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Definitions :

$(E,d)$ a metric space is finished-compact if any covering of $E$ by open, we can extract a finite subcover

$(E,d)$ is $\aleph_0$-compact if for any infinite covering of $E$ by open, we can extract a countable subcover

Remark :

we can imagine what's $\aleph_i$-compactness.

we known the space $(E,d)$ with $\aleph_0$-compactness is exactly the space $(E,d)$ separable.

Question :

What is we known about the $\aleph_i$-compactness for $i\in \mathbb N^*$ ?

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    $\begingroup$ Why do you call $\mathbb{N}$-compact what everybody calls "compact". Especially meaning distinct things for $\mathbb{N}$-compact and $\aleph_0$-compact is confusing. Actually, defining $\alpha$-compact as the condition that from every covering there exists a subcover of cardinal $<\alpha$ (rather than $\le\alpha$) gives a more general definition, for which $\aleph_0$-compact just means compact. $\endgroup$
    – YCor
    Commented Oct 16, 2019 at 10:25
  • $\begingroup$ Last and not least, the question is way too vague. $\endgroup$
    – YCor
    Commented Oct 16, 2019 at 10:26
  • $\begingroup$ Yes $\mathbb N$-compact is a classical compact, with a $\alpha$-compact we can extract a subcover of cardinal $\leq \alpha$, so $\aleph_0$-compact is the seperable space for (E,d) metric fr.wikipedia.org/wiki/Espace_séparable $\endgroup$
    – Dattier
    Commented Oct 16, 2019 at 10:44
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    $\begingroup$ If I understand your question correctly, these concepts are well-known in the set-theoretic topology community, usually phrased with terminology like, e.g., "Initially $\omega$-compact" (= countably compact) or "finally $\omega_1$-compact" (=Lindelof). The Handbook of Set-Theoretic Topology is a good source, even if it is 35 years old! $\endgroup$
    – Todd Eisworth
    Commented Oct 16, 2019 at 11:33
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    $\begingroup$ Anyway, the $\le$-definition misses such a notion as "every cover has a subcover of cardinal $<\aleph_\lambda$" for each limit ordinal $\lambda$. $\endgroup$
    – YCor
    Commented Oct 16, 2019 at 13:39

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I think you are asking about cardinal functions on metric spaces. First, the property you call $\aleph_0$-compact is more usually known as Lindelöf. As you say, it is well-known that for a metric space $E$, it is Lindelöf iff it is separable (iff it is second-countable iff it is c.c.c.). Of course, for topological spaces in general, no implication holds in either direction.

For metric spaces, this equivalence holds for the generalization of these properties to cardinals. A reference for this is the Handbook of Set-Theoretic Topology, edited by Kunen and Vaughan.

The density (sometimes called density character to distinguish it from other meanings of density) $d(X)$ of a topological space $X$ is the minimal cardinality of a dense subset (this exists by well-orderedness of cardinals). So "$X$ is separable" is the same as $d(X) = \aleph_0$. The Lindelöf number $L(X)$ is the smallest infinite cardinal $\kappa$ such that for each open cover there is a subcover of cardinality $\leq \kappa$. So "$X$ is Lindelöf" is the same as $L(X) = \aleph_0$. (These definitions come from pages 11-12 of the aforementioned book.)

Now, for a metric space $E$, $d(E) = L(E)$ (see Theorem 8.1 (c) on pages 32-33), which is the generalization that I think you are looking for. Unfortunately, the proof given there treats the notion of a "$\sigma$-discrete base" and its existence in any metric space as well known, so I'll provide a little more information.

In the Handbook, $\sigma$-discrete is defined in chapter 9 (on page 350) and what you need to do to get a $\sigma$-discrete base in a metric space is to take the open cover of $E$ by open balls of radius $2^{-n}$ (for each $n \in \mathbb{N}$) and take a $\sigma$-discrete refinement of it using Theorem 2.8 of Chapter 9 (page 357). I suppose the reason the existence of a $\sigma$-discrete base is considered standard is that it is one direction of Bing's metrization theorem, and can be found in textbooks such as Engelking's General Topology, Theorem 4.4.3.

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  • $\begingroup$ this theorem is knowing : ***SB theorem :***(E,d) is separable iff (for all A⊂E and card(A)>card(N) then A have an accumulation point) ? @Robert Furber $\endgroup$
    – Dattier
    Commented Oct 16, 2019 at 12:34
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    $\begingroup$ Under the definitions in the Handbook, what you're saying is "$E$ is separable iff $E$ has countable spread" (or $s(E) \leq \aleph_0$). In Theorem 8.1 (c) it is also shown that $s(E) = d(E) = L(E)$. $\endgroup$
    – Robert Furber
    Commented Oct 16, 2019 at 12:36
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    $\begingroup$ @Dattier The spread $s(E)$ is the supremum of the cardinalities of discrete subsets. So if $A \subseteq E$ has $|A| > s(E)$, it must have an accumulation point. $\endgroup$
    – Robert Furber
    Commented Oct 16, 2019 at 12:38

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