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Axiom of Choice and Metric Spaces.

  1. Jul 19, 2012 #1
    So the axiom of choice is confusing to me, apperently there is a distinction between the exsistence of an element and the actual selection of an element?

    I'm confused as to how much the axiom of choice is needed in elementary metric space theorems?

    As an example, is the Axiom of Choice needed to prove the equivlent definitions of compactness for metric spaces?(Covering compact = sequentually compact = complete + totally bounded)

    What about the Baire Category Theorem? This PDF(http://math.berkeley.edu/~jdahl/104/104Baire.pdf [Broken])claims that the Axiom of choice is not needed to prove it in complete, totally bounded metric spaces but it appears the person still makes implict use of it.

    I guess I am confused in general as to when it is implicitly used in a proof.(as in the statement "select a s_n from each U_n.....".
     
    Last edited by a moderator: May 6, 2017
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  3. Jul 19, 2012 #2

    micromass

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    The axiom of choice is a quite technical condition. It shows up in the following contexts:

    Take [itex]A_i,i\in I[/itex] a collection of nonempty sets. Then we can find a "sequence" [itex](x_i)_{i\in I}[/itex] where [itex]x_i\in A_i[/itex].

    Such a statement requires the axiom of choice in general, but not always. If we can find specific elements in [itex]A_i[/itex], then it is not needed.
    For example, say that we are given [itex]A_n=[0,1/n][/itex]. Then all the [itex]A_n[/itex] contain 0. Thus we can take [itex]x_i=0[/itex]. This does not require choice since each element is well-defined.

    If you read my blog: https://www.physicsforums.com/blog.php?b=2951 [Broken] then you can find more thoughts on the matter.

    Let me give an example. We can make the following definition:

    Let A is a set in a metric space (X,d) and let x in X. We call x a "closed point" of A if each ball B(x,r) intersects A.

    We can now prove that if x is a closed point in A, then there exists a sequence [itex](x_n)_n[/itex] in A such that [itex]x_n\rightarrow x[/itex].

    Indeed, for each integer n>0 holds that [itex]A\cap B(x,1/n)[/itex] is nonempty. By the axiom of choice, we can find a sequence [itex](x_n)_n[/itex] such that [itex]x_n\in A\cap B(x,1/n)[/itex]. This is the converging sequence we're looking for.

    Without the axiom of choice, we had no way of concluding this, since we had no actual rule of constructing the set.

    However, if we are given in the above that [itex]x\in A[/itex], then [itex]B(x,r)\cap A[/itex] contains a very specific point, namely x. So in that case, we can take [itex]x_n=x[/itex] and we have no need of the axiom of choice.
     
    Last edited by a moderator: May 6, 2017
  4. Jul 21, 2012 #3
    I'm confused.

    Consider a collection of nonempty sets U, and let S E U.

    Can we say there exists an element in S without the axiom of choice?

    For countable collections it seems like you should be able to inductivly construct a sequence.
     
  5. Jul 21, 2012 #4

    micromass

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    Yes, that is just one set.

    You are able to pick as many elements as you like, but you can never write the sequence.

    For example, we have a theorem that says: if X is infinite, then there is an injection [itex]\mathbb{N}\rightarrow X[/itex].

    To prove this theorem, we need the axiom of choice because it might not be true without it.

    Now, you could reason as folows:
    X is nonempty, so we can certainly take x0 in X.
    [itex]X\setminus \{x_0\}[/itex] is nonempty, so we can take [itex]x_1\in X\setminus \{x_0\}[/itex]
    [itex]X\setminus \{x_0,x_1\}[/itex] is nonempty, so we can take [itex]x_2\in X\setminus\{x_0,x_1\}[/itex]

    So we can find as many elements as we can. But at any stage, we always have finitely many elements. We cannot just say that we have an infinite sequence, since we never constructed one (all we have is a finite sequence of arbitrary length). We need the axiom of choice to make an infinite sequence.

    To put the axiom of choice in more context. The rules for making sets if one does not have the axiom of choice are the following (simplified):
    • The empty set is a set
    • If X is a set, then the set of subsets of X is a set (= the power set)
    • If X is a set and if P is a property, then [itex]\{x\in X~\vert~P\}[/itex] is a set.
    • If X is a collection of sets, then [itex]\bigcup X[/itex] is a set.
    • If X,Y are sets, then [itex]\{X,Y\}[/itex] are sets.
    • The natural numbers [itex]\mathbb{N}[/itex] are a set.
    • If X is a collection of sets, and if F is a "rule" that associates with [itex]x\in X[/itex] a unique set [itex]Fx[/itex], then [itex]\{Fx~\vert~x\in X\}[/itex] is a set

    These are the only rules allowed for making sets.

    Now, try it yourself. If [itex]X[/itex] is infinite, try to make an injection [itex]\mathbb{N}\rightarrow X[/itex] using the above rules. You won't be able to do it. That's why we need the axiom of choice.
     
  6. Jul 21, 2012 #5
    If I may ask a question: why the axiom of choice controversial? It seems reasonable to me.

    Nevermind: I just saw the link to the blog; I'll read that first.
     
    Last edited: Jul 21, 2012
  7. Jul 21, 2012 #6
    I'm having a hard time seeing the distinction between an infinite sequence and the ability to construct a sequence of any finite length which we could expand.

    Also wouldn't the property rule work to construct it? Also since each set is is selectable, it seems like the rule which governed the selection of each set would almost give you your function, but maybe this is also equivelent to the axiom of the choice.

    Anyway to put the question in more context, in page 4 of the PDF I linked at the top the author argues he is avoiding the axiom of choice by giving a rule but I am having a difficult time seeing the distinction between what he did and the axiom of choice, since his construction is for an arbitrary but finite length.
     
    Last edited: Jul 21, 2012
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