# Precompact set of metric space has fundamental sequence.

So, let's define "precompact set".

Precompact set $A$ in a metric space is set, if its closure is compact in $X$. From using a test compactness, the set of points of a complete metric space is precompact if and only if it's totally bounded. The set $A$ of points of a metric space $\left (X,\rho \right )$ is called totally bounded if for any $\varepsilon>0$ in M there exists a finite $\varepsilon$-net.

Okey, but I don't understand the fact, which I must prove.

M $\subset$ $\left (X,\rho \right )$ is precompact $\Leftrightarrow$ $\forall (x_{n}) \in M$ $\Rightarrow$ $\exists$ $(x_{n_{k}})$ - fundamental

What should I use for proof this statement? Can I use tests compactness for compact set? But I don't see difference between compact and precompact set of metric space..

• It does not make sense to ask whether a space $(X, \rho)$ is precompact. The definition is about subsets of a metric space, so you want some "precompact inside $X$". – Crostul Oct 12 '16 at 23:03
• Sorry, I make mistake. I have corrected. – Dimatywe1 Oct 12 '16 at 23:06
• Do you know that in a metric space $X$ a compact set $M$ is the same as sequentially compact (every sequence in $M$ has a subsequence convergent to a point of $M$)? With this characterization, the proof is straightforward. – Crostul Oct 13 '16 at 7:50