2

Is every subset of a precompact set also precompact?

It seems the answer is yes if I am using the definition: $S$ is precompact iff for every $\epsilon>0$ it can be covered by a finite number of balls of radius $\epsilon$.

This is due to the fact that smaller subset of a precompact set can certainly be covered by a finite number of balls.

Is there any trick here?

Silvia Ghinassi
  • 4,707
yoyostein
  • 19,608

1 Answers1

2

As mentioned in the comments, a set is precompact if its closure is compact. As closed subsets of compact sets are compact, it follows that every subset of a precompact set is also precompact.

The definition you use is the one of total boundedness. The result you claim is also true for totally bounded sets, but the two concepts are equivalent only in complete metric spaces. In general, total boundedness is weaker (see here).

Community
  • 1
Silvia Ghinassi
  • 4,707
  • I'm sorry to comment on an answer after it has been almost 2 years, since it was posted, but I don't understand the proof.

    Being precompact (= totally bounded) is an intrinsic property. Now if $X$ is precompact and $A$ is a subset, then your claim seems to be that $\bar{A}$, the closure of $A$ is compact, as it is a subset of a compact space. But which is the latter compact space? X need not to be and it makes no sense to speak about $\bar{X}$ (without using further results at least)

     – kesa Feb 09 '18 at 14:48
  • @kesa Let $X$ be a complete metric space and $A \subset X$ precompact. Since $A$ is precompact we know that $\overline{A}$ is compact. Let $B \subset A$. Since $B \subset A$ we have $\overline{B} \subset \overline{A}$. Since $\overline{B}$ is a closed subset of a compact set, it is compact, making $B$ precompact. – ViktorStein Jul 24 '19 at 22:16