I have been considering about $A+\alpha\sim A$ when $\omega\le\alpha If there is $\alpha$ s.t. $\omega\le\alpha However by a counterexample showed by Brian M. Scott in this question, this path is obstructed. So could anyone give me a hint?
$A+\alpha\sim A$ when $\omega\le\alpha
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$\begingroup$
elementary-set-theory
cardinals
axiom-of-choice
infinity
1 Answers
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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.
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0Oh, it is surprising there is a shortcut. Thanks a lot. – 2012-12-02