# Proof that convex open sets in $\mathbb{R}^n$ are homeomorphic?

This is an exercise from Kelley's book. Could someone help to show me a proof?

It seems very natural, and it is easy to prove by utilizing the arctan function in $\mathbb{R}^1$. Thanks a lot.

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Maybe you could try to prove that any open, convex set in $\mathbb{R}^n$ is homeomorphic to an open ball. – Siminore Jul 2 '12 at 10:06
I know this is not an answer, and not an answer of mine. Anyway, I think the answer is contained in this discussion. It's funny, because it seems that many book authors consider this theorem popular but hard to prove. And they do not prove it! – Siminore Jul 2 '12 at 10:57
I am confused by the nlab page. The argument for starshaped open sets doesn't seem to be that straightforward, because of jumps in the distance function. See my answer below and the discussion. (I get around this problem because I use more than starshapedness.) Moreover, the nlab page talks about compact open sets, and the only example that comes to my mind is the empty set. They might mean open sets with compact closure. But there is still something off. Also unbounded starshaped open sets can be homeomorphic to open balls. – Stefan Geschke Jul 4 '12 at 14:03
@Stefan: are we looking at the same page? I assume you are talking about the sentence: "we can prove that the closure of any open star-shaped region is homeomorphic to the n-disk iff it is compact" which is certainly true. The $n$-disk in the quote is the closed disk (see the definition on the page). – Willie Wong Jul 4 '12 at 14:58
But you are right of course. Convexity is better than star-shaped and you do use it nicely in your post. – Willie Wong Jul 4 '12 at 15:01

3rd edit: I have now typed things up in a slightly more streamlined way:

How about this: Call a set $U\subseteq\mathbb R^n$ star-shaped if there is a point $x\in U$ such that for all lines $g$ through $x$, $g\cap U$ is a connected open line segment. Every convex set is star-shaped. Translating $U$ if necessary we may assume that $x=0$. Now use arctan to map $U$ homeomorphically to a bounded set. This set is still star-shaped around $x=0$.

Now scale each ray from $0$ to the boundary of the open set appropriately, obtaining a homeomorphism from the open set onto the open unit ball.

Edit: I am very sorry that I didn't reply to the comments earlier. It seems that my answer was too sketchy. I think we agree that we can assume that the origin is an element of the original open set $U$. As Henning Makholm points out, we use arctan applied to the radius in Polar coordinates to map our potentially unbounded set $U$ to a bounded open set $V$.

Now we show that in any direction the distance of the boundary of $V$ from the origin depends continuously on that direction.
Let $v$ be a direction, i.e., let $v$ be a vector in $\mathbb R^n$ of length $1$.

First case: the ray from the origin in the direction of $v$ is contained in $U$. In this case $U$ is unbounded in the direction of $v$. Since $U$ is open, there is $\varepsilon>0$ such that the $\varepsilon$-ball around the origin is still contained in $U$. Since $U$ is convex, the convex hull of the union of the ray in the direction of $v$ and the open $\varepsilon$-ball is also contained in $U$. Let us call this convex open set $C$.

For every $N>0$ the set of elements of $C$ that have distance at least $N$ from the origin is open. It follows that the set of directions $v'$ such that the ray in direction $v'$ has elements of $C$ of distance at least $N$ from the origin is open.

But this implies that the map assigning to each direction $v$ the distance of the boundary of the transformed set $V$ to the origin in that direction is actually continuous in all the directions in which the original set $U$ is unbounded.

Second case: the ray from the origin in the direction $v$ is not contained in $U$, i.e., $U$ is bounded in the direction of $v$. We have to show that the distance of the boundary of the transformed set $V$ in some direction $v'$ is continuous in $v$.
But since arctan is continuous, it is enough to show the same statement for the distance to the boundary of the original set $U$.

So, let $d$ be the distance of the boundary of $U$ from the origin in the direction of $v$. Let $\varepsilon>0$. Choose $x\in U$ in the direction $v$ such that $d-|x|<\varepsilon/2$. For some $\delta>0$, the $\delta$-ball around $x$ is contained in $U$. The set of directions $v'$ such that the ray in that direction passed through the open ball of radius $\delta$ around $x$ is an open set of directions containing $v$. It follows that on an open set of directions containing $v$ the distance of the boundary of $U$ is at least $d-\varepsilon$.

Now let $x$ be a point on the ray in the direction of $v$ with $|x|>d$. If we can show that $x$ is not in the closure of $U$, then there is an open ball around $x$ that is disjoint from $U$ and we see that there an open set of directions that contains $v$ such that for all directions $v'$ in that set the distance of the boundary in direction $v'$ from the origin is at most $|x|$. This shows that the map assigning the distance of the boundary of $U$ to the direction is continuous in $v$ and we are done.

