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If $A$ is an infinite set and $B$ is denumerable, $A$ is equipotent with the union $A\cup B$.

How to prove this?

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What is your definition of "infinite set"? (Seriously; there are a couple of ways of defining it, and some lead to easy proofs of this, others less so). Also, are you working with the Axiom of Choice? – Arturo Magidin May 7 '12 at 19:46
Yes im working with the AC, and the book im studying defined infinite set as a set that is not equipotent with a natural number n – Katlus May 7 '12 at 19:48
Have you proven that every infinite set has a denumerable subset? – Arturo Magidin May 7 '12 at 19:49
Ya i proved that – Katlus May 7 '12 at 19:50
With the axiom of choice, $|A|$ is some cardinal and $|B| = \aleph_0$. $|A \cup B| \leq |A| + |B| = \text{max}(|A|, |B|) = |A|$ since $|B| = \aleph_0$ is the smallest infinite cardinal. – William May 7 '12 at 19:51
up vote 2 down vote accepted

Let $B'=B\setminus A$. If $B'$ is finite, then biject $B$ with $A\cap B$, and use that bijection to biject $A\cup B = A\amalg B'$ with $A$ (where $\amalg$ denotes a disjoint union).

If $B'$ is infinite, let $D$ be a denumerable subset of $A$. Then biject $D\amalg B'$ with $D$, and use that bijection to get a bijection of $A\cup B = A\amalg B'$ with $A$.

(This requires you to prove that the union of two disjoint denumerable sets is denumerable; I trust you know how to do that)

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I proved it.. But this problem precedes the problem 'union of two denumerable sets is denumerable'. Do you have any easier proof? – Katlus May 7 '12 at 20:56
@Katlus: I suspect there are ways of doing it, but I can't think of any off the top of my head; you could break up the cases with $B'$ infinite according to whether $B\cap A$ is finite or infinite, but I don't think it will be any simpler – Arturo Magidin May 8 '12 at 1:30

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