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The entire questions goes like this:

Let $(X, \rho)$ be a metric space and suppose $K$ and $F$ are nonempty disjoint subsets of $X$ with $K$ compact and $F$ closed. (a) Prove there is a $\delta >0$ such that $\rho(x,y) \geq \delta$ for all $x \in K$ and $y \in F$. (b) Show that part (a) may fail if $K$ is closed, but not compact.

Part (a) is answered for the most part at the link below:

Show that exists $\delta >0$ such that:$d(x,y)\geq \delta$

I am having difficulty with part (b) though, as I can't seem to come up with any proof or counterexample. Any tips or hints would be greatly appreciated!

obewanjacobi
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    In addition to the examples and answers already given, let $X=\Bbb R$ with the usual metric $\rho (x,y)=|x-y|.$ Let $K=\Bbb N$ and $ F={n+(n+1)^{-1}: n\in \Bbb N}.$ Then $K,F$ are closed and disjoint but $\inf {\rho (x,y):x\in F,y\in K}=0.$ – DanielWainfleet Dec 17 '17 at 05:14

4 Answers4

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Try to reverse engineer a bit. You want to have $x_n$ and $y_n$ in $K$ and $F$ respectively so that $d(x_n,y_n) \to 0$. If $K$ were compact then $x_n$ would have some convergent subsequence $x_{n_k}$ and this requirement would force $y_{n_k}$ to converge to the same limit, which would necessarily be in both $K$ and $F$. So you need:

  1. $x_n$ is a sequence in a closed set $K$ and has no convergent subsequence
  2. $y_n$ is a sequence in a closed set $F$ and has no convergent subsequence
  3. $K \cap F = \emptyset$
  4. $d(x_n,y_n) \to 0$.

Some specific hints:

  1. You can do this in $\mathbb{R}$, you don't need to go to any weird spaces where Bolzano-Weierstrass fails or anything like that.
  2. It suffices to take $K,F$ to consist of just the sequences $x_n,y_n$ themselves.
  3. In view of hint 1 and 2, it is useful to think about what sequences in $\mathbb{R}$ have no convergent subsequences.
Ian
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  • You can take any strictly increasing sequence $(x_n){n\in \Bbb N}$ in $ \Bbb R$ with no upper bound and let $0<y_n-x_n<\min (1/n, x{n+1}-x_n)$. – DanielWainfleet Dec 17 '17 at 05:21
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Take $$K:=\bigcup_{n\in\Bbb{N}}[2n,2n+1]\text{ and } F:=\bigcup_{n\in\Bbb{N}}[2n+1+1/n,2n+2-1/n]$$.

Nightgap
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Here's an easy visual example in $\mathbb{R}^2$ (with the usual metric): Consider $F$ to be the $x$-axis and $K$ to be the graph of $\frac{1}{x}$ below:

enter image description here

Even though $F$ and $K$ are both closed, can you see how they still get closer and closer as $x$ becomes larger? In other words, for any $\delta > 0$, you can find a pair of points from $K$ and $F$ that are within $\delta$ of one another. Try showing that fact rigorously.

balddraz
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In $\mathbb{R}^2$ with its usual $\rho$, take $K$ as the $x$-axis (that is $\mathbb{R}\times\{ 0 \}$) and $F$ as the graph of the exponential function (that is $F=\{(x,y)\in\mathbb{R}^2 \mid y=e^x\}$). They are clearly non-empty and mutually disjoint. Both $K$ and $F$ are closed, but neither is compact (they are unbounded).

Can you see why their distance is zero so that a strictly positive $\delta$ of the desired kind does not exist?

For a different example, take $X=[-1,1]\setminus\{0\}$ with the metric inherited from $\mathbb{R}$, that is $\rho(x,y)=|y-x|$. $X$ has finite diameter $2$. The sets $K=\left[-1,0\right)$ and $F=\left(0,1\right]$ are closed in $X$. Can you prove that? They also satisfy the other requirements.

Jeppe Stig Nielsen
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  • This example clears up what I'm looking for. Thanks very much. The exponential is a great way to show it in $\mathbb{R}^2$ with ease! – obewanjacobi Dec 17 '17 at 02:39