Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In his proof "Any bounded nonempty subset $\mathbb{R}$ has a least upper bound in $\mathbb{R}$", Enderton showed that the least upper bound is just $\bigcup A$. Since by definition $x\subseteq \bigcup A\quad \& \quad x\in A$ . In the case $x=\bigcup A\quad \& \quad x\in A$, $x$ becomes the largest element in $A$, but this cannot be true since the Dedekind Cut is a subset of $\mathbb{Q}$ that must have no largest element. How do you explain that?

share|cite|improve this question
up vote 4 down vote accepted

You are confusing between sets and sets of sets.

$A$ is a set of sets of rational numbers. While Dedekind cuts have no upper bound, there might very well be an upper bound to a set of Dedekind cuts.

Note that $x,y\in A$ the $x\subseteq y$ or $y\subseteq x$. Then $\bigcup A=\{z\in\mathbb Q\mid \exists x\in A(z\in x)\}$. That for itself is a Dedekind cut.

In particular, $\bar A=A\cup\{\bigcup A\}$ is a set of Dedekind cuts, and it is simple to see that every $x\in A$ is a subset of $\bigcup A$, as well $\bigcup A$ is a subset of itself. Therefore ordered by inclusion, $\bar A$ has a maximal element.

share|cite|improve this answer
Thank you. You are an excellent teacher. – user11750 Aug 16 '11 at 22:52
@user11750: Thanks :-) – Asaf Karagila Aug 16 '11 at 22:58

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.