# Union of balls is convex

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?

• Don't you mean $X\subset \mathbb R^n$, rather than $\in$? – paul garrett Sep 16 at 21:51
• Yes, I'm sorry. – Geaquinto Sep 17 at 0:29

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

• Thanks a lot!!! – Geaquinto Sep 17 at 0:29
• Please, see my answer. – Jean Marie Sep 17 at 6:49

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.