Prove the set of matrices with one Jordan block is not dense in $M_n(\mathbb{C}).$

You can prove both fairly elegantly with the following observation:

A continuous function which is constant on a dense set is constant on the entire space.

To show SL$(2, \mathbb{C})$ isn't dense, what function could you use to get a contradiction? (Hint: You use it in the definition of the special linear group.)

Matrices with only one Jordan block is maybe a little harder, but the function $M_3(\mathbb{C}) \to \mathbb{C}$ taking a matrix to the discriminant of its characteristic polynomial works.

Assume that $A , A_k \in M_n(\mathbb C)$, $A_k \to A$ (that is, $(A_k)_{ij} \to A_{ij}$ for all $i, j$) and $A_k$ has one Jordan block $J_{\lambda_k}$ (with eigenvalue $\lambda_k$) Then $p(\lambda) := \det (A-\lambda I)$ satisfies

$$p(\lambda)=\lim_{k\to \infty} p_k(\lambda).$$

But $$\det (A_k - \lambda I) = \det (J_{\lambda_k} - \lambda I) = (\lambda_k -\lambda)^n$$

This implies that $p(\lambda)$ also has $n$-identical eigenvalue. Thus some elements in $M_n(\mathbb C)$, for example those with distinct eigenvalue, does not lie in the closure of the set of matrix with one Jordan block.