Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.
  1. Is it possible to prove without Continuum Hypothesis that for every uncountable subset $S$ of $\mathbb{R}$ there is a real number $x$ that splits it into two parts of the same cardinality, i.e. $\left|S \cap (-\infty,x)\right|=\left|S \cap (x,\infty)\right|$?
  2. (if the answer to the first question is no) Is this statement equivalent to Continuum Hypothesis?
share|improve this question

2 Answers 2

up vote 22 down vote accepted

No, the statement cannot be proven in ZFC without assuming continuum hypothesis or something similar. In fact, it is equivalent to the statement that there are finitely many cardinalities between $\aleph_0$ and $2^{\aleph_0}$, so it is strictly weaker than the continuum hypothesis.

Suppose that there were infinitely many such cardinalities, then you can let $S=\bigcup_{n=1}^\infty S_n$ where $S_n\subseteq(0,1/n)$ has cardinality $\aleph_n$ to obtain a contradition.

On the other hand, if there are only finitely many such cardinalities, then $f(x)=\vert S\cap(-\infty,x)\vert$ must achieve its maximum, say $\aleph_n$ ($n > 0$). If $x_0$ is the infimum of the $x\in\mathbb{R}$ such that $f(x)=\aleph_n$ then $S\cap(x_0,\infty)$ has cardinality $\aleph_n$. Choosing $y_k\in\mathbb{R}$ decreasing to $x_0$, the cardinality of $S\cap(y_k,\infty)$ must be $\aleph_n$ for large enough $k$, otherwise $S\cap(x_0,\infty)=\bigcup_k(S\cap(y_k,\infty))$ is a countable union of sets of cardinality less than $\aleph_n$, so is of cardinality less than $\aleph_n$, giving a contradiction. So, $S\cap(-\infty,y_k)$ and $S\cap(y_k,\infty)$ are both of cardinality $\aleph_n$ for $k$ large enough.

share|improve this answer

Suppose that the continuum is larger than $\aleph_\omega$. Choose a subset $S_n$ of $(n,n+1)$ of cardinality $\aleph_n$ and let $S=\cup_{n=1}^\infty S_n$. Then for each $x$, $S\cap (-\infty,x)$ has cardinality smaller than $S\cap (x,-\infty)$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.