Krull dimension greater than or equal to small inductive dimension for Noetherian topological spaces

Question

I am trying to prove Krull dimension and the small inductive dimension coincide for any Noetherian topological space $X$. The inequality Krull$(X) \le$ ind$(X)$ holds for all topological spaces. It is the other direction ind$(X)\le$ Krull$(X)$ that troubles me. By Noetherian induction we can take a minimal closed set $A$ violating this (and we can assume $A=X$). If we can show $X$ is irreducible we can easily finish the argument (just take a point $p$ and a neighborhood $U$ whose boundary has small inductive dimension ind$(X)-1$. Then since the boundary is not $X$, we can take a chain of irreducible closed sets of length ind$(X)-1$ in $\partial U$ and add $X$ to this chain.) But how to show $X$ must be irreducible? Can someone give a hint?

Answer 1

For any open $U\subseteq X$, $\partial U$ is a closed subset of $X$ with empty interior, and therefore it cannot contain any irreducible component of $X$ (any irreducble component has nonempty interior, namely the complement of the union of all the other irreducible components). So, $\partial U$ must have strictly smaller Krull dimension than that of $X$, and therefore also smaller inductive dimension than the Krull dimension of $X$ by the induction hypothesis.