Tangent Bundle topology is Hausdorff I am reading Lee's ''Introduction to Smooth Manifolds'' and it states that the tangent bundle TM has a natural topology and smooth structure that makes it a 2n-dimensional manifold.
I get most of the proof, my only problem is to check the Hausdorff condition because it does not specify which topology they use, so I thought of one that works, but I do not know if it is the usual one.
Having the specified topology does not bother most of the proof since you have a projection: $\pi: TM \to M$ defined as $\pi(p,X)=p$
And just take open sets as $\pi^{-1}(U)$ where $U$ is open in $M$. This works well for most of the proof, but when you want to check the Hausdorff condition in two points in the same fiber $(p,X),(p,Y)$.
So to avoid this I can define the open sets as the collection of sets:
$(U,\{V_x\}_{x \in U}):=\{ (x,v) : x \in U, v \in V_x \}$ where $U$ is an open set of $M$ and $\{V_x\}_{x \in U}$ a collection of open sets $V_x \subset T_xM$ where many but not all can be the empty set.
Is this right? I checked and it does not change the smooth structure since the local charts $(U,\phi^{\sim})$ can be seen as $((U, \{T_xM\}_{x\in M}),\phi^{\sim})$ in my topology. Thanks in advance.
 A: You know that each chart $\phi : U \to V \subset \mathbb{R}^n$ for $M$ induces a canonical bijection $\tilde{\phi} : TM \mid_U = p^{-1}(U) \to V \times \mathbb{R}^n$. We define $W \subset TM$ to be open if $\tilde{\phi}(W \cap p^{-1}(U))$ is open in $V \times \mathbb{R}^n$ for all $\phi$.
It is an easy exercise to show that this is in fact a Hausdorff topology.
A: I guess the theorem that you mention is Proposition 3.18 on Lee's book.
For any smooth chart $(U_{\alpha},\varphi_{\alpha})$ of $M$, we have a map $\widetilde{\varphi}_{\alpha} : \pi^{-1}(U_{\alpha}) \to \Bbb{R}^{2n}$ defined as
$$
\widetilde{\varphi}_{\alpha} (v_p) = (x^1(p),\dots,x^n(p),v^1,\dots,v^n) .
$$
This map is a bijection onto its image $\varphi_{\alpha}(U) \times \Bbb{R}^n$.
The topology defined on $TM$ is the one that generated by $\{\widetilde{\varphi}_{\alpha}^{-1}(V) : \forall \alpha \in A, V\subseteq \Bbb{R}^{2n} \text{ is open}\}$. You can check directly that the collection above indeed generate a topology.
To show the Hausdorff property, let $p \in U \subseteq M$, where $U$ is the domain of some smooth chart $(U,\varphi)$ of $M$.  And $v_p,w_p \in \pi^{-1}(p) \subseteq \pi^{-1}(U)\subseteq TM$, be a pair of points that lie in the same fiber.
With this topology, the map $\widetilde{\varphi} : \pi^{-1}(U) \to  \Bbb{R}^{2n}$ is a homeomorphism onto its image. We can write $\widetilde{\varphi}(v_p) = (\hat{p},v)$, with $\hat{p} = \varphi(p)$. So let $\widetilde{\varphi}(v_p)=(\hat{p},v)$ and $\widetilde{\varphi}(w_p) = (\hat{p},w)$ be their images and $v,w \in \Bbb{R}^n$ be two distinct vectors. To obtain disjoint neighbourhoods of $v_p$ and $w_p$, just take disjoint neighbourhoods $B_1,B_2 \subseteq \Bbb{R}^n$ of $v$ and $w$ resp. and maped back the disjoint open sets $\varphi(U)\times B_1$ and $\varphi(U)\times B_2$ to $\pi^{-1}(U)$. So $\widetilde{\varphi}^{-1}(\varphi(U) \times B_1)$ and $\widetilde{\varphi}^{-1}(\varphi(U) \times B_2)$ are the desired neighbourhoods for $v_p$ and $w_p$.
