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January 9th, 2014, 02:43 AM   #1
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Axiom of Choice

I do some exercises in set theory. Among them, there is one that says:

Prove, using the Axiom of Choice, the following:

a)
b)

It is clear to me that point a) requires AC (to "define" the function f(x)). However, I am not sure why is it required for b)?
I solve b) as:



My questions are:
1) Is my solution correct?
2) Does my solution require the AC? In some implicit way maybe? Maybe it is required for proving the implication ?
3) If not, why use AC? Maybe there is a more elegant way to solve b) using AC (or point a) for that matter, which requires AC)?

Note 1: this is not a homework so don't tell me to go and ask the lecturer. I don't have one, that's why I am asking the question here.
Note 2: my question concerns only point b). Point a) is clear enough.
smieci is offline  
 
January 9th, 2014, 01:12 PM   #2
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Re: Axiom of Choice

Quote:
Originally Posted by smieci
I do some exercises in set theory. Among them, there is one that says:

Prove, using the Axiom of Choice, the following:

a)
b)
Perhaps others are smarter but I have no idea what "" is supposed to represent!

Quote:
It is clear to me that point a) requires AC (to "define" the function f(x)). However, I am not sure why is it required for b)?
I solve b) as:



My questions are:
1) Is my solution correct?
2) Does my solution require the AC? In some implicit way maybe? Maybe it is required for proving the implication ?
3) If not, why use AC? Maybe there is a more elegant way to solve b) using AC (or point a) for that matter, which requires AC)?

Note 1: this is not a homework so don't tell me to go and ask the lecturer. I don't have one, that's why I am asking the question here.
Note 2: my question concerns only point b). Point a) is clear enough.
HallsofIvy is offline  
January 10th, 2014, 08:40 AM   #3
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Re: Axiom of Choice

Quote:
Originally Posted by HallsofIvy
Perhaps others are smarter but I have no idea what "" is supposed to represent!
I assume that is just a family of sets labeled by two indices.

Quote:
Originally Posted by smieci
I solve b) as:



My questions are:
1) Is my solution correct?
Yes.

Quote:
Originally Posted by smieci
2) Does my solution require the AC? In some implicit way maybe?
I don't think so.

Quote:
Originally Posted by smieci
Maybe it is required for proving the implication ?
No, this definitely does not require AC. It is provable in first-order logic (even constructive one).

Quote:
Originally Posted by smieci
3) If not, why use AC? Maybe there is a more elegant way to solve b) using AC (or point a) for that matter, which requires AC)?
The only thing that comes to mind is that to prove you can define a constant function using the whose existence is guaranteed by , and then use the right-to-left direction of a). However, this direction does not use AC.
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Evgeny.Makarov is offline  
January 11th, 2014, 01:27 AM   #4
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Re: Axiom of Choice

Evgeny, thank you for the answers. Yes, is any family of sets labeled by two indices. Good idea to use right-to-left a) to prove b). Maybe this is what was meant by the author of the book? I agree with you, however, that AC is not required in this case also.

Thanks for your help, and best regards
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