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Maybe this is a trivial question. I see that every infinite cardinal is necessarily a limit ordinal, but is the converse true ?

Asaf Karagila
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fyusuf-a
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    Being a limit ordinal is an inherent property: an ordinal is limit if and only if it has no largest element. On the other hand, being a cardinal (an initial ordinal) is not quite so inherent: whether or not an ordinal is initial depends on the existence of a bijection with a smaller ordinal, and we can use forcing methods to “add” such a bijection without disturbing the internal (order) structure of any ordinal. – tomasz May 01 '14 at 18:46

2 Answers2

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No, $\omega + \omega$ is a limit ordinal. Its cardinality is $\omega$

Ross Millikan
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  • What lead me to my question was this : I do not understand why $\omega+\omega$ is limit. – fyusuf-a May 02 '14 at 00:14
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    It is not a successor. What would be its predecessor? It is the limit of $\omega+1, \omega+2, \omega+3 \dots$ As ordinals, each of these is greater than the previous. Maybe, as Asaf Karagila says, the question is about the contrast between ordinal and cardinal addition. – Ross Millikan May 02 '14 at 02:05
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More generally to Ross' canonical example, if $\alpha$ is any ordinal then $\alpha+\omega$ is the smallest limit ordinal which is strictly larger than $\alpha$. And if $\alpha$ is infinite then $|\alpha|=|\alpha+\omega|$, so $\alpha+\omega$ is not a cardinal.

Note that this is ordinal addition (as in Ross' example), and not cardinal addition.

Asaf Karagila
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