I found the following lemma in Lee's book on smooth manifolds.

Lemma 1.14 (One-Step Smooth Manifold Structure). Let M be a set, and suppose we are given a collection {Uα} of subsets of M, together with an injective map ϕα : Uα → Rn for each α, such that the following properties are satisfied.
(i) For each α, Ueα = ϕα(Uα) is an open subset of Rn.
(ii) For each α and β, ϕα(Uα ∩ Uβ) and ϕβ(Uα ∩ Uβ) are open in Rn.
(iii) Whenever Uα ∩ Uβ 6= ∅, ϕβ ◦ ϕ−1 α : ϕα(Uα ∩ Uβ) → ϕβ(Uα ∩ Uβ) is smooth.
(iv) Countably many of the sets Uα cover M.
(v) Whenever p, q are distinct points in M, either there exists some Uα containing both p and q or there exist disjoint sets Uα, Uβ with p ∈ Uα and q ∈ Uβ.

Then M has a unique smooth manifold structure such that each (Uα, ϕα) is a smooth chart.

Is condition (v) different than saying that the collection $\{U_{\alpha} \}$ is a covering of $M$?

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    $\begingroup$ When you say $\{U_{\alpha}\}$ is a covering of $M$, your $\alpha$ might belong to an uncountable set. Condition $(iv)$ is stronger than that, as it says, $\alpha$ comes from a countable set. $\endgroup$ – Ajay Kumar Nair Mar 1 at 18:02
  • $\begingroup$ @AjayKumarNair my bad, I wanted to say "condition (v)" not "condition (iv)", I eddited. $\endgroup$ – roi_saumon Mar 1 at 18:30
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    $\begingroup$ It is different. There could be $U_{\alpha}, U_{\beta}$, $p\in U_{\alpha},q\in U_{\beta}$, $U_{\alpha}\cap U_{\beta}\neq\phi$. $\endgroup$ – Teebro Prokash Mar 1 at 19:48
  • $\begingroup$ @TeebroProkash thanks, I get it now! $\endgroup$ – roi_saumon Mar 2 at 17:38

Condition 5 is there to make sure that the set is given an Hausdorff topology (https://en.m.wikipedia.org/wiki/Hausdorff_space)


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