Neighborhood of a compact subset of $\mathbb{R}^n$

Question

I found this question on a real analysis textbook:

Let $K\subset U\subset\mathbb{R}^n$, where $K$ is compact and $U$ is open. Show that there is an $\varepsilon > 0$ such that $x\in K$, $y\in\mathbb{R}^n$, $|x-y| < \varepsilon$ $\implies [x, y] \subset U$ (Where $[x,y]$ is the line segment between $x$ and $y$, both included)

I don't see how this is not trivial: if $x\in K\subset U$ and $U$ is open, then by definition of openness there is an open ball with radius $\varepsilon$ with this property. But the solution set of the book uses the fact that $\mathbb{R}^n\setminus U$ is closed and that the distance between this set and $K$ is actualy achieved by a pair of points.

Answer 1

I would argue as follows. For each $x \in K$, there exists some $\varepsilon_x > 0$ such that the open ball $B(x, 2\varepsilon_x)$ is contained within $U$. The existence of such a ball follows from the fact that $U$ is open.

The balls $B(x, \varepsilon_x)$ form an open cover of $K$. $K$ is compact, so there exists a finite subcover $B(x_1, \varepsilon_{x_1}), \dots , B(x_n, \varepsilon_{x_n})$ that covers $K$.

Take $\varepsilon = \min(\varepsilon_{x_1}, \dots , \varepsilon_{x_n})$.

If $x' \in K$, then $x'$ is contained within $B(x_i, \varepsilon_{x_i})$ for some $i$.

Now suppose that $y$ is any point in $\mathbb R^n$ such that $| x' - y | < \varepsilon$. If $z$ is any point on the line segment $[x', y]$, then $|x' - z| < \varepsilon$.

It follows that $$|z - x_i| \leq | x' - z| + |x' - x_i| < \varepsilon + \varepsilon_{x_i} < 2\varepsilon_{x_i}.$$ Thus $z \in B(x_i, 2\varepsilon_{x_i})$.

But $B(x_i, 2\varepsilon_{x_i}) \subset U$. So $z \in U$.