Closed Convex sets of $\mathbb{R^2}$

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

-
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
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

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: $\emptyset$, $\{0\}$, $[0,1]$, $[0,+ \infty)$, $\mathbb{R}$, $[0,1]^2$, $\mathbb{R} \times [0,1]$, $\mathbb{R} \times [0,+ \infty)$, $\mathbb{R}^2$.

-