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

I have been considering about $A+\alpha\sim A$ when $\omega\le\alpha<h(A)$, in which $h(A)$ is the Hartogs number of $A$.

If there is $\alpha$ s.t. $\omega\le\alpha<h(A)$, we can get $\alpha \le A$ and $A$ is Dedekind-infinite. So if $|A| \ne |A|+|\alpha|$ then both $|\alpha|$ and $|A|$ $<|A|+|\alpha|$. Note that they are all Dedekind-infinite, so if it can be proven that every Dedekind-infinite set cannot be split into two smaller Dedekind-infinite sets then $|A| = |A|+|\alpha|$ holds.

However by a counterexample showed by Brian M. Scott in this question, this path is obstructed.

So could anyone give me a hint?

share|cite|improve this question
up vote 2 down vote accepted

Note that $\alpha+\alpha\sim\alpha$ for any infinite ordinal.

Therefore if $A>\alpha\geq\omega$ we can write $A\sim\alpha\cup B$ for some $B\subseteq A$ disjoint from $\alpha$. Then $\alpha+|A|=\alpha+\alpha+|B|=\alpha+|B|=|A|$, as wanted.

The requirement that $\alpha\geq\omega$ is clear because finite ordinals do not have the property $n+n\sim n$, and so it is false.

share|cite|improve this answer
Oh, it is surprising there is a shortcut. Thanks a lot. – Popopo Dec 2 '12 at 13:07

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.