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This question is from Goldrei's Classic Set Theory:

Let $\lambda$ be an ordinal. If $\cup\lambda=\lambda$, then $\lambda$ is a limit ordinal.

But what if $\lambda=\varnothing$? I think I am doing something wrong by writing $\cup\varnothing=\varnothing$.

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    $\begingroup$ You’re not doing anything wrong. It’s frequently useful to treat $0$ as a limit ordinal, since it’s not a successor ordinal, and that definition does so. $\endgroup$ – Brian M. Scott Feb 22 '16 at 7:47
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    $\begingroup$ I've always taken $0$ to be a limit ordinal, and I've never encountered a case where it is of any use to consider it not to be one. What you write is correct; the empty union is the empty set! $\endgroup$ – Clive Newstead Feb 22 '16 at 18:09
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One often defines a limit-ordinal as an ordinal which is not $\emptyset$ yet satisfies your mentioned properties. So no, $\emptyset$ is not a limit ordinal, though some authors may use it as a limitordinal since it in some sences behave as one, and it may make some proofs easier.

See also the wikipedia page.

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The definition that I know is that all ordinals are successor ordinals, limit ordinals or 0. I believe this answers your question.

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  • $\begingroup$ Is it the one given in Goldrei's Classic Set Theory ? $\endgroup$ – user228113 Feb 22 '16 at 18:06

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