The nilpotent cone is closed but not compact

Idea. Characteristic polynomial $t^n$ means that all complex eigenvalues are $0$, which is the same as nilpotency.

Step 1: $\chi_A=t^n\Rightarrow A$ nilpotent. By the Cayley-Hamilton theorem, $\chi_A(A)=0$, i.e. $A^n=0$.

Step 2: $A$ nilpotent $\Rightarrow\chi_A=t^n$. If $A^m=0$ and $Av=\lambda v$ with $v\ne0$ ($\lambda\in\mathbb{C}$), then $0=A^mv=\lambda^mv$ gives $\lambda=0$. So all complex roots of $\chi_A$ vanish and, $\chi_A$ being monic of degree $n$, $\chi_A=t^n$.

Step 3: closed. $\mathcal N=\chi^{-1}(\{0\})$ is the preimage of a point under a continuous map (Problem 76), hence closed. Equivalently $\mathcal N=\{A:A^n=0\}$ is the zero set of polynomial maps.

Step 4: unbounded. For $s\in\mathbb{R}$ let
$$N_s=\begin{pmatrix}0&s\\0&0\end{pmatrix}\oplus 0_{n-2}.$$
Then $N_s^2=0$, so $N_s\in\mathcal N$, and $\|N_s\|=|s|$ (it maps $e_2\mapsto se_1$ and kills the other basis vectors). Letting $s\to\infty$ shows $\mathcal N$ is unbounded; hence it is not compact, although it is closed.

Remark. The nilpotent cone is a *cone*: $tN\in\mathcal N$ for all $t\in\mathbb{R}$ if $N\in\mathcal N$. It has empty interior (Part 2) and is singular at $0$. For $n=1$ the fibre is $\{0\}$ (a point), which explains the hypothesis $n\ge2$.