Bounded and compact sets using Heine-Borel, distance to a set The distance between a point $x \in \mathbb{R}^n $ and a set $E \subset \mathbb{R}^n$ is defined as $d(x,E)=\inf\{\ d(x,e) : e\in E \} $, where $d(x,e)=\|x-e\|$. 
If $E \subset \mathbb{R}^n$ is bounded, I need to prove that $E_a = \{\ x \in \mathbb{R}^n : d(x,E) \leq a  \}$ is compact.
I'm approaching this problem using the Heine-Borel Theorem, i.e., I'm trying to prove that $E_a$ is bounded and closed. I've already proven that $E_a$ is closed, but I'm having some trouble with the bounded part, I'm trying to find $x_0 \in \mathbb{R}^n $ and $r>0$ such that $d(y,x_0) < r $ for al $y \in E_a$.
 A: You've stated that $E_a$ is closed, but I'll prove it here anyways, for the sake of completeness. If we have a sequence $\{x_n\} \subseteq E_a$ converging to a point $x$, then we can find a corresponding sequence $\{y_n\} \subseteq E_a$ for which $\|x_n - y_n\| \le a + \epsilon$; that is, $y_n$ is "close enough" to $x_n$. There is an $N$ such that
$$\|x_N - x\| < \epsilon$$
Given this, we have that
$$\|x - y_N\| \le a + 2\epsilon$$
from the triangle inequality, implying that $d(x, E) \le a + 2\epsilon$; since $\epsilon$ was arbitrary, we can conclude that $d(x, E) \le a$, so that $x \in E_a$.
Next, some idea of why this set is bounded: The set $E_a$ contains only points close to $E$, and every point in $E$ is close to $0$ (where the degree of closeness depends on just how bounded $E$ is). Draw a picture for this part: If $M$ is a bound for the set $E$, then $E$ lies in a ball of radius $M$ around the origin; since we know nothing about the structure of $E$, we don't really lose much by working with the ball instead. Adding a distance of $a$ only makes the ball a bit bigger, now having radius $M + a$, which is still finite.
To formalize this, notice that if $x \in E_a$, we can choose $y \in E$ so that $\|x - y\| \le a + 1$. Then again using the triangle inequality,
$$\|x\| = \|x - y + y\| \le \|x - y\| + \|y\| \le a + 1 + M$$
for all $x$. In fact, a slight modification to this argument means that you can actually conclude $\|x\| \le a + M$.
