Retract from compact convex set without a point into its boundary.

Question

$X \subset \mathbb{R}^n$ is a compact and convex set with non-empty interior

Let $x_0$ be an interior point from X, I should prove the boundary $\partial X$ is a deformation retract from $X \setminus \{x_0\}$

Looks pretty intuitive as it easy to see it for the unitary ball without its center and $\mathbb{S}^1$, but I can`t solve it at this more abstract context.

Answer 1

It is enough to show that X is homeomorphic to a closed ball $\overline{B(x,r)}$ in $\mathbb{R}^n$. Since $X$ is compact and convex, it is a closed, bounded and path-connected (since the straight line connecting any two points lies completely in $X$) region with non-empty interior. So let $x_0\in \text{int}X$ be such that there exists any open ball $B(x_0,2r)$ inside $X$ for some $r>0$. So we have the closed ball $C=\overline{B(x_0,r)}$ inside $X$. Now for any $y\in \partial X$, let us take $\lambda_y(t)=ty+(1-t)x_0; t\in [0,1]$ to be the straight line joining $x_0$ and $y$. The map $f:\lambda_y\to C$ defined by $f(t)={t\lambda_y(t_0)+(1-t)x_0}$ (where $t_0\in (0,1]$ satisfies the equation $||\lambda_y(t)-x_0||=r$) scales the line $\lambda_y$ to a smaller line segment of length $r$ such that $y$ goes to the point lying on $\lambda_y$ at a distance $r$ from $x_0$ and $x_0$ goes to $x_0$. Since $X$ is convex and bounded every point in $X$ lies on such a unique line segment connecting $x_0$ and some point on the boundary. So we can define a map from $X$ to $C$ such that each line segment connecting $x_0$ to some boundary point gets scaled to a radial line of the closed ball $C$. This map can be shown to be continuous (easy task but takes some effort) and also the map is bijective. Since $X$ is compact and $C$ is Hausdorff we get that $X$ and $C$ are homeomorphic by such a map that maps $\partial X$ homomorphically onto $\partial C$. Since $\partial C$ is the deformation retract of $C- \{x_0\}$, we get that $\partial X$ is the deformation retract of $X-\{x_0\}$.

Answer 2

Without loss of generality we can assume that $x_0=0$. Let $B(0,r)\subset X$ ($B(x,r)$ is an open ball centered in $x$ with radius $r$).

For any $y\in\mathbb R^n\neq\{0\}$ define $\Lambda(y)=\{\lambda\in[0,\infty) | \lambda y\in X\}$.

• From convexity we see that if $\lambda\in\Lambda(y)$ then $\lambda'\in\Lambda(y)$ for $0\leq \lambda'\leq\lambda$.
• From closedness of $X$ we get that $\Lambda(y)$ is closed (continuous preimage of $X$). From boundedness of $X$ we get that $\Lambda(y)$ is bounded: $$\forall (x\in X): \|x\|\leq A\implies \forall (\lambda\in\Lambda(y)):|\lambda|\leq \frac A{\|y\|}.$$
• From the condition $B(0,r)\subset X$ we get that $[0,r/\|y\|]\subset \Lambda(y)$.

Therefore this uniquely defines a function $\lambda\colon \mathbb R^n\setminus\{0\}\to (0,\infty)$ such that $\Lambda(y) = [0,\lambda(y)]$.

Since $$\lambda(y)y\in X\text{ and }\forall (y>\lambda(y)):\lambda y\notin X$$ we get that $\lambda(y)y\in\partial X$.

For any point $x\in X$ we define the set (a 'cone') $$C(x) = \bigcup_{0\leq \lambda<1}(1-\lambda)B(0,r)+\lambda x.$$ Of course it's an open subset of $X$. Particularly, $$\forall(y\in\mathbb R^n)\ \forall(\lambda<\lambda(y)): \lambda\cdot y\in\mathrm{int}(X).$$ Thus, $\lambda(x)=1$ for $x\in\partial X$ and $$x\neq 0\implies x\in X \iff \lambda(x)\geq 1.$$

Assume it's proven that $\lambda$ is a continuous function. Define the function by the formula $$f(x)=\lambda(x)\cdot x,\ x\in X\setminus\{0\}.$$ It's a continuous function $f\colon X\setminus\{0\}\to\partial X$, $f(x)=x$ for $x\in\partial X$ and $$H(x,t) = ((1-t)+t\lambda(x))\cdot x$$ is a continuous homotopy $H\colon X\setminus\{0\}\times[0,1]\to X\setminus\{0\}$.

For the completness of the proof, let's show the continuity of $\mathbb R^n\setminus\{0\} \ni y\mapsto \lambda(y)$.

Look at the picture. The orange disk is $r\cdot D = D(0,r)$.

Let $x,y\in \mathbb R^n\setminus\{0\}$. Take any $t\in(0,1)$. Then $$z+t\cdot r D = t\cdot D(0,r) + (1-t)\lambda(x)x \subset X$$ (green disc).

Now assume $\|y-x\|\leq \frac{tr}{(1-t)\lambda(x)}$. Then $y\in x+\frac{tr}{(1-t)\lambda(x)}D $ (purple disc). Then $$(1-t)\lambda(x)y \in z+tr D\subset X\implies \lambda(y)\geq (1-t)\lambda(x).$$

This shows that $x\mapsto \lambda(x)$ is lower semicontinuous. To prove the upper semicontinuity we have to restrict our considerations to the open sets $\mathbb R^n\setminus c\cdot D$. If $\|x\|,\|y\|> c$ then $\lambda(x),\lambda(y)\leq M:=A/c$ and therefore $$\|y-x\|\leq \frac{tr}{(1-t)M} \implies \lambda(x)\geq (1-t)\lambda(y).$$ Since $\lambda$ is usc on every such set, it's usc on the sum, so on $\mathbb R^n\setminus\{0\}$.