Talk:Ordinal number

The definition of multiplication is not the conventional one, and will probably cause confusion.

The definition given in the article was Cantor's in 1883. He changed it in 1887 -- see Michael D. Potter, "Sets : an Introduction", Clarendon press 1990, p 120.

The point is that one wants a^(b+c) = a^b >< a^c. a >< b should be read "a, b times" (as Potter suggests.)

Fixed. AxelBoldt 16:32, 2 Feb 2004 (UTC)

What is an ordinal set?

The definition of set ordinality (John von Neumann) relies on "set containment". This does not seem to be defined anywhere.

"set containment" is the same as the subset relationship. So for example, the set {1,3} is contained in {1,2,3}. AxelBoldt 16:32, 2 Feb 2004 (UTC)

Perhaps some math person could provide an explination.

It might help answer the question: Is {1,2} an ordinal set?

No, of course! As nought isn't a member of its, it isn't transitive, then it isn't ordinal.

uncountable ordinals

I wonder whether in ZF there is a proof the class of all countable ordinals is a set. If no, why every text about such a matter takes it trivial that there are some uncountable ordinals, therefore a least one, that it must be less or equal (depending on Continuum Hypothesis) than the power set of naturals; neither greater than nor uncomparable to it?

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Doesn't

Ordinals which don't have an immediate predecessor can always be written as a limit of smaller ordinals

contradict

no sequence of elements in ω1 has the element ω1 as its limit

? --SirJective 11:58, 4 Dec 2003 (UTC)

Both of these statements are true and (therefore) there is no contradiction. The crucial word in the second statement is sequence. The first statement would be false if we inserted "as a limit of a sequence of smaller ordinals". A sequence here is understood as a countable collection of things. I'll clarify this in the article. AxelBoldt 16:32, 2 Feb 2004 (UTC)

Thank you. The explanation now given in the article (together with the article net) makes this point clear to me, an I will soon change the german article to reflect this. --SirJective 23:17, 2 Feb 2004 (UTC)

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I certainly don't object to having the von Neumann definition of ordinal in this article; but it would be useful to have it preceded by a more naive- axiomatic. Somethinh along the lines of the following (written in TeX)

axiomatic description. Note that the collection of ordinal numbers do not form a set.

$\alpha$ is an {\em ordinal} is written $\opr{On}(\alpha)$. In addition, there is a binary predicate $\sqsubset$ defined for pairs of ordinals.

\begin{enumerate} \item \label{linear-ordering-property}$\sqsubset$ is a strict linear ordering on the ordinals. This means $\sqsubset$ is transitive: $\alpha \sqsubset \beta$ and $\beta \sqsubset \gamma$ implies $\alpha \sqsubset \gamma$, linear: for every $\alpha, \beta$ one of $\alpha \sqsubset \beta$, $\alpha = \beta$ or $\beta \sqsubset \alpha$ holds and irreflexive: $\alpha \not\sqsubset \alpha$. \item $\sqsubset$ is well-founded. For any non-empty set $A$ of ordinals there is an $a \in A$ such that for all $x \in A$ either $x = a$ or $a \sqsubset x$. We refer to such an element $a$ as a least element of $A$. \item Given an ordinal $\alpha$, there is a set whose members are precisely the ordinals $\beta$ such that $\beta \sqsubset \alpha$. \item There is no set whose members are all the ordinal numbers. \end{enumerate} Write $\alpha \sqsubseteq \beta$ iff $\alpha \sqsubset \beta$ or $\alpha = \beta$. Let $\opr{On}_\alpha=\{\beta: \beta \sqsubset \alpha\}$. By (3) above, $\opr{On}_\alpha$ is a set.

Clearly, least elements of non-empty sets of ordinals are unique.

An {\em initial ordinal segment} is a {\em set} $V$ all of whose members are ordinals and such that for all ordinals $x,y$, if $x \sqsubset y$ and $y \in V$ then $x \in V$. For example, for any ordinal $\alpha$, $\opr{On}_\alpha$ is an initial segment. $\emptyset$ is an initial segment. \begin{prop} \label{recursion-theorem} Suppose $\mathcal{U}$ is an initial ordinal segment, $A$ an arbitrary set, $\mathrm{I}_{\mathcal{U}}$ the set of all functions $g$ with values in $A$ such that $\domain(g)$ is an initial ordinal subsegment of $\mathcal{U}$ with $\domain(g) \neq \mathcal{U}$ and $\mathcal{E}:\mathrm{I}_{\mathcal{U}} \rightarrow A$ an arbitrary function. Then there is a unique function $f: \mathcal{U} \rightarrow A$ such that \begin{equation} \label{recursive-def-prop} f(x)=\mathcal{E}(f | \opr{On}_x) \mbox{ for every $x \in \mathcal{U}$} \end{equation} \end{prop}

CSTAR 22:42, 19 May 2004 (UTC)

Cantor normal form

Does Cantor normal form apply to uncountable ordinals? The article states that it applies to all ordinals > 0. If so, what is the normal form of ω₁? -- Fropuff 01:31, 2005 Mar 8 (UTC)

It's <math>\omega_1=\omega^{\omega_1}<math>. May seem strange, but it is true. 157.181.80.93 17:29, 31 May 2005 (UTC)