The number of subsets of cardinality less than $\kappa$ of a cardinal $\kappa$ is $\kappa$

Question

Let $L=\bigcup_{\alpha \in Ord} L_\alpha$ be Godel's constructible universe and thus $L \models GCH$. Let $\kappa$ be an infinite cardinal and $S:=\{A \subseteq \kappa : \#A < \kappa \}$. Is it true that $L \models \#S=\kappa$? ($\#S \ge \kappa$ trivially.)

My reasoning: Assuming $V = L$, we have $A \in L_\kappa \iff A$ bounded in $\kappa$ (i.e. $\exists \beta < \kappa. A \subseteq \beta$). If $A \subseteq \kappa$, $\#A < \kappa$ and the cofinality $cf(\kappa)=\kappa$, then $A$ needs to be bounded. Hence $A \in L_\kappa$, but $\#L_\kappa = \kappa$ and so $\#S=\kappa$ when $\kappa$ is a regular cardinal.

Answer 1

This is not true, even assuming $\sf GCH$, unless we also add that $\kappa$ is regular.

If $\kappa$ is regular (and assuming $\sf GCH$ of course), we have that $2^{<\kappa}=\kappa$, and then by this argument the result follows.

If $\kappa$ is singular, e.g. $\operatorname{cf}(\kappa)=\lambda<\kappa$, then there are $\kappa^\lambda$ subsets of size $\lambda$, but Koenig's theorem tells us that $\kappa<\kappa^\lambda$.

Answer 2

Not necessarily. Let $\kappa$ be a successor and $\lambda$ be the cardinal preceeding $\kappa$. $\kappa$ has a subset of size $\lambda$, which has $2^\lambda$ subsets of size $\lt \kappa$. We know it is consistent that $2^\lambda \gt \kappa$