So it remains to show that $x$ is not in the closure of $U$. We assume it is. Let $y$ be the point on the ray in direction $v$ that satisfies $|y|=d$. $y$ is on the boundary of $U$. Then $|y|<|x|$. Let $B$ be an open ball around the origin contained in $U$. We may assume that $x\not in B$. Now I am waving my hands a bit, but I think this should be clear: There is some $\varepsilon>0$ such that when $|x-z|<\varepsilon$,
then $y$ is in the convex hull $C_z$ of the union of $B$ and $\{z\}$. Now choose $z\in U$ such that $|x-z|<\varepsilon$. Now $y\in C_z$ and by the convexity of $U$, $C_z\subseteq U$. But $C_z\setminus\{z\}$ is an open neighborhood of $y$. This shows that $y$ is not on the boundary of $U$, a contradiction.

This finishes the proof of "continuity of scaling". I hope this was comprehensible.

2nd edit: It doesn't seem clear why the map $d$ assigning to each direction $v$ the distance of the boundary of $V$ from the origin in direction $v$ is well defined. For $v\in\mathbb R^n$ of length $1$ (i.e., $v$ a direction) let $$d(v)=\inf\{|x|:x\in\mathbb R^n\setminus V\wedge x/|x|=v\}.$$ Since $V$ is bounded, this is well defined. I take that as the definition of "the distance from the origin of the boundary of $V$ in the direction of $v$". In the "so it remains to show"-paragraph above it is shown that no point $x$ in the direction of $v$ with $d(v)<|x|$ is in the closure of $V$.
(Strictly speaking, I prove this for $U$ instead of $V$, and only in the case that the ray in direction $v$ is not contained in $U$. But if the ray is not contained in $U$, the arctan transformation preserves the property that we have shown for $U$. If the ray is contained in $U$, then $d(v)$ is $\pi/2$ and the point $\pi v/2$ is the unique point on the boundary of $V$ in the direction of $v$. The problem with $U$ is that it is unbounded and hence we might be taking the infimum over the empty set. In this case the distance would be $\infty$.) It follows that the unique point $y$ in the direction of $v$ with $|y|=d(v)$ is in the boundary of $V$. Hence every ray through the origin intersects the boundary of $V$ in exactly one point.

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Why is this scaling continuous? – akkkk Jul 2 '12 at 10:40
Stefan - I am no expert here, but... did you mean "use arctan to map the boundary of U homeomorphically to a bounded set"? For example, surely if we use arctan to map the unit disc to [0,2pi] then we don't get a bijection. Also, the arctan from the unit circle to [0,2pi] is not a homeo because of the discontinuity at either 0 or 2pi. – Adam Rubinson Jul 2 '12 at 10:52
This approach may be useful. Roughly speaking, you fix a point (an "origin") of $U$, and then you deform $U$ "radially" around the point. It is rather clear that you can do this continuously, but it might be tricky to write down formulae. – Siminore Jul 2 '12 at 10:52
@Adam: The idea is to use polar coordinates and do $(r,\theta)\mapsto(\tan^{-1}(r),\theta)$ to map the entire plane homeomorphically to an open ball. Note that the arctan applies to the radial coordinate -- and it's not important that it is arctan in particular, just that it is strictly increasing from 0 towards a horizontal asymptote. $r\mapsto 1-\frac{1}{r+1}$ would work just as fine. – Henning Makholm Jul 2 '12 at 11:00
Well, I don't think proofs in topology are made by moving hands in the air, even though I heard similar remarks before. In particular algebraic topology was invented to prove some things rigorously that are intuitively rather clear. I admit that my first answer was too short. I hope my extended answer is satisfying. Some points might need more clarification, but my answer is now already rather long as it is. – Stefan Geschke Jul 4 '12 at 10:47

Here is a fairly complete proof for the existence of a diffeomorphism in the star-shaped case. This was gleaned from old, incomplete answers (not on math stackexchange) to this question. Getting a diffeomorphism is only slightly harder than getting a homeomorphism.

The idea is to push points outwards radially relative to the star point so that the boundary reaches infinity and all problems with non-smoothness of the original boundary disappear.

First, if the region is unbounded, map it diffeomorphically to a bounded region by pushing points in radially relative to the star point using a nice dilation function. Almost any $C^\infty$ diffeomorhism from $[0,\infty)$ to $[0,1)$ will do. Make it equal to the identity map near $0$ to avoid complications with differentiability at the star point. After this, we may assume that the region is bounded.

One idea that doesn't quite work in any obvious way is to use $r(x)$ = radius in direction $x$ (where $x$ is in the unit circle) to push radially outwards using a suitable modification of $r$, since $r$ may be discontinuous.

