Inaccessible cardinal and von Neuman Hierarchy I read a couple of days ago that if $k$ is inaccessible, then $V_k=H(k)$, where $V$ is the von Neumann hierarchy and $H(k)$ the class of sets that are heredititarily of cardinality $< k$. My question would be: is there any singular $k$ for which $V_k=H(k)$? I managed to show that if $K$ is regular then $V_k=H(k)$ iff $k$ inaccessible or $k=\omega$, but I'm clueless about an example?

You do not need the regular part. If $\kappa > \omega$, $V_\kappa = H_\kappa$ if and only if $\kappa = \beth_\kappa$. (Kunen \textit{Set Theory} 78.)

Define $a_0 = \aleph_0$. Define recursively, $a_{i + 1} = \beth(a_i)$. Let $\alpha = \lim_{i< \omega} a_i$. $\beth(\alpha) = \alpha$. $\alpha$ has cofinality $\omega$. So $\alpha$ is a singular cardinal with $\beth(\alpha) = \alpha$.

I hope it is correct now.