I want to verify whether the statement in the title is indeed true, i.e, if $\vert X\vert=\kappa \geq \aleph_0$ then the collection of subsets of cardinality strictly less than $\kappa$ is also $\kappa$?
It relies in the following propositions:
- For all $0<\lambda< \kappa$, $\kappa^\lambda< 2^\kappa$ and $\kappa\cdot \kappa=\kappa$.
- There are $\kappa$ cardinalities lesser than $\kappa$. I think this derives from the definition of an ordinal.
- $\vert\{ A\subseteq X: \vert A \vert< \vert X\vert \} \vert= \Big\vert \sqcup_{\lambda<\kappa} \{ A\subseteq X: \vert A \vert=\lambda \}\Big\vert = \sum_{\lambda<\kappa} \kappa ^ \lambda < \big( 2^\kappa \big)^{\kappa}=2^{\kappa\cdot \kappa}= \vert 2^X\vert. $
I am unsure about $(1)$, but it seems natural that $2^\lambda< 2^\kappa$ if $\lambda<\kappa$, but I know that in the world of set theory this might be the case. I have a suspicion that the claim is true even if (1) isn't, but I would welcome any counter-example, whether it be for (1) or the whole claim.
First, your restatement of the title is not the same as the title since less than $2^\kappa$ is not the same as at most $\kappa$.
Second, for any infinite $\lambda<\kappa$, we have that $\kappa^\lambda$ is the cardinality of the subsets of $\kappa$ of cardinality $\lambda$ (see: Proving that for infinite $\kappa$, $|[\kappa]^\lambda|=\kappa^\lambda$). Therefore, any failure of (1) (i.e., $\lambda<\kappa$ and $\kappa^\lambda=2^\kappa$) will constitute a counterexample to the question. Now, it is certainly possible for this to happen. For example, by Easton's Theorem it is consistent that there are regular $\lambda<\kappa$ with $2^\lambda=2^\kappa$. In a different direction, if you assume GCH then, for any singular $\kappa$, if $\lambda=\text{cof}(\kappa)$ then $\lambda<\kappa$ and $\kappa<\kappa^\lambda$, hence $\kappa^\lambda=2^\kappa$ by GCH at $\kappa$.