Timeline for Independence of the countable axiom of choice
Current License: CC BY-SA 3.0
12 events
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| Mar 15, 2015 at 0:40 | history | edited | Noah Schweber | CC BY-SA 3.0 |
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| Mar 15, 2015 at 0:38 | comment | added | Noah Schweber | Yes, exactly: without the axiom of choice, it is possible that a countable union of countable sets - in fact, a countable union of finite sets - might not be countable! | |
| Mar 15, 2015 at 0:33 | comment | added | Ilias kaperonis | For the exercise :We need to work with sequences cause we know that the union of these sequence is also a sequence therefore the union is countable.Am i right? | |
| Mar 14, 2015 at 23:59 | comment | added | Ilias kaperonis | Haven't solve your exercise yet.But i will and comment you.Mine problem with the hint is that a sequence is basically a function which means a binary relation(a subset of the product of A and omega(sets of natural numbers).The the choice fuction will give us a sequance for every A_{n}.So what next we use the image of that sequance?And we "make" the union of all those images?Have we done something "illegal"?(Excuse i am new) | |
| Mar 14, 2015 at 23:42 | history | edited | Noah Schweber | CC BY-SA 3.0 |
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| Mar 14, 2015 at 23:41 | comment | added | Noah Schweber | A fun exercise - in the argument I gave above, it would not obviously work if we replaced $A_n$ by the set of subsets of $A$ of size $n$ - it is important that we work with sequences, or we need a little bit extra justification. Why? | |
| Mar 14, 2015 at 23:39 | comment | added | Noah Schweber | BTW, for basic exercises in proofs involving choices, mathstackexchange is maybe a better fit. | |
| Mar 14, 2015 at 23:38 | comment | added | Noah Schweber | Yes, that hint is basically the whole problem - given an infinite set $A$, for each $n$ the set $A_n=\{X: X$ is an injection from $n$ into $A\}$ is nonempty, so - by countable choice - the family $\{A_n: n\in\omega\}$ has a choice function. What is this choice function? Well, it's essentially a sequence of sequences of distinct elements of $A$, of increasing length. Concatenating these sequences gives an infinite sequence of elements of $A$ - a single element might occur multiple times in this sequence, but infinitely many elements will occur. Delete repetition, and we're done. | |
| Mar 14, 2015 at 23:34 | comment | added | Ilias kaperonis | I have a hint but probably i am missing something obvious..Let's assume that A is infinite set then we construct the set of all one-to-one sequences in A of length n.We do so for every n and we name the previous set as A_{n} for each n.Then the family of all A_{n} is a set,let's call it S.Using a choice fuction on S we obtain a countable subset of A.The exact statment of CAC i am using is "Every countable family of nonemplty sets has a choise fuction".And F is a choice fuction for the family of non empty sets S iff F(X) is in X (for all X in S). | |
| Mar 14, 2015 at 23:26 | comment | added | Ilias kaperonis | OK!That helped a lot thank you!I am new in set theory studies and i am more intersted on weaker forms of the axiom of choice,equivalent forms and the consistency of the AC.Plus i recently found a proof for the sentence "every finite set has a choice fuction"(in ZF).So the question occurs about the Countable axiom of choice(CAC) is there any proof that CAC is not provable in ZF(okay independence needs to verify that ZF doesn't disprove AC).You seem well informed on the matter and i whould like a bit more help.How does the CAC implies that every infinite set has a countable subset? | |
| Mar 14, 2015 at 20:49 | comment | added | Noah Schweber | You may also be interested in the following: another question mathoverflow.net/questions/104450/road-to-solovays-land; Timothy Chow's article arxiv.org/abs/0712.1320; Cohen's account projecteuclid.org/download/pdf_1/euclid.rmjm/1181070010; and many others. | |
| Mar 14, 2015 at 20:38 | history | answered | Noah Schweber | CC BY-SA 3.0 |