I was reading Knapp's Lie groups beyond an introduction and, in the first pages, he shows that the set of all tangent vectors to a given closed linear group $G$ at the identity matrix, that is $$\mathfrak{g} = \{c'(0)\;|\; c\colon \mathbb{R}\to G \mbox{ is a smooth curve with } c(0)=I\} $$ is a Lie algebra taking $[X,Y] = XY-YX$ (and it's usually called the linear Lie algebra of $G$). In order to prove that the Lie braket is well defined he states that $\mathfrak{g}$ is closed, but I don't follow. How could I prove that $\mathfrak{g}$, as a subspace of the matrix space, is topologically closed? Any hints?
Another way of proving what he's trying to prove is that, by taking $X=c'(0)$ and $Y=d'(0)$ in $\mathfrak{g}$, proving that $XY = c'(0)d'(0)$ is in $\mathfrak{g}$ (thus, $XY-YX = [X,Y]\in \mathfrak{g}$ by being a linear combination of elements in $\mathfrak{g}$). The other question would then be: given $c(t)$ and $d(t)$ smooth curves, can I find a smooth curve $f(t)$ such that $f'(t) = c'(t)d'(t)$?, I tried doing integration by parts but I end in a $0=0$ type of scenario.