Let $k$ be an algebraically closed field. For any $k-$variety, $V$, let $t(V)$ denote the set of all irreducible, closed subsets of $V$. Irreducible here means that a set $V$ cannot be written as a union of two closed subsets of $V$.
Is my claim, $$t(V) = \{\{p\}^-\mid p\in V\}$$ correct?
Here, $S^-$ denotes the closure of the set in $V$. Is it correct to assume that not all points of a $k$-variety are closed?
I think this is true because if $F\in t(V)$ is closed and irreducible, then it is a subset of $V$, say $F = \bigcup \{q\}$ for some $q\in V$. If we take the closure on both sides, we get $$F = \bigcup \{q\}^-$$ and if there is any $q,r$ such that their closures are not equal to $F$, then clearly, I can write $F$ as a union of closed subsets (as $F\cap \{q\}^- = \{q\}^-$ is closed in $F$)
Thus, we must have, for all $q$ in $F$, that $\{q\}^- = F$ which shows $$t(V) \subset\{\{p\}^-\mid p\in V\}$$
The other inclusion holds as $\{p\}^-$ is clearly closed and also irreducible as $\{p\}$ is a singleton and thus irreducible.