Why the tangent bundle is Hausdorff?

Solution 1:

First, let me fix some terminology (taken from Lee's book).

Let $M$ be a smooth $n$-dimensional manifold. A coordinate chart on $M$ is a pair $(U, \varphi)$ where $U$ is an open subset of $M$ and $\varphi : U \to \tilde{U}$ is a homeomorphism from $U$ to an open subset $\tilde{U} = \varphi(U) \subset \mathbb{R}^n$. The set $U$ is called a coordinate domain.

Now to the situation at hand. As $M$ is Hausdorff, and $p \neq q$, there are disjoint open sets $U, V \subset M$ with $p \in U$ and $q \in V$. As you note, there is no guarantee that $U$ and $V$ will be coordinate domains. While that is true, we can take them to be coordinate domains without loss of generality. Let $(X, \varphi)$, $(Y, \psi)$ be charts (i.e. $X, Y$ are coordinate domains) with $p \in X$ and $q \in Y$ (note, we do not require $X$ and $Y$ to be disjoint).

Claim: The sets $U' = U\cap X$, $V' = V\cap Y$ are coordinate domains with $p \in U'$ and $q \in V'$.

As $U'$ and $V'$ are open, and the restriction of a homeomorphism to an open subset is a homeomorphism, $(U', \varphi|_{U'})$ and $(V', \psi|_{V'})$ are charts; that is, $U'$ and $V'$ are coordinate domains. As $p \in U$ and $p\in X$, $p \in U'$; likewise, as $q \in V$ and $q \in Y$, $q \in V'$.

So, without loss of generality, we can take $U$ and $V$ to be coordinate domains (if they aren't, pass to subsets $U' \subseteq U$, $V' \subseteq V$ which are, and call these sets $U$ and $V$ respectively).

Solution 2:

$TX= \{(p,v) : p \in X, v \in T_{p}X \}$ since $X$ and $T_{p}(X)$ are connected Hausdorff spaces ( $T_{p}(X)$ is a subspace of manifold $X$ ) Hence $TX$ is a connected Hausdorff space