Proof $U_\epsilon (x) \cap U_\epsilon(y) \neq 0 \Leftrightarrow \Vert x-y\Vert < 2\epsilon$

Question

Let $x,y \in \mathbb{R}^n$ and $\epsilon > 0$.

How can one prove that

$$U_\epsilon (x) \cap U_\epsilon(y) \neq 0 \Leftrightarrow \Vert x-y\Vert < 2\epsilon$$

I tried to search for a proof online via google and wasn't able to find a proof via Approach0 either.

Can someone show us how to prove this?

Answer 1

Suppose $z \in U_\epsilon (x) \cap U_\epsilon(y)$ exists. Then

$\|z-x\| < \epsilon$ and $\|z-y\|< \epsilon$ by definition of the balls.

So $$\|x-y\|=\|(x-z) + (z-y)\| \le \|z-x\| + \|z-y\| < \epsilon + \epsilon = 2 \epsilon$$

For the reverse suppose $\|x-y\| < 2 \epsilon$. Define $z=\frac12(x+y)$, the midway point between $x$ and $y$. Then $\|z-x\| = \|z-y\|=\frac12\|x-y\| < \varepsilon$ and so $z \in U_\epsilon (x) \cap U_\epsilon(y)$.

Answer 2

Hint: If $z\in U_\varepsilon(x)\cap U_\varepsilon(y)$, then$$\lVert x-y\rVert\leqslant\lVert x-z\rVert+\lVert z-y\rVert<2\varepsilon.$$