Re: Set Theory question

Jules a écrit :
aatu.koskensilta@xxxxxxxxx wrote:
Stephen J. Herschkorn wrote:
thoughtcube@xxxxxxxxx wrote:

Let (A, <=) be an ordered (i.e. totally ordered) set. Then there exists
a series a_n of elements of A, for which: for any element a in A, there
is some a_n that is larger than it, i.e. a <= a_n.
The claim is false. No such series exists in omega-1.
Right. But interestingly, if choice fails it's possible that every
cardinal has cofinality omega.

--
Aatu Koskensilta (aatu.koskensilta@xxxxxxxxx)

"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus

I have a couple of questions. First, one poster (Aatu Koskensilta)
wrote "But interestingly, if choice fails it's possible that every
cardinal has cofinality omega." Did you mean that every ordinal has
cofinality omega? If not, then how is the cofinality of a cardinal
defined?


Cardinals *are* ordinals (the smallest ordinal of given cardinality) in this approach



Also, someone wrote that, without choice, the collection of all
well-orderings of N is a well-ordered set of ordinality omega-1. I
thought choice was neccesary to show that the collection of all
countable ordinals was totally ordered.


No, we are not exactly speaking of ordinals, but of well-orderings on N. It is easy to define order on such well orderings (O_1<=O_2 iff there exists a map from N to N , strictly increasing with respect to the orderings (or even stronger, such a map whose image is an initial segment of O_2)), and the proof that <= (or more precisely the relation on the quotient set) is a well-ordering is easy, and doesn't necessite choice.




.

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