Why is this subset closed I try to understand the following statement:
Let $L(\mathbb{R}^n,\mathbb{R}^m)$ denote the vector space of linear maps from $\mathbb{R}^n$ to $\mathbb{R}^m$. Now the statement is that the set
$S := \{\phi \in L(\mathbb{R}^n,\mathbb{R}^{m-k})~|~ \text{rank}(\phi)<m-k \}$ is a closed subset in $L(\mathbb{R}^n,\mathbb{R}^{m-k})$.
Can anyone tell me why that is? Thanks!

Here is a hint.  Consider the special case $m=n$ and $k=0$.  Then $S$ consists of square matrices with rank less than $n$.  This is the same as the set of matrices with zero determinant.  Since the determinant is a continuous function on $L(\mathbb{R}^n,\mathbb{R}^n)$ (I assume this is true in your chosen topology), and $S$ is the inverse image of the point $0$, then $S$ is closed.
Now in general, you can still use the determinant, or many determinants, to determine rank.  
