Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can some one please list the closed convex sets in $\mathbb{R^2}$ up to homeomorphism. How many of them are compact

share|improve this question
3  
I expect the answer is: Empty set, point, line segment, closed disk, half plane, entire plane. And $4$. (If I felt inclined to try to give a detailed proof, I'd post this as an answer, but I don't.) –  Harald Hanche-Olsen Mar 15 '13 at 8:17
    
There is also the band $\{(x,y) : 0 \leq y \leq 1 \}$. –  Seirios Mar 15 '13 at 10:02
1  
I think the spaces in question are: $\emptyset$, $\{0\}$, $[0,1]$, $[0,+ \infty)$, $\mathbb{R}$, $[0,1]^2$, $\mathbb{R} \times [0,1]$, $\mathbb{R} \times [0,+ \infty)$ and $\mathbb{R}^2$. Notice that the spaces whose the interior is nonempty are a cartesian product of closed convex sets in $\mathbb{R}$ (I don't know whether it is useful). –  Seirios Mar 15 '13 at 10:24
    
I am conviced that for any convex set $C \subset \mathbb{R}^2$ with $\overset{\circ}{C} \neq \emptyset$, $C \simeq \pi_x(C) \times \pi_y(C)$ with $\pi_x : (x,y) \mapsto x$ and $\pi_y : (x,y) \mapsto y$; think about a square growing inside $C$. Does anyone know wether it is true? –  Seirios Mar 15 '13 at 10:45
add comment

1 Answer 1

up vote 1 down vote accepted

The arguments below are inspired by Geometry I, Marcel Berger (chapter 11).

Let $C \subset \mathbb{R}^2$ be a closed convex set.

Case 1: Suppose $\overset{\circ}{C} = \emptyset$. Then $C \subset D$ for some line $D \subset \mathbb{R}^2$. It is not difficult to deduce that $C$ is either homeomorphic to $\emptyset$, $\{0\}$, $[0,1]$, $[0,+ \infty)$ or $\mathbb{R}$.

From now on, suppose $\overset{\circ}{C} \neq \emptyset$.

Case 2: Suppose $C$ bounded. The construction is classical, for example see here. So $C \simeq [0,1]^2$.

Case 3: Suppose $C$ contains a line $D$. By considering parallels of $D$ inside $C$, you find that $C$ is either a band, an half-plane or the whole plane; in particulier, $C$ is homeomorphic to either $\mathbb{R}^2$, $\mathbb{R} \times [0,+ \infty)$ or $\mathbb{R} \times [0,1]$.

Case 4: Suppose $C$ does not contain any line and $C$ is not bounded. Then $C$ contains a circle $S$. For any $y \in S$, let $R(y)$ be the half-line from the center of $S$ through $y$. Because $C$ is not bounded, $M= \{ y \in S : R(y) \subset C \}$ is nonempty, and because $C$ does not contain any line and is convex, $\bigcup\limits_{y \in M} R(y)$ is a semi-cone.

Without loss of generality, suppose $O \in M$. For $x \in C$, let $R'(x)$ be the half-line from $O$ through $x$. Define $\eta(x)=||S(x)||$ where $R'(x) \cap \partial M=\{S(x)\}$ and $\delta(x)=d(O,R'(x) \cap \partial C)$ (if $R'(x) \cap \partial C= \emptyset$, set $\delta(x)=+ \infty$).

Lemma: $\delta : \mathbb{R}^2 \to [0,+ \infty]$ and $\eta : \mathbb{R}^2 \to [0,+ \infty)$ are continuous.

The proof is essentialy geometric, based on the convexity of $C$: suppose by contradiction the existence of a sequence $(y_n)$ converging to some $y \in C$ such that $\delta(y_n)$ (resp. $\eta(y_n)$) does not converge to $\delta(y)$ (resp. $\eta(y)$). More details can be found in Marcel Berger's book.

Now, set $h(x)= \left\{ \begin{array}{cl} x \cdot \eta(x)/\delta(x) & \text{if} \ ||x||<\delta(x)<+ \infty \\ x & \text{if} \ \delta(x)=+ \infty \\ S(x) & \text{if} \ \delta(x)=||x|| \end{array} \right.$.

In fact, $h$ turns out to be a homeomorphism from $C$ to $M$. However, $M \simeq \mathbb{R} \times [0,+ \infty)$.

To conclude:

Theorem: Let $C \subset \mathbb{R}^2$ be a closed convex set. Then $C$ is homeomorphic to one the following spaces: $\emptyset$, $\{0\}$, $[0,1]$, $[0,+ \infty)$, $\mathbb{R}$, $[0,1]^2$, $\mathbb{R} \times [0,1]$, $\mathbb{R} \times [0,+ \infty)$, $\mathbb{R}^2$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.