Distances between points and sets

Question

In doing some old exam questions I came across this. The results are all excruciatingly obvious intuitively, but I have no idea how to formalize a proof for any of them.

Let $(M,d)$ be a metric space. For $A\subseteq M, A \ne \emptyset$ and $x\in M$ we define the distance from x to A as $$dist(x,A):=\inf\{d(x,a):a\in A\}$$ Show the following:

a) If $x,y\in M$ and $r:=\frac{d(x,y)}{2}$, then $dist(x,U_r(y))\ge r$

b) Let $A_1, ..., A_i$ be non-empty subsets of M and let $x\in M$ then $$dist(x,\bigcup_{k=0}^i A_k) = \min \{dist(x,A_j): j\le i\}$$

c) If $K\subseteq M$ is compact and $x\in M\setminus K$, then $dist(x,K)>0$

My best attempt at (a) was using that obviously for every $z\in M\setminus U_r(y)$ $$d(y,z)\ge r$$ but then i still need to prove $x\in M\setminus U_r(y)$

I'm stumped on (b) and (c). I would be grateful for any hints on how to solve these kinds of problems in general as well.

Answer 1

Hints: a) uses the triangle inequality, b) uses that the infimum of a finite set is the minimum of that set, and c) proceed by contradiction using an open cover.

For a), you want to use the idea that $d(x,a) + d(a,y) \leq d(x,y)$ is true for every $a \in U_r(a)$ (which I assume is the open ball of radius $r$ centered at $a$).

In particular, manipulate it to the form $d(x,b) \geq d(x,y) - d(b,y)$. \begin{align*} dist(x,U_r(a)) &= \inf \{d(x,b) : b \in U_r(a)\} \\ &\geq \inf \{d(x,y) - d(b,y): b \in U_r(a)\} \\ &= \inf \{2r - d(b,y): b \in U_r(a)\} \\ &= 2r - \sup \{d(b,y): b \in U_r(a)\} \\ &\geq r \text{.} \end{align*} (We can't finish with "${} = r$" because we do not know that the point of the ball that would be closest to $x$ is actually in the metric space. All we know is that the closest point is a distance ${} \leq r$ from $y$.)

For b) use that the infimum of the union of finitely many sets is the minimum of their infima: $\inf(A \cup B) = \min\{\inf A, \inf B\}$. (Warning: need not be true for an infinite collection of sets. Consider $\cup A_k$, $A_k = \{1/k\} \subseteq \mathbb{R}$.)

In detail, \begin{align*} dist(x,\bigcup_{k=0}^i A_k) &= \inf\{d(x,b) : b \in \bigcup_{k=0}^i A_k\} \\ &= \min\{\inf\{d(x,b) : b \in A_k\} : k \leq i\} \\ &= \min\{dist(x, A_k) : k \leq i\} \text{,} \end{align*} as desired.

For c), cover $K$ by open balls of radius $r$ (part a) might help us subsequently) and note that since $K$ is compact, finitely many balls suffice (part b) might help us subsequently). So if $dist(x,K) = 0$, it is always in this cover.

In particular, suppose $dist(x,K) = 0$, let $r > 0$, and let $A = \bigcup_{k=0}^i U_r(x_k)$ be a (finite sub-)cover of $K$ by open balls of radius $r$. By parts a) and b), there is a ball in the union defining $A$ containing $x$. Let $y_r$ be a point in that ball. Repeating, with a decreasing sequence of $r$, $r_i$, we get a sequence $y_{r_i}$ converging to $x$ with each $y_{r_i} \in K$. Since in a metric space compact is equivalent to sequentially compact, and $x$ is a convergent of $y_{r_i}$, $x \in K$. With this contradiction, we conclude $dist(x,K) > 0$.

Answer 2

a) Take an arbitrary $z \in U_r(y)$ and play with the triangle inequality.

b) For bounded sets $A$ and $B$, $\inf A \cup B = \min (\inf A, \inf B)$. Apply induction.

c) For a closed subset $A \subset M$, $d(x, A)=0$ iff $x \in A$.