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Let $X \subset \mathbb{R}^p$ , $X$ is star convex if there exists a point in $X$ such that for any y in $X$, the line segment joining x and y is contained completely in $X$.

A set $X \subset \mathbb{R}^p$ is contractible if it can be reduced to one of its points, say P, by a continuous deformation.

Prove if $ X \subset \mathbb{R}^p $ is star convex then X is contractable.

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I will be able to answer this question in 3-7 days, hopefully. –  Cornelius Johnson Mar 27 '13 at 20:44
    
Hint to what? You have just given two definitions. Are you by any chance alluding to some mysterious relationship between them ? –  Georges Elencwajg Mar 27 '13 at 20:50
    
Yes, read the title, I will restate the question for you –  Cornelius Johnson Mar 27 '13 at 20:51
    
Try defining a homeomorphism that contracts the star domain to its centre $y$. To do this you may want to use the fact that for any $x$, the line segment joining $x$ and $y$ is contained in the star domain. –  Eckhard Mar 27 '13 at 20:53
    
Consider the straight-line homotopy from the identity function on $X$ to the constant function on $X$ whose constant value is $x$. This will be the required deformation. –  J. Loreaux Mar 27 '13 at 20:58
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2 Answers

How can we continuously deform the segment from $x$ to $y$ into the singleton $x$? If we find a nice interpolative way to do so, and apply this uniformly across the whole star domain, then we should get a continuous deformation to the star center $x$.

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Let $X\subset \mathbb{R}^n$ be a star convex set. Let $x\in X$ be the associated "star point." Consider the function $f:X\times[0,1]\to X$ given by

$$f(y,t) = xt + y(1-t). $$

Then for all $y\in X$, $t\in[0,1]$, $f(y,t)$ lies on the line segment joining $y$ to $x$, and therefore $f(y,t)\in X$. Furthermore, $f(y,0)=y$, and $f(y,1)=x$. Thus $f$ is a deformation of $X$ onto the point $x\in X$. (Oh, I forgot to mention, it is trivial that $f$ is continuous).

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