Let $V$ be a 0-dimensional variety. If $Z$ is a proper closed subset of $V$, then $Z$ is discrete?
I want use this in a proof, but I couldn't justify so made me wonder if it's true. Can someone help me? I'm in an introductory course of algebraic geometry, I still don't know many things. I have the definition of $k$-Zariski topology and results such as: Hilbert Nullstellensatz, Hilbert Basis Theorem, some results on Noetherian spaces/varieties. But affine variety has not yet been defined and some things I see being used all the time.
Definition. Let $k$ be a field, and let $C$ be an algebraically closed field containing $k$. If $S$ is a subset of $k[x_{1},...,x_{n}]$, then the zero set of $S$ is $$Z(S) = \lbrace (a_{1},...,a_{n}):f(a_{1},...,a_{n})=0\quad\mathrm{for\;all}\;f\in S\rbrace.$$
Definition. Let $k$ be a field, and let $C$ be an algebraically closedd field containing $k$. Then a set $V \subset C^{n}$ is said to be a $k$-variety if $V=Z(S)$ para algum conjunto $S$ of polynomials in $k[x_{1},...,x_{n}]$.