# The Hausdorff Condition in the Smooth Manifold Chart Lemma

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.

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.