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Here is the lemma stated in Lee's book, the second edition.

LEMMA. Let $M$ be a set and $\{U_\alpha\}_{\alpha\in J}$ be a collection of subsets of $M$, along with maps $\varphi_\alpha:U_\alpha\to\mathbf R^n$, such that the following properties are satisfied:

(i) $\forall \alpha\in J$: $\varphi_\alpha$ is a bijection between $U_\alpha$ and an open subset $\varphi_\alpha(U_\alpha) \subset\mathbf R^n$.

(ii) $\forall \alpha,\beta\in J$: the sets $\varphi_\alpha(U_\alpha\cap U_\beta)$ and $\varphi_\beta(U_\alpha\cap U_\beta)$ are open in $\mathbf R^n$.

(iii) $\forall\alpha,\beta\in J$: $U_\alpha\cap U_\beta\neq \emptyset \quad \Rightarrow \quad \varphi_\beta\circ\varphi_\alpha^{-1}:\varphi_\alpha(U_\alpha\cap U_\beta)\to \varphi_\beta(U_\alpha\cap U_\beta)$ is smooth.

(iv) Countably many of the sets $U_\alpha$ cover $M$.

(v) $ \left. \begin{array}{c} p,q\in M\\ p\neq q \end{array} \right\} \quad \Rightarrow \quad \left\{ \begin{array}{c} \exists \alpha\in J\text{ such that } p,q\in U_\alpha,\quad\text{ or}\\ \exists \alpha,\beta\in J\text{ such that } p\in U_\alpha, q\in U_\beta \text{ and } U_\alpha\cap U_\beta=\emptyset \end{array} \right. $

Then $M$ has a unique manifold structure such that each pair $(U_\alpha,\varphi_\alpha)$ is a smooth chart.

We define the topology on $M$ by taking all sets of the form $\varphi_\alpha^{-1}(V)$, with $V$ an open subset of $\mathbf R^n$.


I feel like the condition (v) in the lemma is redundant but I am not sure about that. The condition (v) is used to provide the Hausdorff property of $M$. But it seems like we can show that $M$ is Hausdorff even without the condition (v). Here is my reasoning.

Let $p,q\in M$ and $p\neq q$.

By the condition (iv), there exist some $\alpha,\beta \in J$ s.t. $p\in U_\alpha$ and $q\in U_\beta$. If $\alpha = \beta$, then by the condition (i), we can find two disjoint open subsets of $U_\alpha$ to separate $p$ and $q$ by the Hausdorff property of $\mathbf R^n$. So we may assume that $\alpha \neq \beta$ and $U_\alpha \, \cap U_\beta \neq \emptyset$.

Let $A:= U_\alpha \setminus \overline{U_\beta}$, $B:=U_\beta \setminus \overline{U_\alpha}$, $C:= U_\beta \,\cap U_\alpha$.

Note that $A,B,C$ are all open by the definition of the topology on $M$ and the condition (i).

Case $1$: $p\notin B$ and $q\notin B$.

In this case, $A$ and $C$ separate $p$ and $q$. We are done.

Case $2$: $p\in B$ and $q\in B$.

In this case, it reduces to the situation that $p,q \in U_\alpha$. And we are done again.

Case $3$: One of $\{p,q\}$ is in $B$ and the other one is not.

W.L.O.G, assume that $p\in B$ and $q\notin B$.

In this case, we must have $q\in C$. But $B$ and $C$ are disjoint, which implies that $B$ and $C$ separate $p$ and $q$. So we obtain the Hausdorff property of $M$ again.

$\tag*{Q.E.D}$

I think there may be something wrong in my proof but I can't see it. Thanks for help.

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In a manifold like the line with two origins (classic non-Hausdorff example), we can have that for $0$ and $0'$ $\alpha \neq \beta$ and $0 \in U_\alpha$ and $0' \in U_\beta$ with $U_\alpha \cap U_\beta$= $U_\alpha \setminus \{ 0 \} = U_\beta \setminus \{0'\}$. So $U_\alpha \setminus \overline{U_\beta}= \emptyset$ and vice versa. So $A=B= \emptyset$ but $C$ contains neither origin, in your notation.

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  • $\begingroup$ Sorry for my late reply. It's a really impressive counterexample! I've checked that this manifold does satisfy the conditions (i) - (iv). But I just want to double check with you again. It does satisfy the conditions (i) - (iv) right? $\endgroup$
    – Sam Wong
    Feb 8 at 5:04
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    $\begingroup$ @SamWong yes it shows we cannot omit (v). $\endgroup$ Feb 8 at 5:52

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