Irreducible finite Kolmogorov spaces

Question

Is it true that irreducible finite Kolmogorov spaces of Krull dimension 1 are homeomorphic iff they are equipotent? I am trying to imagine how the specialization order would look like and it seems to me that it is true,

Answer 1

Yes. Note that if $x$ is maximal in the specialization order, then $\{x\}$ is open. So if our space $X$ is irreducible, it must have exactly one maximal element $x$ in the specialization order (if there were more than one, they would give disjoint nonempty open sets). So every element $y\in X$ satisfies $y\leq x$. Moreover, if $y,z\in X$ are distinct and distinct from $x$, $y\not\leq z$, since otherwise $y\leq z\leq x$ would make the dimension greater than $1$. So if $X$ has $n$ elements, this completely determines the specialization order up to isomorphism: there is a greatest element $x$, and the other $n-1$ elements are below $x$ and incomparable to each other.