A lemma in Kunen's (2011) set theory states that if $\gamma$ is a limit ordinal with $\kappa=\text{cf}(\gamma)>\omega$, then $\text{add}(\mathsf{nonstat}_\gamma)$ $=$ $\text{cov}(\mathsf{nonstat}_\gamma)$ $=$ $\text{non}(\mathsf{nonstat}_\gamma)$ $=$ $\kappa$.
(Where $X\notin\mathsf{nonstat}_\gamma\iff X\text{ is stationary}$ in $\gamma$)
He says that this lemma follows easily from the definitions along with the three facts:
The intersection of any family of fewer than $\kappa$ club subsets of $\gamma$ is club.
There is a family of $\kappa$ club subsets of $\gamma$ whose intersection is empty.
If $\mathcal{I}$ is an ideal on $A$ with $[A]^{<\omega}\subseteq\mathcal{I}$, then $\text{add}(\mathcal{I})\le\text{non}(\mathcal{I})\le|A|$, $\text{add}(\mathcal{I})\le\text{cov}(\mathcal{I})\le|A|$, and $\text{add}(\mathcal{I})$ is regular.
From these, I am able to see that $\text{add}(\mathsf{nonstat}_\gamma)=\text{cov}(\mathsf{nonstat}_\gamma)=\kappa$. I was also able to deduce that any stationary $X\subseteq\gamma$ will have $\text{sup}(X)=\gamma$, which implies that $\text{non}(\mathsf{nonstat}_\gamma)\ge\kappa$.
But I can't think of a stationary $X\subseteq\gamma$ with cardinality $\kappa$. (When $\gamma$ is regular, I could simply take $X=\gamma$, but that won't work when $\gamma$ is singular).
So my question is: What stationary subset of $\gamma$ has cardinality $\kappa$ (when $\gamma$ is singular)?
Any help is appreciated.
-Thanks
If $\gamma$ is singular of cofinality $\kappa$, then there is a normal function $f\colon\kappa\to\gamma$ such that $\sup\operatorname{rng}(f)=\gamma$.
You can check and see that $\operatorname{rng}(f)$ is indeed a club in $\gamma$, and its cardinality is indeed $\kappa$. Now since clubs are stationary sets...