Talk:Freiling's axiom of symmetry

message: The page says:
:Given a function f in A, and some arbitrary real numbers x and y, it is generally held that x is in f(y) with probability 0, i.e. x  is not in f(y) with probability 1. 
How is this probability counted? My intuity would be that x is in f(y) with chance 0.5, since my method of getting a random subset of real numbers would be by giving each real number chance 0.5 to be in the set. Apparently there is another way of defining 'random' set of real numbers being used here, but which is it? Andre Engels 14:11, 27 Jan 2004 (UTC)
replies: text: f(y) is not a "random" set of real numbers at all (the article says nothing about f being random).  However, it is countable; this means that f(y) covers a very small fraction of all possible real numbers and that a "random" real number will be in f(y) (or any other countable set) with probability 0. Cwitty 22:15, 27 Jan 2004 (UTC)
replies: text: Oops... I missed that 'countable' in the definition. Thanks. Andre Engels 00:50, 29 Jan 2004 (UTC)

message: I seem to recall that this appeared in the Journal of Symbolic Logic in about 1985.  I'll add the reference if I find it, unless someone beats me to it. Michael Hardy 23:34, 23 Jan 2005 (UTC)
thread_title: Reference?

message: The article claims the argument is based on "Stewart Davidson"'s intuition.  Who is he?  --Aleph4 18:04, 28 May 2006 (UTC)
replies: text: Never mind.  Stuart Davidson is mentioned in the abstract of Freiling's article. --Aleph4 19:13, 28 May 2006 (UTC)
thread_title:  Stewart Davidson  

message: If we replace "countable" by any other statement which implies "of Lebesgue measure zero" we will still get probability = 1 that x is not in f(y) and that y is not in f(x). I do not see any contradiction with the axiom of choice. Leocat 21:26, 21 October 2006 (UTC)
replies: text: Fix a well-ordering of the continuum <. Let f map its argument to the set of all smaller elements under <. Now for x and y, either x is in f(y) or y is in f(x). So by AXI it must be that there is some x for which f(x) is not of zero Lebesgue measure, even though its cardinal less than the continuum. Technically, I don't see any contradiction with the axiom of Choice, though. If we assume that f(x) is always a measurable set, we can repeat the above argument on f(x) and thereby set up an infinite descending sequence of ordinals. Ben Standeven 05:37, 8 April 2007 (UTC)
thread_title: Where is the contradiction with the axiom of choice?

message: Since the discussion in the section "connection with random forcing" is intuitive, like much of Freiling's argument, it isn't possible to establish properties like countable additivity from axioms. The formal way to argue in set theory is to do something equivalent to Solovay forcing. But having said that, an intuitive property implies that the dart throw measure is countably additive.
The important thing to realize is that the dart throws can be used simultaneously to find the measures of a collection of disjoint sets and that of the disjoint union, by throwing the dart at the interval and seeing which of the sets it landed in. Either it landed in one of the sets, or else it landed in the complement, and since the sum of the probability of all the possibilities has to equal one, the measure of the complement of the disjoint union must equal 1 minus the sum of the measures of the disjoint sets. This is countable additivity. The probability of landing in the complement is 1 minus the probability of landing in the set.
Despite that, I just wrote a bogus argument! I believe it is easy, but I made a mistake just now.128.84.241.136 (talk) 00:05, 8 January 2008 (UTC)
replies: text: Well, one other way you can prove that is to define the measure "explicitly", i.e. outer measure rather than Borel measure.  Also, (I think) ZF doesn't prove that Lebesgue measure is countably additive, because (I think) it is consistent that the continuum is a countable union of countable subsets.
JumpDiscont (talk) 17:27, 30 September 2009 (UTC)
thread_title:  Countable Additivity of a Dart-Throw Measure 

message: It sure would be nice to have an expanded explanation of why the argument in the second bullet in the section "Objection to Freiling ..." leads to a contradiction. Even one more sentence.
I'm sure the statement that it leads to a contradiction is true, because I trust the Wikipedia math author's, but, I sure don't see it.DeaconJohnFairfax (talk) 21:17, 16 July 2008 (UTC)
thread_title:  "Objection to Freiling" section needs more words. 

