Let $k \le \min\{m,n\}.$ Let $M_k=\{A:rank~A\le k\}\subset Mat_{m\times n}(K)$ and $S_k=\{A:rank~A=k\}\subset Mat_{m\times n}(K)$. Show that $\bar S_k=M_k.$ ($K$ is not finite field. $K=\mathbb{R} ~ or ~\mathbb{C}$)
My attempt: I have proved this statement:
Let $ F:\mathbb{A^1 \to A^n}$ be a polynomial map. Suppose that $F(t) \in S$ for infinitely many $t \in \mathbb{A^1}$. Then $F(\mathbb{A^1})\subset \bar S$.
I have a hint that I need to use the above statement (or similar ideas) to do the proof.
One direction $\bar S_k \subset M_k$ is easy. For another direction, my idea is to find a polynomial map $F$ such that the image of $F$ = $M_k$. Then we can use the above statement. But I did not find such a polynomial map and I doubt that such a polynomial does not exist. So, I got stuck on this problem.
Let $A\in M_k$. We use the definition of rank as the maximum number of linearly-independent columns. Up to a change of basis on the target, we may assume that the first $l < k$ columns are linearly independent and each further column is in the span of the first $l$. Select $k-l$ linearly independent vectors not in the span of the first $l$ columns. Then the matrix given by adding $t$ times each of these columns to the corresponding $(l+1)^{th},(l+2)^{th},\cdots,k^{th}$ column is in $S_k$ for all nonzero values of $t$, but is exactly equal to our original $A$ when $t=0$. This shows the result.