How to show $X=\{A\in\mathcal{L}(\mathbb R^m, \mathbb R^n); \textrm{Ker}(A)=\{0\}\}$ is open in $\mathcal{L}(\mathbb R^m, \mathbb R^n)$? How to show $X=\{A\in\mathcal{L}(\mathbb R^m, \mathbb R^n); \textrm{Ker}(A)=\{0\}\}$ is open in $\mathcal{L}(\mathbb R^m, \mathbb R^n)$? 
Here $\mathcal{L}(\mathbb R^m, \mathbb R^n)$ is the set of all linear applications between $\mathbb R^m$ and $\mathbb R^n$. 
 A: Notice if $m>n$, then $X=\varnothing$ is open.  So suppose $m\le n$.
Identify $\mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$ with $\mathbb{M}_{n\times m}(\mathbb{R})$.  Notice that for $A\in\mathbb{M}_{n\times m}(\mathbb{R})$, $\ker(A)=0$ if and only if rank$(A)=m$ if and only if the determinant of some $m\times m$ submatrix of $A$ is nonzero.  This shows that $X^c$ is the intersection of the inverse images of $0\in\mathbb{R}$ of polynomial functions defining the determinants of $m\times m$ submatrices of $A$.  Polynomials are continuous, and $\{0\}$ is closed, hence $X^c$ is closed, and $X$ is open.
For further concreteness, there are $\binom{n}{m}$ square submatrices of size $m$.  If we consider a matrix of variables $x_{ij}$, then we obtain $\binom{n}{m}$ polynomial functions $p_1,\ldots,p_{r}\in\mathbb{R}[x_{ij}]$ which represent the determinants of each $m\times m$ submatrix, where $r=\binom{n}{m}$.  Each $p_i$ defines a map $\mathbb{M}_{n\times m}(\mathbb{R})\rightarrow \mathbb{R}$ simply by evaluating at the entries of a real matrix.  If $A\in X^c$, then $A$ must map to $0\in\mathbb{R}$ under all of the $p_i$.  Hence, we have
$$X^c=\bigcap_{i=1}^{r}p_i^{-1}(0)$$
A: Here is a alternative with sequences and local compactness. 
Let us prove that the complement of $X$, i.e.
$$
X^c=\{A:\mathbb{R}^m\longrightarrow\mathbb{R}^n\;;\;\mbox{Ker}\;A\neq\{0\}\}
$$
is (sequentially) closed. 
So assume $A_k$ is a sequence of $X^c$ which converges to $A$ for the operator norm.
For every $k$, there exists a unit vector $\|x_k\|=1$ in $\mathbb{R}^m$ such that $A_kx_k=0$. By local compactness of the finite-dimensional normed vector space $\mathbb{R}^m$, there exists a subsequence $x_{n_k}$ which converges to $x$. By continuity of the norm, we have $\|x\|=1$ (or for short, we could have used the compactness of the unit sphere directly). Now 
$$
\|Ax\|=\|A(x-x_{n_k})+(A-A_{n_k})x_{n_k}\|\leq \|A\|\|x-x_{n_k}\|+\|A-A_{n_k}\|\|x_k\|
$$
$$
=\|A\|\|x-x_{n_k}\|+\|A-A_{n_k}\|\longrightarrow 0.
$$
Hence $Ax=0$ and $A$ is not injective, i.e. $A$ belongs to $X^c$.
