Showing a set is convex in $\mathbb R^n$ I want to show that If $B[X,\epsilon] = \bigcup _{x \in X} B[x,\epsilon]$ is convex for some $\epsilon$ then $X$ is convex.
I could show the reverse implication but not this one
 A: In the plane let $X$ be the unit circle, and let $\varepsilon>1$. Then $B[X,\varepsilon]$ is a closed disk (of radius $1+\varepsilon$), hence convex, but the circle $X$ (interior not included) is not convex. (Note that each $B[x,\varepsilon]$ denotes a closed ball, or disk.) 
One obtains a different answer when $X$ is closed, $n\ge1$ and one says 
"for every $\varepsilon$ " (as shown further below). The condition that $X$ is closed is essential. Indeed, if $X$ is not required to be closed, then let $X$ be the deleted unit disk (removing the origin) then $B[X,\varepsilon]$ is convex for every $\varepsilon>0$, yet $X$ is not convex. 
Assume that $X\subset \Bbb R^n$ is closed and $B[X,\varepsilon]$ is convex for every $\varepsilon>0$. We will show that 
$X$ is convex. Proof. Take any $x\in \mathrm{conv} X$ (the convex hull of $X$). 
For every $\varepsilon>0$ we have that $\mathrm{conv} X\subseteq 
\mathrm{conv}B[X,\varepsilon]=B[X,\varepsilon]$, hence $x\in B[X,\varepsilon]$ 
and there is $x_\varepsilon\in X$ with $x\in B[x_\varepsilon,\varepsilon]$. 
Notice that $\lim\limits_{\varepsilon\to0}x_\varepsilon=x$, hence $x$ belongs to the closure of $X$, but since $X$ is closed we have that $x\in X$. Since $x$ was an arbitrary point in $\mathrm{conv} X$ it follows that $\mathrm{conv} X\subseteq X$, hence $X$ is convex. 
A: False, just check this example in $\mathbb{R}^2$. The rectangle $X$ of height $1u$ is not convex but $B[X,1]$ is convex.