message: The introduction says that AX is equivalent to the negation of CH.  There's a section showing one direction of this: AX implies the negation of CH, shown indirectly by the contrapositive: CH implies negation of AX.  The article ought to show the other direction, if only for completeness.  I'm having trouble seeing this on my own. Perhaps this is because I'm trying to show that the negation of one of the axioms, which amounts to the existence of hypothesized peculiar set or function, implies the other axiom, which says something universal about all sets or functions.  In other words, I don't see how to jump from one bad object to an assertion about a huge bunch of objects.  I'd be really interested to see this.  DRLB (talk) 17:07, 25 June 2009 (UTC)
replies: text: Here's a hint: First show that not AX implies that there is a total order on the reals where each element has countably many predecessors; then show that this implies the continuum hypothesis. (The former part requires a cardinality argument, the latter part requires the Axiom of Choice.)Ben Standeven (talk) 22:06, 15 July 2009 (UTC)

text: Hey DRLB, it's been up there for a while, I'm sure you've seen it by now. It's now been more generalised. Note that it is not uncommon in mathematics that the existence of one thing can open up the gateway to the existence of loads of things (typically the machinery behind this is Zorn's lemma). —drusus null 00:28, 10 January 2011 (UTC)
thread_title:  How does negation of CH show that AX fails? 

message: How about adding an example to the first section: 
Let x = 1/2 and let y = π/4. Let B be the subset of A for which f(x) = {1/2, 1/4, 1/8, ...} and for which f(y) = {1/2 + π/10, 1/4 + π/10, 1/8 + π/10, ...}. Then x is not an element of f(y) because all elements of f(y) are irrational and x is not. Likewise, y is not an element of f(x) because all elements of f(x) are not irrational while y is. Thus for the subset B of A the statement of AX is provably true.
This decidedly would have made it immediately clear to me what AX says. [I was the one complaining about the notation above.] 
A second example:
The statement "For every f in A, there exist x and y such that x is not in f(y)" is simply true. This is because f(y) is by definition countable and can't possibly contain the uncountably many real numbers x in [0,1]. The problem arises when we in addition require that y is not in f(x) for the same x and y.
YohanN7 (talk) 01:30, 2 September 2009 (UTC)
thread_title:  Example 

message: There are a couple of questionable things here. First is that given an arbitrary sequence x:\omega\to[0,1]\,, then for any subset S\subseteq[0,1]\,, there is no guarantee that \frac{|\{i<n:x_{i}\in S\}|}{n}\, will converge (in fact given an S\,, one can find continuum many sequences where this won't converge). Second is that, even given a sequence x:\omega\to[0,1]\, for which \mu_{x}(S):=\operatorname{lim}_{n}|\{i<n:x_{i}\in S\}|/n\, exists for all S\subseteq[0,1]\,, there is no guarantee that this defines a countably additive measure, let alone translation invariant. I actually agree with the intuition that this should work, but in keeping with what is the case, one can only at best say "it seems likely that this sets up a well-defined, countably additive, rotation-invariant… probability measure".
—drusus null 01:02, 10 January 2011 (UTC)
thread_title:  Connection with Solovay's forcing 

message: There is an interesting recent text by Hamkins dealing with this problem.  Is it worth including it? Tkuvho (talk) 13:05, 12 January 2014 (UTC)
thread_title:  Text by Hamkins 

message: I will edit it in if there are no objections.
Assume that 2^{\kappa} \neq \kappa^{+}. Then 2^\kappa > \kappa^+. Let A be a subset of 2^\kappa such that |A|=\kappa^+. There are at most \kappa^+ \cdot \kappa = \kappa^+ elements m of 2^\kappa such \exists y \in A. m \in f(y). Let x be an element of 2^\kappa such that this is not true. Likewise, there are only \kappa elements n of A such that n \in f(x). Let y be an element of A such that this is not true. Therefore x \notin f(y) and y \notin f(x). Therefore Freiling's axiom is true.
(Credit goes to Eric Wofsey (as part of another argument).)
TheKing44 (talk) 21:05, 17 November 2017 (UTC)
thread_title:  Simpler proof of Part (ii) 

message: I heard about Freiling's axiom in the PhD thesis of the Dutch logician Michiel van Lambalgen, which was published a year or two before Freijling's paper. Looks like several brilliant people had the same idea about the same time. https://www.illc.uva.nl/Research/Publications/Dissertations/HDS-08-Michiel-van-Lambalgen.preamble.pdf Richard Gill (talk) 01:35, 20 March 2021 (UTC)
thread_title:  Earlier work by Michiel van Lambalgen?