I was just learning about set theory and came across this exercise. It does not specify whether we work with CH as an axiom. The statement is easy to prove if CH is assumed, but I'm not sure where to start (if it is provable at all) if I have to work without CH.
Let $\kappa$ be a cardinal with $\omega\leq\kappa\leq 2^\omega$. The following are equivalent:
(1) For all $X\subseteq\mathbb{R}$, if $|X|=\kappa$, then there is $q\in\mathbb{Q}$ such that $|X\cap(-\infty,q)|=|X\cap(q,\infty)|=\kappa$.
(2) $\text{cf}(\kappa)>\omega$.
Any hint would be appreciated. Thank you.
It does not require $\mathsf{CH}$ at all.
As usual, we prove an equivalence by treating each direction separately. It's easy to show that $\neg(2)\implies\neg (1)$: given $\kappa\in[\omega,2^\omega]$ with countable cofinality, we can whip up a sequence of sets of reals $X_i$ ($i\in\omega$) such that $X_i\subseteq [i,i+1)$, $\vert X_i\vert<\kappa$, and $\vert \bigcup_{i\in\omega}X_i\vert=\kappa$. But then $\bigcup_{i\in\omega}X_i$ is a clear counterexample to $(1)$.
The other direction is more interesting, so I'll just provide a sketch. Suppose $\kappa\in[\omega,2^\omega]$ and $X\subseteq\mathbb{R}$ have cardinality $\kappa$. If $X$ didn't satisfy $(1)$, then every rational $q\in\mathbb{Q}$ would lie in one of the following two sets $$L=\{q\in\mathbb{Q}: \vert(-\infty,q)\cap X\vert<\kappa\}\quad\mbox{or}\quad U=\{q\in\mathbb{Q}: \vert (q,\infty)\cap X\vert<\kappa\}.$$ Note that every element of $L$ is less than every element of $U$, and each set is nonempty, so we get a Dedekind cut. Do you see how to use this to build a cofinal sequence in $\kappa$ of ordertype $\omega$ (think about "walking towards the cut" ...)?