a sequence of compact converging to another compact

Question

Let $K$ be a compact and $A_{n}$ the set of points having a distance equal to $\dfrac{1}{n}$ to $K$. Suppose the interior of $K$ is empty. Can we say that for each $x$ in $K$ there is a sequence $(y_{n})$ in $A_{n}$ that converges to $x$ as $n$ approaches infinity?

Answer 1

Yes, that is true. I'm assuming that $K \subseteq \mathbb R^k$ for some $k$ (from the real-analysis tag) and use single bars($|$) to denote the distance. I'll note that assuming some general topology might break down the problem completely, as for example all but at most one of the $A_n$ are empty for a discrete topology.

1) For each point $y \notin K$, the distance to $K$ is not $0$ (otherwise, $y$ was a limit point of $K$ and $K$ thus not closed and not compact). Since the distance is a continuous function, $f_y(x) = |y-x|$ actually attains a minimum value over compact $K$. So there is (at least) one point $D(y) \in K$ such that $|y - D(y)| \le |y-x|, \forall x \in K$.

2) For each point $z$ on the line segment $\overline{yD(y)}$, $D(y)$ is also the nearest point to $K$: If there was a point $w \in K$ with $|w-z| < |D(y) - z|$, then we'd have $$|y-w| \le |y-z| + |z-w| < |y-z| + |D(y) - z| = |y - D(y)|$$ contradicting the construction of $D(y)$ above.

3) Fix a point $x \in K$ and choose any $\epsilon > 0$. Consider the open ball around $x$ with radius $\frac{\epsilon}{4}$. In it is at least one point $y \notin K$, as otherwise $x$ was an interior point of $K$. Construct a point $D(y)$ as explained in step 1). Obviously $|y - D(y)| < \frac{\epsilon}{4}$, because $|y - x| < \frac{\epsilon}{4}$. Hence $|x-D(y)| < \frac{\epsilon}{2}$.

4) Set $N=\max(\lfloor\frac4{\epsilon}\rfloor+1, \frac1{|y - D(y)|})$, so $\frac1n < \frac{\epsilon}{4}$ and $\frac1n \le |y - D(y)|$ for all $n \ge N$.

5) For all $n \ge N$ let $z_n$ be the point on the line segement $\overline{yD(y)}$, with distance $\frac1n$ from $D(y)$. This works because of 4). Because of 2) we know that the distance of $z_n$ to $K$ is the distance of $z_n$ to $D(y)$, so we get $z_n \in A_n$ for $n \ge N$. Also because of 4) we know that $|x-z_n| \le |x-D(y)| + |D(y) - z_n| < \frac{\epsilon}{2} + \frac{\epsilon}{4} = \frac{3\epsilon}{4}$.

Let's recap: For any $x \in K$ and any $\epsilon > 0$ we have found an integer $N$ such that we have points in $A_n$ that are nearer than $\frac{3\epsilon}{4}$ to $x$ for all $n \ge N$.

Let's now define a sequence of points $(y_n)$ with $y_n \in A_n$ that converges to our chosen $x$. Let $d_n = \inf \{|a-x|: a \in A_n\}$. By construction of $A_n$, we have $d_n \ge \frac1n$. Chose $y_n \in A_n$ such that $|y_n - x| \le \frac43 d_n$.

Let's look at the recap above: For any given $\epsilon > 0$, take any $n \ge N$ and we have points $z_n \in A_n$ with $|x-z_n| < \frac{3\epsilon}{4}$. By the definition of $y_n$, we now know that $|x-y_n| \le \frac43 |x-z_n| <\epsilon$. In other words $\lim_{n \to \infty}y_n = x$.