precompact compact differece

Question

the definition of precompact on a metric space (X,d) shall be: $\forall r >0 \exists F\subset X \ \mathrm{ finite, s.t. } \ X= \cup_{x\in F} B_r(x).$

As all open sets in (X,d) are balls, why is that condition not sufficient, and we also need completeness?

Further, can one define precompactness in the above sense also on topological spaces?

Answer 1

To see an example why this is not sufficient, note that $(0,1)$ is precompact with this definition. This does not depend on the topology alone, but also on the metric; the homeomorphic space $\mathbb R$ is not precompact as a metric space. Sometimes a metric space with this property is called "totally bounded."

Answer 2

You can define "total boundedness" (as this property is more commonly known) for uniform spaces (which generalise certain aspects of metric spaces), but not for general spaces. There the term precompact is used for "the closure is compact", whihh can be a useful notion, especially in locally compact spaces. In a complete metric space a set that is precompact (totally bounded) in "your" sense indeed has a compact closure, e.g. in $\Bbb R^n$ this holds. But two sets can be homeomorphic topologically and one precompact and the other not. So it's not a purely topological notion.

And precompact means there is a finite subcover for "boring" open covers ( balls where all radii are the same $r$!) while compactness means all open covers (WLOG we can test compactness by covers by balls, but the radii can vary by ball for compactness). This is why $(0,1)$ is precompact but not compact..