Let $X = \mathbb{R}^{m+n}$ be the space.
Define $$F_m(X) = \{(v_1, v_2, ..., v_m) : v_i \in X, \{v_1, v_2, ..., v_m\} \ \mbox{is linearly independent}\} \subseteq \mathbb{R}^{m+n} \times ... \times \mathbb{R}^{m+n}$$ is the set of all $m$-frame in $\mathbb{R}^{m+n}$.
I suppose that it is (well-known) that $F_m(X)$ can be given a smooth manifold structure, and it is called $\textbf{Stiefel manifold}$.
The only remark from the book I read is that the set $F_m(X)$ is open in $\mathbb{R}^{m+n} \times ... \times \mathbb{R}^{m+n} = \mathbb{R}^{m(m+n)}$, and since any open subset of a smooth manifold is again a smooth manifold which gives that $F_m(X)$ is a smooth manifold.
The question here is how to see that $F_m(X)$ is open ? I try to think of its as a preimage of some continuous functions, but I am not so sure.