A codimension of 0 with no irreducible components

I found this problem where I think there is a mistake but am not 100% sure:

Let $Z$ be a closed subset of a topological space $X$. If $Z$ is irreducible we call codim($Z,X$) as the supremum of lengths of the chains of irreducible closed subsets of $X$ which contains $Z$: $$Z\subset Z_0\subsetneq Z_1\subsetneq\dots\subsetneq Z_n.$$ For a general closed subset $Z$ we define codim$(Z,X)=$inf$_Y$codim$(Y,X)$ where $Y$ ranges over the irreducible components $Z$.

The second question asks: Prove that codim$(Z,X)=0$ if and only if $Z$ contains an irreducible component $X$.

Now for one direction I find no issues. If $Z$ contains an irreducible component $Z_{-1}$ then because of the maximality the longest chain from $Z_{-1}$ to $X$ is $Z_{-1}\subsetneq X$, so codim$(Z,X)=0$.

But in the other direction I think I have a counterexample:

First let $A=k[x_{\mathbb{N},\mathbb{N}}]$, $\mathfrak{p}_i=\langle x_{\mathbb{N},i}\rangle$ and $S=A\smallsetminus\bigcup_{i}\mathfrak{p}_i$.

Let $X=S^{-1}A$.

Now in $X$ let $\mathfrak{q}_i=\langle x_{i,0},\dots,x_{i,i}\rangle$ and $Z=\bigcup_i\mathfrak{q}_i$.

Wouldn't the codim$(Z,X)=0$ despite $Z$ having no irreducible components?