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Suppose I have a set along with a relation which is a total order. The set has no minimum element. Now, I want to create an infinite decreasing sequence of elements in this set. For this I have to use axiom of choice. But I'm not able to figure from which family of sets should I pick the elements so that I can use AoC.

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3 Answers

up vote 8 down vote accepted

For each $x\in X$, let $L_x = \{y\in X \mid y\lt x\}$. Take the family $\{L_x\}_{x\in X}$. Let $f$ be an element of $\prod_{x\in X}L_x$.

Now pick $x_1\in X$. Use the recursion theorem to define a function $g\colon\mathbb{N}\to X$ by letting $g(1) = x_1$, and letting $g(n+1) = f(g(n))\in L_{g(n)}$. The sequence you want is $x_i=g(i)$.


In fact, you don't need all of AC to do the above, it suffices to use the Axiom of Dependent Choice which is implied by, but is not equivalent to, the full Axiom of Choice. Showing that you actually need ADC for the general result (as opposed to showing that it suffices to use it) requires arguments along the lines of Asaf's answer.

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Two downvotes, within a minute of each other, neither with an explanation. Someone's voting with their liver... –  Arturo Magidin Sep 26 '11 at 16:21
    
Maybe today is the day of sudden downvotes? math.stackexchange.com/questions/67397/… –  Ilya Sep 26 '11 at 16:27
    
@Gortaur: If it is the same individual, it should be caught by the system as an anomaly (if he does it enough times...) –  Arturo Magidin Sep 26 '11 at 16:36
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Arturo: I believe that DC is more than needed for the existence of such sequence. Axiom of countable choice seems more than enough too. I will have to check "The Bible" for more information. –  Asaf Karagila Sep 26 '11 at 17:36
    
@Asaf, interesting point. I've started two comments now that were supposed to prove that countable choice is not enough, but I found bugs in both at the last moment. Please share your findings if you find anything definite. –  Henning Makholm Sep 26 '11 at 18:09
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Consider all the ordinals which can be embedded into the order inversely, that is order type forming decreasing chains in the linear order.

Let $\alpha$ be the supremum of all ordinals. If $\alpha<\omega$ then there is a minimal element; if $\alpha>\omega$ then we can inversely-embed $\omega$ and have the chain.

If $\alpha=\omega$, then we can use Zorn's lemma as follows:

Take all the decreasing chains ordered by extension. Every chain is bounded since the union of such chain gives a decreasing sequence. Therefore there exists a maximal element. Since we know that a finite chain can be extended by another element (otherwise a minimal element exists) the maximal element has an infinite length.

As Arturo remarked this can be done with Axiom of Dependent Choice, which is a restricted version of Zorn's lemma for partial orders whose height is $\omega$.


To see that the axiom of choice is needed, take the Basic Cohen model, that is add a Dedekind-finite set of reals. This set is totally ordered, it is unbounded and it has no countable subset. In particular we cannot find a countable sequence in this totally ordered set.


The exact choice strength is actually the one suggested by Arturo in the comments as a fragment of the Principle of Dependent Choice.

Let $\operatorname{DC}_{\operatorname{lin}}$ be the assertion that every linear ordering in which every finite chain can be extended has an increasing $\omega$-sequence.

Clearly, $\operatorname{DC}\Rightarrow\operatorname{DC}_{\operatorname{lin}}$, as $\operatorname{DC}$ speaks on non-linear orders as well.

First note that this is obvious that $\operatorname{DC}_{\operatorname{lin}}$ implies that an ordering which is not bounded from below has a decreasing $\omega$-sequence. Inverse the order, as every finite (increasing) sequence can be extended (otherwise we hit a maximum) we have an $\omega$-sequence; inverse the order again and there we have it.

On the other hand, if $\operatorname{DC}_{\operatorname{lin}}$ does not hold then there is a linear order such that every finite increasing sequence can be extended but there is no $\omega$-sequence. In particular the inverse of such order will be a linear ordering with no least element, that has no decreasing $\omega$-sequence.

Question I: Suppose ZF+Exist an amorphous set. Since amorphous sets cannot be linearly ordered they cannot be used as counterexamples of $\operatorname{DC}_{\operatorname{lin}}$.

Can there be a model of ZF+Exists an amorphous set+$\operatorname{DC}_{\operatorname{lin}}$? Can we relax the requirement of amorphous to an infinite Dedekind-finite set?

