexity appears in the statement of the Brouwer fixed point theorem. There is an excellent discussion of this here. – Jan Bohr Mar 28 '18 at 18:37

If I haven't made a mistake, the following characterization works:

A subset $X$ of $\mathbb{R}^n$ is homeomorphic to a convex set if and only if it is homeomorphic to the union of an open disc with a subset of its boundary.

The "if" direction is pretty clear, and here's a sketch of a proof of the "only if":

Without loss of generality, assume $X$ is bounded (we can map all of $\mathbb{R}^n$ to the unit sphere homeomorphically).

First, if $X$ is contained in a hyperplane of dimension $n-1$, proceed by induction on $n$. Otherwise, the set must contain at least $n+1$ points that are not contained in a hyperplane of dimension $n-1$, so by convexity we can see there's an open ball in $X$ (contained in the convex hull of these points).

Choose a point $p$ in this ball. We can think of $X$ as a union of intervals emanating from $p$, such that their only point of intersection is $p$. We can identify this set of intervals with $S^{n-1}$, call this mapping $s$. Essentially here we are making $p$ the "center" point, and drawing all the line segments from it to the other points in the set $X$, and a point in $S^{n-1}$ corresponds to a "direction" of a segment coming out of $p$.

Now consider the map $d:S^{n-1}\rightarrow \mathbb{R}$, which takes each "direction" to the length of the segment from $p$ emanating in that direction. This map is convex and therefore continuous. You can thus apply a mapping to $X$ by shrinking each point's distance to $p$ proportionally to the full length of the interval it occupies, and this map is a homeomorphism.

Under this transformation, each interval is transformed into either a closed or half-open interval emanating from $p$ of length 1. Since $p$ was an interior point of $X$ such intervals emanate in all directions, so the set is exactly an open ball together with a subset of its boundary.

• Firstly, I find this to be an important first step, so thank you for the response. This still begs the question, however (at least in my mind) is there a simple characterization of these subsets? – Bar Alon Mar 26 '18 at 18:47
• I've given you the entire class of topological spaces that can be convex in $\mathbb{R}^n$. I don't think you'll get any simpler than "the set is basically a disc". Trying to further classify them is basically asking to classify subsets of spheres. – MartianInvader Mar 26 '18 at 22:39