I have just started of learning about Manifolds from Milnor's book.I am stuck on following exercise,give me some idea.

Let $U$ be an open subset of a manifold $X$. Show that for any $x \in U$ the tangent space at $x$ to $U$, i.e., $T_x(U)$ is same as $T_xX$.

• How does Milnor define $T_xX$? Where did you get stuck applying his definition? – Neal Oct 10 '14 at 5:50
• @Neal Definition:Suppose that $X$ sits in $R^n$ and that $f: U \rightarrow X$ is a local parametrization around $x$, where $U$ is an open set in $R^k$ and assume that $f(0) = x$ for convenience. Define the tangent space of $X$ at $x$ to be the image of the map $df_o: R^k \rightarrow R^n$ – Arpit Kansal Oct 10 '14 at 6:16

As you likely know, the topology of a manifold embedded in $\Bbb{R}^n$ is the subspace topology. This means that a set $A$ is open in $X$ if and only if there is a set $B$ open in $\Bbb{R}^n$ such that $A = B \cap X$.
First note that $T_x U \subseteq T_x X$ in the most straightforward way: Given a local diffeomorphism $f: W \to V$ with $W$ open in $\Bbb{R}^k$, $V$ open in $U$, and $f(0) = x$, we can identify $T_x U = \operatorname{im} df_0$. But, we can also consider $V$ as an open set in $X$, so the same parametrization allows us to define $T_x X = \operatorname{im} df_0$.
If, on the other hand, we begin with $f: W \to V$ a local parametrization around $x \in X$, then we must restrict $f$ to $f' = f|_{f^{-1}(V \cap U)}$ so that its a diffeomorphism onto $V \cap U$. The derivative has a local definition so $T_x U = df'_0 = df_0 = T_x X$.