To avoid discontinuities, start instead with $d(x)$ = distance from $x$ to the complement of the region, where $x$ is now in the region. It is an easy exercise that $d$ is continuous. $d$ doesn't have the mapping properties that we need. Adjust it by averaging the inverse of it along rays from the star point: $$e(x) = \int_0^1 1/d(s + t*(x-s))dt,$$ where s is the star point. The final mapping (if we only want a homeomorphism) is: $$m(x) = e(x)*(x-s).$$

The main points to show for $m$ being a homeomorphism are: $m$ is one to one because $\|m\|$ is strictly increasing on each ray (because $1/d > 0$); the image of $m$ is the whole of $\mathbb R^n$ because $e(x)$ approaches $\infty$ as $x$ approaches the boundary along each ray ($1/d$ certainly does, and a simple estimate shows that integration preserves this); the inverse is continuous because the map is proper (the dilation factor is bounded from below).

Minor modifications turn $m$ into a $C^\infty$ diffeomorphism. The above gives a continuous $d$ which is strictly positive and approaches $0$ sufficiently rapidly at the boundary. An easy partition of unity argument gives a strictly positive $C^\infty$ $d$ that is smaller than the continuous $d$ so it has the same boundary property. $e$ and $m$ are then obviously also $C^\infty$. The inverse of $m$ is $C^\infty$ for much the same reason that $m$ is one to one: the partial derivative of $\|m\|$ along each ray is invertible because it is $1/d > 0$. The full derivative at $s$ is $Dm(s) = e(s)1_n$, where $1_n$ is the identity map on (the tangent space of) $\mathbb R^n$. The full derivative at points other than $s$ is best calculated in polar coordinates $(r,\theta$) where $r = \|x-s\|$ and $\theta$ is in the unit sphere: $$m(r,\theta) = (r*e(r,\theta), \theta);$$ $$Dm(r,\theta) = \left( \begin {array} {ccc} 1/d(r,\theta) & 0 \\ * & 1_{n-1} \end {array} \right).$$ These are obviously invertible everywhere, so the inverse of $m$ is $C^\infty$.

By the Riemann mapping theorem, the diffeomorphism can be chosen to be analytic in dimension $2$. I don't know if this is true in any dimension. There is no obvious way to adapt the above to give an analytic $m$ even in dimension $2$.

The integration trick is from an answer by Robin Chapman in mathforum in 2005. I couldn't find any better answer on the internet. But this answer says to use any smooth strictly positive function that vanishes on the complement for $d$. This has the following flaw: $1/d$ might not approach $\infty$ at the boundary rapidly enough for its radial integrals to also approach $\infty$. For example, let the region be the unit disk and $d(x) = (1-\|x\|^2)^\frac 1 2$.

This question is an exercise in many books. I first saw it in Spivak's Differential Geometry Volume 1 (exercise 25 p.322). But later, Spivak gives a proof for a much easier case (Lemma 11.17 p.592 is for when $r$ is $C^\infty$; Lemma 11.18 p.593 shows that $r$ is always lower-semicontinuous (Spivak says upper-smicontinuous, but this is backwards); the preamble to Lemma 11.19 p.593 says that proving the general case is "quite a feat", so Lemma 11.19 only proves an isomorphism of (De Rham) cohomology). But the integration trick seems to make the feat small.

It is exercise 8 p.20 in Hirsch's Differential Topology. This exercise is starred, but has no hints.

Other answers give further references to textbooks. Most books don't prove it. Some say that it is hard and others give it as an exercise. It is exercise 7 p.86 in Brocker and Janich's Introduction to Differential Topology. This exercise gives a hint about using the flow of a radial vector field. It becomes clear that the integration trick is a specialized version of this general method.

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There is a pitfall which seems easy to forget, so I think it is worth adding some clarity. This text is not an answer, but a problem with an intuitive proof.

Below is an image of a bounded set in $\mathbb{R}^2$ (of course, the set is simply the inside of my curve), which I must admit is not convex, where scaling from the origin does not result in a ball (or at least not homeomorphically). In this case, the set is star-shaped, but not from any point. I drew it so that the origin is a center for the star-shaping, but only just. You can see that the "distance function" is either not a function or not continuous.

If you want to prove this question by scaling the boundary, please resolve my issue. I would try this:

Assume that the distance function, when defined as the supremum of all distances to the boundary in some direction (making it a function), has an incontinuity. Draw an $\varepsilon$-ball around the origin, prove that there is a point in that ball which is not a center for star-shapedness of the set, contradicting convexity.

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Note that I don't claim that my proof works in the starshaped case. It works for sets that are obtained from convex open sets by the "arctan-transformation". – Stefan Geschke Jul 4 '12 at 14:04
Oh I do believe the theorem holds, but if your "scaling" proof also holds for star-shaped sets in general, then your proof must have shortcomings. – akkkk Jul 4 '12 at 14:06
It doesn't, as your example shows. See the linked pdf-file. This is a bit clearer than my answer above, even though I am still convinced that my answer above is correct. – Stefan Geschke Jul 4 '12 at 18:26