TITLE: How to prove that the collection of rank-k matrices forms a closed subset of the space of matrices?
QUESTION [14 upvotes]: Let $M$ be the space of all $m\times n$ matrices. And $C=\{X\in M|rank(X)\leq k\}$ where $k\leq \min\{m,n\}$. How to proof that C is a closed subset of M?

REPLY [3 votes]: Zev Chonoles's answer is totally fine, but since we don't need to care about redundancy, we may simply look at $$f:\mathbb R^{m\times n}\to\mathbb R^{m!n!}\;,\;\;\;A\mapsto\left(\det\left(A_{\pi_1(i)\pi_2(j)}\right)_{1\:\le\:i,\:j\:\le k}\right)_{\substack{\pi_1\:\in\:S_m\\\pi_2\:\in\:S_n}}$$ instead, where $S_n$ denotes the symmetric group. Then, $f$ maps $A$ to the determinants of the submatrix given by the first $k$ columns and rows for every possible permutation of the columns and rows of $A$. And clearly $C=f^{-1}(\{0\})$.