5

In Berkeley Problems in Mathematics (collection of their previous qualifying exam problems), problem 4.1.11 asks us to prove:

"Let $X$ and $Y$ are nonempty subsets of a metric space M. Define $$d(X, Y)=\inf\{d(x,y) | x\in X, y\in Y\}.$$

  1. Suppose $X$ contains only one point $x$ and $Y$ is closed. Prove $$d(X, Y)=d(x,y)$$ for some $y\in Y$.

  2. Suppose $X$ is compact and $Y$ is closed. Prove $$d(X, Y)=d(x,y)$$ for some $x\in X$ and $y\in Y$."

However, at least problem 2 seems to be false according to this old thread: If $A$ is compact and $B$ is closed, show $d(A,B)$ is achieved, where the counterexample using the space $M=\{0\}\cup (1, 2)$ was used.

Despite this, the Berkeley book provides a proof of the two above statements and I'm not sure of where these proofs fail. Could someone help me spot the error in this proof that fails, especially for a situation like the counterexample given in the old thread?

The Berkeley proof:

"1. Let $X = {x}$ and $(y_n)$ be a sequence in $Y$ such that $|x - y_n| < d(X, Y) + 1/n$. As $(y_n)$ is bounded, passing to a subsequence, we may assume that it converges, to $y$, say. As $Y$ is closed, $y \in Y$ and, by the continuity of the norm, $|x - y| = d(X, Y).$

  1. Let $(x_n)$ be a sequence in $X$ such that $d((x_n), Y) < d(X, Y) + 1/n.$ As $X$ is compact, by the Bolzano-Weierstrass Theorem [Rud87, p. 40], [MH93, p. 153], we may assume, passing to a subsequence, that $(x_n)$ converges, to $x$, say. We then have $d(X, Y) = d({x}, Y)$ and the result follows from Part 1."
Bundle_Time
  • 53
  • I'm curious; perhaps at the beginning of the chapter, is it stated that metric spaces are assumed finite-dimensional? It's definitely a possibility that this error was missed, but there might be a subtle statement somewhere in the book that all spaces are considered finite dimensional unless otherwise stated (I find it unlikely, to be honest, though). – Clayton Mar 22 '19 at 20:14
  • 1
    I see now. The chapter where I pulled this from is called "Topology of $\mathbb{R}^n$. It does not explicitly say that we are assuming anything about the spaces given in the problems, but I guess we are supposed to take them to be subsets of $\mathbb{R}^n$. – Bundle_Time Mar 23 '19 at 03:10
  • I’d say that is perfectly reasonable, given the title of the chapter. In that case, the problem is correct :) – Clayton Mar 23 '19 at 10:49

1 Answers1

3

The issue seems to be that they are assuming the Bolzano-Weierstrass property in the proof of $(1)$ to conclude the bounded sequence $(y_n)$ converges; for instance, if you look at the example with $M=\{0\}\cup(1,2)$, the the sequence $(y_n)$ could be given by $y_n=1+1/(n+1)$, but you see that this won't actually converge to an element of $M$ even though the sequence is bounded.

Therefore it seems the issue can be remedied by forcing the assumption that $M$ has the Bolzano-Weierstrass property, but for metric spaces $M$ we have

$$\text{$M$ has the Bolzano-Weierstrass property}\iff\text{$M$ is sequentially compact}\iff\text{$M$ is compact},$$

so really we should just be assuming from the start that $M$ is compact (note $M$ is not compact in the example $M=\{0\}\cup(1,2)$).

TY Mathers
  • 19,533
  • Ah, nice catch! My guess is they wanted this to be in $\mathbb{R}^n$ so closed+bounded gives compactness. Thanks! – Bundle_Time Mar 22 '19 at 20:02
  • @Art No problem. – TY Mathers Mar 22 '19 at 20:03
  • @AlexMathers: In a finite-dimensional setting, does there exist a complete, connected counterexample? (The motivation is that we only need $A$ closed and $B$ compact in $\Bbb R^n$, I believe.) – Clayton Mar 22 '19 at 20:04
  • @Clayton No idea to be honest, I'd be interested in an example as well – TY Mathers Mar 22 '19 at 20:09
  • @AlexMathers: Here is a question that more-or-less confirms that it needs to be finite-dimensional. It isn't exactly the same, but instead does it for a point; it should be straightforward to generalize to a compact set from a point. – Clayton Mar 22 '19 at 20:11