Conversation started Mar 10, 2012 at 18:07.
Kannappan Sampath
Kannappan Sampath
Mar 10, 2012 18:07
In the most recent answer you had written on AC, you wrote that: The axiom of choice asserts that if we have a collection $X$ of nonempty sets, then we can choose from each one. What does "choose" mean? It means that there is a function whose domain is $X$ and for every $a \in X$ we have that $f(a) \in a$.
Asaf Karagila
Asaf Karagila
Yes.
Kannappan Sampath
Kannappan Sampath
But, once I had asked a question on MO about AC which I subsequently deleted. There was a comment from Ryan Reich that said this is a misunderstanding. We can always choose elements, even infinitely many of them from a set. What is guaranteed by AC is, we can form a set from them. So, I don't see if what you write and this are the same. 
If the set is nonempty, it is always possible to pick one or even finitely many elements from it, and this does not invoke AC but is merely a consequence of the rules of logic. AC says that given a set, you can pick elements from each of its members, and from those elements form another set. It's not the picking that's a problem, but the forming. – Ryan Reich Sep 19 at 22:49
Asaf Karagila
Asaf Karagila
What? Link me to that question of yours.
Wait.
You said infinitely, but quoted finitely.
Kannappan Sampath
Kannappan Sampath
@AsafKaragila Oh, sorry the quote is right.
Asaf Karagila
Asaf Karagila
He's wrong there.
Consider $X$ as a collection of nonempty sets. Now consider $\bigcup X$, by the axiom of union it is also a set. Every choice function is a function into $\bigcup X$ and its range defines a subset. By replacement axiom we have that this is a subset of $\bigcup X$.
You also have subsets of $\bigcup X$ which are not the range of a choice function (or transversal sets as Brian calls them).
The axiom of choice (or its fragments) tell you how many choice functions exist and thus how many (partially) transversal sets exist.
I feel, however, that I'm not answering your real question.
Kannappan Sampath
Kannappan Sampath
Mar 10, 2012 18:18
Can we do a bit slowly? I haven't yet digested your third post
Asaf Karagila
Asaf Karagila
Well, you have about ten minutes. :-)
Kannappan Sampath
Kannappan Sampath
What are "partially" transversal sets?
Asaf Karagila
Asaf Karagila
Sets that would correspond to choice functions on a proper subset of $X$.
For example singletons, which chose from only one member of $X$.
Kannappan Sampath
Kannappan Sampath
So, Partially transversal sets are images of Choice functions restricted to some proper subset of $X$?
Asaf Karagila
Asaf Karagila
It's an ad-hoc definition. But yes, this is what I meant.
Kannappan Sampath
Kannappan Sampath
Mar 10, 2012 18:24
OK. I think I got this. So, how does this help?
Asaf Karagila
Asaf Karagila
I'm not sure anymore what your question is, I think I got lost in my own words.
What was that deleted question of yours on AC, by the way? (if you have link I could read it there...
Kannappan Sampath
Kannappan Sampath
@AsafKaragila Do deleted questions exist on MO?
Asaf Karagila
Asaf Karagila
Oh. I don't have sufficient reputation on MO yet. ;-)
Kannappan Sampath
Kannappan Sampath
I have a .txt file that contains the question, an answer and the comment. Shall I link it here?
Asaf Karagila
Asaf Karagila
Sure. Is that your question to me now as well?
Hm, it seems my time is up. I will be back in a bit, I need to put the chicken on the stove.
Kannappan Sampath
Kannappan Sampath
Mar 10, 2012 18:28
@AsafKaragila Sure. : )
@AsafKaragila No. My question now is: Does picking elements from a collection of sets require AC or not? As you wrote in the recent answer, seems yes. But, Ryan seems to say no.
(Of course, collection is non-empty.)
Asaf Karagila
Asaf Karagila
How many do you want to pick?
I wrote this and that about finitely many choices without AC.
Kannappan Sampath
Kannappan Sampath
I want to pick finitely many, say.
Looks like Ryan talks about non-empty sets (collection), we can pick finitely many elements (sets) from it but we cannot pick elements from these sets without AC, generally. Am I right in making this statement?
@Asaf Sorry, here is the ping and my question is just above this. ^
Please ping me, I am another tab. : )
Asaf Karagila
Asaf Karagila
Mar 10, 2012 18:45
@KannappanSampath No, you're not right. You can always pick finitely many elements from finitely many sets. The sets only need to be nonempty, and they could be finite or infinite. If you want to choose from infinitely many sets at once then you need the axiom of choice.
Kannappan Sampath
Kannappan Sampath
@AsafKaragila I think I am understanding. : )
Let me think about this and get back. Thanks for your time.
 
Conversation ended Mar 10, 2012 at 18:47.