Question II: Can we have a model of ZF+$\operatorname{DC}{\operatorname{lin}}$+$\lnot\operatorname{AC}\omega$?

That is a model in which ZF holds, $\operatorname{DC}_{\operatorname{lin}}$ holds but there is a countable family of sets without a choice function.

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Not sure which part of this is supposed to do it with only countable choice. For "if x is linearly ordered and infinite, take a countable subset of it", how can you be sure that the countable subset you take still has no minimum element? (For example, what if the original set were $\mathbb R$ and the countable subset your axiom gives you happens to be $\mathbb N$?) –  Henning Makholm Sep 26 '11 at 18:33
    
@Henning: I have to think about it. While the argument itself is good, my intuition tells me it is not true, and if there is a countable subset and no minimal element then we can find one which is decreasing. –  Asaf Karagila Sep 26 '11 at 18:59
    
@Henning: This is a brain teaser, no doubt. Luckily, I am quite free for the next few days and might have some time to sit and think it through. Watch updates as they come! –  Asaf Karagila Sep 26 '11 at 23:03
    
Before falling asleep five hours ago I started coming up with a possible idea. –  Asaf Karagila Sep 27 '11 at 4:07
    
Funny, it was staring at me right in the fact. –  Asaf Karagila Sep 27 '11 at 16:48
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If you want to apply the axiom of choice directly, pay careful attention to what you're choosing:

  • First, you choose an element $x_0 \in X$
  • Second, you chose an element $x_1 \in \{ a \in X \mid a < x_0 \}$

Pay close attention to the second point -- we aren't really choosing an element from a set! Instead, we are choosing from a set-valued function of $x_0$, and so $x_1$ really should be a function of $x_0$!

Now how to apply choice is clear.

  • First, you choose $x_0 \in x$
  • Second, you choose $f_1$ from the subset $S_1$ of functions $X \to X$ satisfying $f(a) < a$
  • Third, you choose $f_2$ from the subset $S_2$ of functions $X \times S_1 \to X$ satisfying $g(a,f) < a$ and $g(a,f) < f(a)$
  • ...

But to make things easier on ourselves, I will simplify:

  • First, you choose $x_0 \in X$
  • Second, you choose $f_1 : X \to X$ satisfying $f(a) < a$
  • Third, you choose $f_2 : X^2 \to X$ satisfying $f(a,b) < a$ and $f(a,b) < b$
  • ...

Now, each of our sets is completely determined. We can show they are non-empty (by invoking induction and choice), and so we can apply choice to the collection.

Finally, our sequence is:

  • $x_0$
  • $x_1 = f_1(x_0)$
  • $x_2 = f_2(x_0, x_1)$
  • ...

EDIT: just to make it explicit, the simplified version defines the sets $S_n$ to be the set of functions $g : X^n \to X$ satisfying $g(\vec{x}) < x_i$ for all $1 \leq i \leq n$.

(Depending on your tastes, you may want to special case $S_0 = X$)

For a choice $g \in S_n$, the value $g(\vec{x})$ is meant to be interpreted as "the choice I would make at at time $n$ if I had previously made the choices $x_0, x_1, \ldots, x_{n-1}$ at times $0, 1, \ldots, n-1$".

Now apply the axiom of choice to get a family of choices $f_n \in S_n$. Under the intended interpretation, our chosen decreasing sequence is

$$x_n = f_n(x_0, x_1, \ldots, x_{n-1})$$

(If you're special casing $0$, then $x_0 = f_0$ instead)

Of course, while this approach is systematic, it did not lead to the simplest result. At some point we might have recognized that our choice at time $n$ could be made to depend only on the choice made at time $n-1$ rather than the entire family of choices made at times $0$ through $n-1$. And if we do that, we might have then realized a single choice would work for all positive times. This would result in Arturo's answer.

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How is this answer different from Arturo Magidin's answer? If you want to apply the Axiom of Choice directly, you have to come up with a family from which you choose elements. It seems to me that really you use the same family as Arturo and the same recursion to define the sequence, just in a slightly less explicit way. –  Stefan Geschke Sep 27 '11 at 5:20
    
My intent was not just to answer the question posed, but to show a thought process that would lead to the answer. –  Hurkyl Sep 27 '11 at 19:46
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Yes, I understand that. But I think the difficulty in this argument is to get the actual choice function from your thought process. And Arturo very cleanly states for which family of set you choose elements. This is somewhat hidden in your answer. –  Stefan Geschke Sep 29 '11 at 5:30
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