I am having trouble to understand what topology is given to the tangent bundle of a smooth manifold that allows it to be a smooth manifold itself. In my understanding, among other things the topology must be second countable and Hausdorff. The definition of the tangent bundle $TM$ of a smooth manifold $M$ I am using is
$TM = \bigsqcup_{p\in M} T_pM$,
that is the disjoint union of all $T_pM$ where $T_pM$ is tangent space at $p$ consisting of all derivations at $p$. Since there is no further specification on what topology this space is given I assume we take the natural disjoint union topology.
However, in that case it seems that $TM$ is not second countable because then every set $(O,p)$ where $O$ is an open subset of $T_pM$ would be open and disjoint from any $(O,q)$ for $q \neq p$. So unless $M$ is countable there would be an uncountable number of disjoint open sets which contradicts second countability.
The only alternative I can think of is using the natural smooth structure of $TM$ as the topology. That is for every open subset $O$ of $M$ the open sets of $TM$ are defined as $\pi^{-1}(O)$ where $\pi$ is the natural projection $TM \rightarrow M$.
But then $TM$ can not be Hausdorff, since any two elements of the same fiber of $\pi$ could not be seperated by open sets.
In conclusion, in both cases $TM$ could not be a manifold, so I must be missing something very obvious. Thus, I would really appreciate it if someone could point out my misconception.
Best Answer
Take some atlas on $M$, and let $U$ be an element of that atlas. Then $TU=\pi^{-1}(U) \cong U \times \mathbb{R}^n$ as a set, so it inherits a topology. Moreover, all these topologies (for different $U$) are compatible with each other, so together they give you a topology on the total space $TM$.
Note that this is very much like your second idea, except that we don't require open sets to contain entire fibres $\pi^{-1}(x)$ -- just open subsets of them.