Originally, my problem was to find a set that is limited but not completely bounded in a non-complete metric space. Then I wanted to think $ (Fin, d _ { infty}) $, from where $ Fin $ is the set of all finite consequences, and $ d _ { infty} $ is the highest metric. Actually, I am looking for a subgroup of $ Fin $ that is limited and not completely limited.

Case 1: If $ (c_0, d _ { infty}) $ is compact, where $ c_0 $ is the set of all null sequences, that is, since the completion of $ Fin $ is this room, $ Fin $ must be completely limited. Therefore, each subset of it is also totally limited. Therefore, this is a bad example of finding an amount that is limited and not completely limited. Then I became curious. What would happen in the general case?

Given sentence $ A $ If the completion in a non-complete space is compact, there is no subset $ V $ from $ A $ so that $ V $ is not completely limited. If the completion is not compact, I might find such a subset. Therefore, I would like to know which of the following complete sequence spaces are compact:

1)$ (c_0, d _ { infty}) $,

2)$ (c, d _ { infty}) $, from where $ c $ is the set of convergent sequences.

3)$ (l _ { infty}, d _ { infty}) $, from where $ l _ { infty} $ is the set of all limited consequences.

4)$ (l_p, d_p) $, from where $ l_p $ is the set of all "p-summable" sequences.

5)$ ( omega, d_ {Fr}) $, from where $ omega $ is the set of all episodes and $ d_ {Fr} $ is the Frechet metric.

In addition, is there a problem with my reasoning? If so, can you explain that to me?

Thanks for any help.