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I'm reading the PCM (The Princteton Companion to Mathematics) article on set theory (IV.22) by Joan Bagaria. I'm trying to understand his comment on difference between cardinal and ordinal numbers.

In set theory, one likes to regard all mathematical objects as sets. For ordinals this can be done in a particularly simple way: we represent 0 by the empty set, and the ordinal number α is then identified with the set of all its predecessors. For instance, the natural number n is identified with the set {0, 1, . . . , n − 1} (which has cardinality n) and the ordinal ω + 3 is identified with the set {0, 1, 2, 3, . . . , ω,ω + 1,ω + 2}.

... cardinal numbers are used for measuring the sizes of sets, while ordinal numbers indicate the position in an ordered sequence. This distinction is much more apparent for infinite numbers than for finite ones, because then it is possible for two different ordinals to have the same size. For example, the ordinals ω and ω+1 are different but the corresponding sets {0, 1, 2, . . . } and {0, 1, 2, . . . , ω} have the same cardinality, as figure 1 shows. In fact, all sets that can be counted using the infinite ordinals we have described so far are countable. So in what sense are different ordinals different? The point is that although two sets such as {0, 1, 2, . . . } and {0, 1, 2, . . . , ω} have the same cardinality, they are not order isomorphic: that is, you cannot find a bijection φ from one set to the other such that φ(x) < φ(y) whenever x < y. Thus, they are the same “as sets” but not “as ordered sets.”

Informally, the cardinal numbers are the possible sizes of sets. A convenient formal definition of a cardinal number is that it is an ordinal number that is bigger than all its predecessors.

Could anyone explain with examples what the sentence in bold means? What could be an example that an ordinal number is equal to or smaller than some of its predecessors?

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    $\begingroup$ Another way to put it is that the cardinal number $\kappa$ of a set $A$ is the least ordinal that is equipotent with $A$. So for every ordinal $\alpha$ less than $\kappa$ in the ordinal ordering there is a surjection $\kappa \to \alpha$ but no injection. Hence $\kappa$ is in some sense "bigger" than $\alpha$ $\endgroup$ – leibnewtz Dec 20 '16 at 18:33
  • $\begingroup$ @leibnewtz: Indeed, that's the word that causes confusion. $\endgroup$ – Jack Dec 20 '16 at 19:50
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$\omega + 1$ is a good example. $\omega + 1$ is the ordinal $\{0, 1, 2, \ldots, \omega\}$; as an ordering, think of it as $\omega$ with one more element at the end. It is greater than $\omega$, because it has $\omega$ as a proper initial segment; for ordinals, that's what "greater" means. But it's not bigger than $\omega$ - there's a bijection between $\omega$ and $\omega + 1$, given by the function $f$ which takes $0$ to $\omega$ and $n$ to $n - 1$ for all $n > 0$.

By contrast, the ordinal $\omega_1$, defined as the first uncountable ordinal, is bigger than all of its predecessors, by definition - if $\alpha < \omega_1$, then $\alpha$ can't be uncountable, so there is an injection from $\alpha$ to $\omega$. But there's no injection from $\omega_1$ to $\omega$, by definition, because that would make $\omega_1$ countable. So $\omega_1$ is a cardinal, often denoted $\aleph_1$.

The key idea here is that the author is using "bigger" to refer to size, not ordering - that is, "bigger" is a statement about whether a certain injection exists, not where an ordinal appears in the standard ordering.

On the other hand, you asked about an ordinal that is "equal to or smaller than some of its predecessors". This can't happen. An ordinal is never equal to its predecessors, because different ordinals are always different - it's like asking whether there's a number that's equal to a different number. And since every ordinal has an injection into all of its successors, ordinals can't decrease in size. The only thing that can happen is the example I've outlined above, where we have an ordinal that's the same size as its predecessor. Note that this means that, for ordinals, "the same size as" and "equal to" do not mean the same thing.

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  • $\begingroup$ As you clearly pointed out in the third paragraph, "bigger" should refer to size, then it is possible that an ordinal is "equal" to some of its predecessors if "equal" is in the sense of size again and an example would be $\omega+1$, which has the same size as $\omega$. Anyway, the reason for the second question is because I was unclear what "bigger" means in the context. Thank you for your answer! $\endgroup$ – Jack Dec 20 '16 at 19:48
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    $\begingroup$ @Jack Yes, but it's worth noting that while "bigger" is not a technical term, "equal" very much is. Most mathematicians will freely use "bigger" and "smaller" loosely, but "equal" always means "actually equal". If you ask "Is $\omega + 1$ bigger than $\omega$?" the response will usually be "What do you mean by 'bigger'?" If you ask "Is $\omega + 1$ equal to $\omega$?" the response will always be "No." I guess what I'm getting at is that it's important that "bigger" changes meaning in context, but "equal" never does. $\endgroup$ – Reese Dec 20 '16 at 20:38
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The ordinal $\omega+1$ is greater than $\omega$ as an ordinal, but they are equipotent. $f\colon\omega+1\to\omega$ defined as $f(\omega)=0; f(n)=n+1$ is a witnessing bijection.

Therefore $\omega+1$ is not a cardinal.

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Not a good choice of words in the definition of cardinal. "Bigger" in what sense?.... Definition : An ordinal $x$ is less than ordinal $y$ iff $x\in y$ iff $x\subsetneqq y.$ A cardinal ordinal $y$ is an ordinal for which there does not exist a bijection from $y$ to any $x\in y.$

There are uncountably many ordinals that are each countably infinite. The $\in$-least of them is $\omega.$ Each of them is a bijective image of $\omega.$

In set theory $|A|$ usually denotes the cardinal of the set $A$, which is the $\in$-least ordinal y for which there exists a bijection fron $A$ to $y.$ For any infinite ordinal $x$ there is a bijection from $x$ to $x+1=x\cup \{x\} ,$ but any such bijection $f$ cannot preserve the $\in$ order. That is, we cannot have $u\in v\implies f(u)\in f(v)$ for all $u,v \in x.$

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