What is "approximate compactness"? What is an example of an approximately compact set?

Question

I read this:

A property of a set $M$ in a metric space $X$ requiring that for any $x\in X$, every minimizing sequence $y_n\in M$ (i.e. a sequence with the property $\rho(x,y_n)\to\rho(x,M)$) has a limit point $y\in M$.

I'm confused with the concept! Can someone give me an example of an approximately compact set? And also an approximately compact set with respect to $A$? Thanks.

Answer 1

It is easy to show that every approximately compact set is closed.

Here's an example of a closed set without that property. Let $X=\Bbb R^\omega$ be the space of maps $f:\Bbb N\to\Bbb R$ such that $f(n)=0$ for almost all $n$, equipped with the norm $||f||=\sum_{n=0}^\infty|f(n)|$. Let $I_n$ be the indicator function of $\{n\}$. Then the sequence $(I_n)_n$ does not have a limit point, thus its image $M$ is a closed (and discrete) subset of $X$. Since $d(0,I_n)=1$ for each $n$, we have $d(0,M)=1$, however, $(I_n)_n$ does not have limit point in $M$, so $M$ is not approximately compact.

Every compact subset of a metric space is approximately compact. For a non-compact approximately compact subset of $X$, take $L=\{f\mid 0\le f(0),\, f(n)=0\forall n>0\}$.

Answer 2

Here’s a simple example of an approximately compact set that isn’t compact.

Let $X=\Bbb N$, with the following metric $d$:

$$d(m,n)=\begin{cases} 0,&\text{if }m=n\\ 1,&\text{if }m,n\in\Bbb Z^+\text{ and }m\ne n\\ 1,&\text{if }m=0\text{ and }n=1\\ 1,&\text{if }m=1\text{ and }n=0\\ 2,&\text{if }m=0\text{ and }n>1\\ 2,&\text{if }m>1\text{ and }n=0\;. \end{cases}$$

It’s not hard to check that $d$ is a metric generating the discrete topology on $X$.

Now let $M=X\setminus\{0\}$; then

$$d(0,M)=\inf\{d(0,n):n\in M\}=d(0,1)=1\;.$$

Suppose that $\langle n_k:k\in\Bbb N\rangle$ is a minimizing sequence for $0$ in $M$; then $\langle d(0,n_k):k\in\Bbb N\rangle\to 1$. But $d(0,m)=2$ for all $m\in M\setminus\{1\}$, so there must be an $\ell\in\Bbb N$ such that $n_k=1$ for all $k\ge\ell$, and therefore $\langle n_k:k\in\Bbb N\rangle\to 1\in M$.

If $m\in M$, then $d(m,M)=0$, and any minimizing sequence for $m$ must be eventually constant at $m$ and therefore converge to $m$.

Thus, $M$ is approximately compact. However, $M$ is an infinite discrete set, so it certainly isn’t compact.