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What is a notation for the minimal ordinal of $\mathbb{R}$?

I know that $\beth_1$ and $\mathfrak{c}$ designate the cardinality of $\mathbb{R}$, and that $\Omega$ denotes the smallest uncountable ordinal, and that $\aleph_1$ denotes the first uncountable cardinal, and that $\Omega$ and $\beth_1$ and $\aleph_1$ are all equipollent if CH is true, but what is a notation for the smallest ordinal equipollent to $\mathbb{R}$ which doesn't implicitly assume any position on CH?

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    I think I learned $\Omega$ from Patrick Suppes' Axiomatic Set Theory? I understood it to denote the first uncountable ordinal.2012-11-02

2 Answers 2

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Assuming the axiom of choice and $\frak c$ is a well-ordered cardinal, it means that $\frak c$ is an initial ordinal, namely it is the least ordinal which can be put in bijection with the real numbers.

It is not uncommon to see $\alpha<\frak c$.

If one wishes to be completely formal and separate the cardinal and ordinal form, one can write:

Let $\aleph_\alpha$ be the cardinality of the continuum. We enumerate $\mathbb R$ as $\{x_\beta\mid \beta<\omega_\alpha\}$ ...

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    Makes sense, thanks.2012-11-02
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There need not be any ordinal equipollent to $\Bbb R$, so there isn't really a standard notation for the least such (at least, none that I'm aware of). Note also that $\omega_1$ is typically used to denote the least uncountable ordinal, except when the reader is expected to have little to no familiarity with ordinals, in which case $\Omega$ is used instead. (Thanks, Brian, for prompting the clarification.)

More generally, we'll typically take $\omega_\alpha$ to be the least ordinal having the cardinal $\aleph_\alpha$. In fact, we'll take $\omega_\alpha$ and $\aleph_\alpha$ to be the same thing. We'll choose one notation over the other depending on whether we're talking about cardinalities or order types, or whether we're using cardinal arithmetic or ordinal arithmetic.

If we assume AC (or at least enough choice so that $\Bbb R$ is well-ordered), then the cardinal of $\Bbb R$--that is $\mathfrak{c}$--can be taken to be an ordinal, in particular the least ordinal equipollent to $\Bbb R$.

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    Fair point, Brian. Thanks.2012-11-02