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According to what I heard of, an ordinal is constructed by taking an union of {$\alpha$} $\cup$ $\alpha$ where $\alpha$ is a predecessor ordinal.

If so, how can there be a set that is not an ordinal?

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By the way, not all ordinals are constructed in this way. That only gives successor ordinals; limit ordinals do not have this form. – Henning Makholm Sep 18 '12 at 10:04
Also, don't confuse the notion of a set being an ordinal with a set being a subset of an ordinal, or a set being in one-to-one correspondence with an ordinal (which is, of course, equivalent to the Axiom of Choice, since it implies that every set can be well-ordered.) – Steven Stadnicki Nov 26 '12 at 18:18
up vote 16 down vote accepted

$\varnothing$ is an ordinal; we call it $0$.

$0\cup\{0\}=\varnothing\cup\{\varnothing\}=\{\varnothing\}=\{0\}$ is an ordinal; we call it $1$.

$1\cup\{1\}=\{\varnothing\}\cup\{\{\varnothing\}\}=\{\varnothing,\{\varnothing\}\}=\{0,1\}$ is an ordinal; we call it $2$.

$2\cup\{2\}=\{0,1\}\cup\{2\}=\{0,1,2\}$ is an ordinal; we call it $3$.

$\{0,2\}$ is not an ordinal: it is not $x\cup\{x\}$ for any ordinal $x$.

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The power set of an ordinal is usually not an ordinal, except for the cases of $0$ and $1$.

For example, $\mathcal P(\omega)$ is not an ordinal.

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An even simpler example of a set that is not an ordinal would be $\{\{\emptyset \}\}$. We have $\emptyset \in \{\emptyset\} \in \{\{\emptyset \}\}$ but $\emptyset \notin \{\{\emptyset \}\}$.

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