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Suppose that $A$ and $B$ are two infinite sets and $|A|<|B|$. The question is that how to prove that $|A∪B|=|B|$. The proof is related to the Axiom of Choice.

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  • $\begingroup$ It is easy to show that $|B|<=|A \cup B|$. I tryed to find a injection from B to $|A \cup B|$. Then with the using of Cantor-Bernstein Theorm, it is proved. However, thus map is not fined yet. $\endgroup$ – Yijun Yuan Oct 18 '14 at 3:40
  • $\begingroup$ The injection is $B \to B \subset A \cup B$ The challenge is to have an injection from $A \cup B$ into $B$. $\endgroup$ – Ross Millikan Oct 18 '14 at 4:35
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$|A\cup B| \leq |A| + |B| \leq |B| + |B| = |B|$ since $|B|$ is of infinite cardinality and $2\cdot \aleph_k = \aleph _k$ for whichever cardinal $|B|$ happens to be.

Then also $|B|\leq |A\cup B|$ by subadditivity.

Hence $|A\cup B| = |B|$ when $B$ is an infinite set and $|A|\leq |B|$

I don't think you need to use the axiom of choice for the proof, though perhaps I have naive notions on how set operations work on higher sized sets.

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    $\begingroup$ $|B| + |B| = |B|$ uses Zorn, so it is using the AC. $\endgroup$ – Aram Oct 18 '14 at 3:52
  • $\begingroup$ Is there a simple way to explain $|B|+|B|=|B|$ without the using of the Zorn( but with AC)? $\endgroup$ – Yijun Yuan Oct 18 '14 at 5:15
  • $\begingroup$ @袁轶君 Imho, using Zorn for this is very simple. You can find a proof in Kaplansky's book about metric spaces. $\endgroup$ – Aram Oct 18 '14 at 19:02
  • $\begingroup$ @Aram Thanks a lot. I've got the answer. :-) $\endgroup$ – Yijun Yuan Oct 26 '14 at 9:34
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By the axiom of choice all infinite cardinals are alephs, so you can use $n+\aleph_0=\aleph_0$ for finite $|A|=n$, and $2\cdot\aleph_\alpha=\aleph_\alpha$ for infinite $|A|=\aleph_\alpha$.

In the absence of the axiom of choice, there could be an infinite cardinal $m$ such that $m\lt2m$. From $m\lt2m$ it follows (this is not trivial) that $2m\lt3m$. Then, if $A,B$ are disjoint sets with $|A|=m$ and $|B|=2m$, we have $|A|\lt|B|$ while $|A\cup B|=m+2m=3m\gt2m=|B|$.

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  • $\begingroup$ "all infinite cardinals are alephs" is based on CH, not AC. If the CH is refused, there is a set A, which $\aleph _{0} <|A|<\aleph _{1}$. $\endgroup$ – Yijun Yuan Oct 18 '14 at 5:18
  • $\begingroup$ @袁轶君 Wrong. If the CH is false, there is a set $A$ such that $\aleph_0\lt|A|\lt|\mathbb R|$. But there is never a cardinal between $\aleph_0$ and $\aleph_1$. $\endgroup$ – bof Oct 18 '14 at 5:34
  • $\begingroup$ Well...It's my fault. $\endgroup$ – Yijun Yuan Oct 18 '14 at 5:48

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