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Sets intersection and the axiom of choice?

  1. Dec 13, 2011 #1
    I'm working on some topology in [itex] \mathbb{R}^n [/itex] problem, and I run across this:
    Given [itex]\{F_n\}[/itex] a family of subsets of [itex] \mathbb{R}^n [/itex], then if [itex]x[/itex] is a point in the clausure of the union of the family, then

    [itex]x \in \overline{\cup F_n}[/itex]
    wich means that for every [itex]\delta > 0[/itex] one has
    [itex]B(x,\delta) \cap (\cup F_n) \neq \emptyset [/itex]

    Now, if I say that because the intersection is a nonempty set, I have a point [itex]y \in \mathbb{R}^k [/itex] such that

    [itex] d(x,y) < \delta \wedge y \in (\cup F_n)[/itex]

    did I use the axiom of choice?

    Because I think it this way, I choose an element [itex] y [/itex] from an infinite (nonumerable) indexed family parametrized by [itex]\delta[/itex] (for each value of [itex]\delta[/itex] I have another intersection).

    So If I don't want to use the AC I can't just say that I can choose an element from that nonempty intersection?
    I'm a little confused :?

  2. jcsd
  3. Dec 14, 2011 #2
    The way you've written it, you've chosen one y from one set. In any event, you can't be invoking the axiom of choice if you can prove that your collection is a set without invoking it.
  4. Dec 14, 2011 #3
    You did not use the axiom of choice. In fact, you've made only one choice. The axiom of choice is only necessary if you want to make an infinite number of choices (and if you can't specify the choices you make).
  5. Dec 14, 2011 #4
    Hi zhentil and micromass.
    Thanks for your answers.

    Maybe I wasn't clear, let me explain again.

    I have an infinte indexed family of nonempty sets, each one of the kind:

    [itex] A_{\delta} = B(x,\delta) \cap (\cup F_n) \neq \emptyset [/itex]
    [itex] A_{\delta} \subseteq \mathbb{R}^n [/itex]
    [itex] A_{\delta} [/itex] is closed.

    So the family is
    [itex] \{ A_{\delta} \}_{\delta \in \mathbb{R}_{>0}}[/itex]

    and for each set in the family I want to choose an element
    [itex] y_{\delta} \in A_{\delta} [/itex]
    they are not necesarily differents elements (but maybe so)

    So I don't think I'm just choosing one element from one set, but that I'm choosing one element from each of the nonumerable family of sets (and I can't specify the choices I make).

    Still didn't I use the AC?
  6. Dec 14, 2011 #5
    This is completely different!!!

    Now you are indeed making an infinite number of choices. So now you are indeed using the axiom of choice!!

    (also, I don't really see why your [itex]A_\delta[/itex] is closed, I don't think this is necessarily the case).
  7. Dec 14, 2011 #6
    Now that I think about it, I'm not even sure what I mean when I say I 'choose' an element.
    I know the family is of nonempty sets, so obviously there is an (at least) an element in each of them, but that didn't mean I 'choose' it, doesn't?

    The problem is I want one to use it in the proof. (and the element I choose depends on [itex] \delta [/itex], so I don't just choose one, but one for every [itex] \delta [/itex])

    I think I'm now even more confused. :confused: :confused: :confused:
  8. Dec 14, 2011 #7
    Hi micromass, thanks for your reply (I wrote the next post while you where writing this)
    It did sound like I was using the AC, but please read the next post, now I'm not even sure what I mean to 'choose' an element, saying it exists is not enough to have 'choosen it', but using it for something else is?

    (my [itex]A_\delta[/itex] is closed because it's part from an exercice where that was one of the hipothesys, and I run across this doubt with the AC). If curious, the exercise is this: to show that if [itex] \{ F_n \} [/itex] is a family of closed subsets of [itex] \mathbb{R}^k [/itex] with [itex] d(F_n, F_m) > 1 [/itex] for all [itex] m,n > 0[/itex] with [itex] m \neq n [/itex], then [itex] \cup F_n [/itex] is closed.

  9. Dec 14, 2011 #8
    Saying that something exists is actually equivalent to choosing it. If it exists, then you can choose it.

    For example, [0,1] is nonempty. So there exists an element in it. That means that I can choose an element in it. Indeed, I can choose the element 1.

    I can also look at [itex]A_i=[0,1][/itex]

    All these sets are nonempty, so I can choose an element in each of these sets. However, can I make a simultanious choice?? That is, can I choose something for all i at once?? Yes I can, I can choose the element 1 for each [itex]A_i[/itex] This does not use the axiom of choice.

    However, when I cannot specify the choice, then I need to use the axiom of choice.

    Maybe you should read my blogs on the axiom of choice: https://www.physicsforums.com/blog.php?u=205308 [Broken] I hope that settles some questions.
    Last edited by a moderator: May 5, 2017
  10. Dec 14, 2011 #9
    OK, but if the [itex]A_\delta[/itex] is closed and bounded (=compact), then there might be ways to make the choice not using AC.

    For example, if [itex]A_i, i\in I[/itex] are compact subsets of [itex]\mathbb{R}[/itex], then I can choose elements [itex]x_i\in A_i[/itex] without invoking AC. Indeed, I can just take


    the maximum exists because of compactness. This works in [itex]\mathbb{R}^n[/itex], but it can be generalized to [itex]\mathbb{R}^n[/itex].
  11. Dec 14, 2011 #10
    Hi micromass,
    If only saying an element exists in an infinite family is 'choosing it', isn't it obvious the AC has to be true? I mean, if I have an infinite colection of nonempty subsets, each of them is nonempty, so an element 'obviously' has to exist in each of them, doesn't?

    I'll surely read your blog on the AC later, I only read the conclusion by now (started backwards)

    On the other hand, the sets [itex] F_n [/itex] in the family in my exercice are closed, but doesn't have to be bounded. Even more, the sets I make a choose isn't from them, but from the intersection of them with an open ball, so they aren't even closed, so the choice is quite complicated indeed (I think).

    Thanks again for your replys, really help me!
  12. Dec 14, 2011 #11
    The point of the AC is not saying that an element exist. The point is that you can make a simultaneous choice. Surely, if I give you sets [itex]A_1,A_2,A_3,A_4,...[/itex] that are nonempty. Surely, you can pick an element from each of them. You can pick an element from [itex]A_{45345455}[/itex] if you want because it is nonempty. The trick is to pick an element from all of them at once.

    The axiom of choice might be obviously true to you (and to many other people), but it cannot be demonstrated from the usual axioms.
  13. Dec 14, 2011 #12
    Hi micromass,
    So, what I'm not sure now, is if the choice I made from the infinite indexed family I have was all at once, or one after another. It is a nonumerable family so there is no "next" set in the family, and I have to chose an [itex] y_\delta [/itex] for each [itex] A_{\delta} [/itex].
    So do you think I'm choosing them all at once? How can I realize?
  14. Dec 14, 2011 #13
    You can't choose them one after another. The only way to do that is by using a form of transfinite induction, this is not what happens here.

    If you reason: I choose one for [itex]A_0[/itex], then I choose one for [itex]A_1[/itex], then I choose on for [itex]A_2[/itex], etc. then you can only make a finite choice. Once you get to something infinite, this reasoning fails.

    So if you want to choose something from an infinite set, then you need to make it all at once.

    I urge you to read part 1 of my blog, that will reveal a lot I think.
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