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Given $X \subset \mathbb{R}^n$ and $\varepsilon > 0$, Let $B(X;\varepsilon)$ be the union of $B(x;\varepsilon)$ (balls with center in $x \in X$ and radius $\varepsilon$).

Show that if X is convex, then $B(X;\varepsilon)$ is convex.

I see that if two elements $x,y$ are inside the same ball, then I have the line segment that connects $x$ and $y$ all inside that ball, because all balls inside $\mathbb{R}^n$ are convex.

But I don't know how to proceed if they aren't inside the same ball.

Any tips?

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    $\begingroup$ Don't you mean $X\subset \mathbb R^n$, rather than $\in$? $\endgroup$ – paul garrett Sep 16 at 21:51
  • $\begingroup$ Yes, I'm sorry. $\endgroup$ – Geaquinto Sep 17 at 0:29
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Let $x,y\in B(X,\epsilon)$. By definition there are $x_0,y_0\in X$ such that $||x-x_0||<\epsilon$ and $||y-y_0||<\epsilon$. Now let $t\in (0,1)$. We have to show that the point $(1-t)x+ty$ is in $B(X,\epsilon)$ as well. Now, since $X$ is convex we know that $(1-t)x_0+ty_0\in X$. Also:

$||((1-t)x+ty)-((1-t)x_0+ty_0)||=||(1-t)(x-x_0)+t(y-y_0)||\leq $

$\leq(1-t)||x-x_0||+t||y-y_0||<(1-t)\epsilon+t\epsilon=\epsilon$

So $(1-t)x+ty\in B((1-t)x_0+ty_0,\epsilon)\subseteq B(X,\epsilon)$.

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  • $\begingroup$ Thanks a lot!!! $\endgroup$ – Geaquinto Sep 17 at 0:29
  • $\begingroup$ Please, see my answer. $\endgroup$ – Jean Marie Sep 17 at 6:49
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In fact your $B(X,\varepsilon)$ is the Minkowski addition $X \oplus B(0,\varepsilon)$ of the convex sets $X$ and $B(0,\varepsilon)$ (ball centered in $0$ with radius $\varepsilon$).

In this context, the answer is immediate: there is a classical theorem (given in the Wikipedia reference) saying that the Minkowski addition of 2 convex sets is itself convex.

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