April 15th, 2017, 03:40 PM  #21 
Senior Member Joined: Jun 2014 From: USA Posts: 363 Thanks: 26  For what? I'm admittedly a novice mathematician at best. I'm a certified public accountant with a Master's in taxation. I'm up to my eye balls in work right now. I don't see how this is that big of a deal or breaks Math though. If that's the case I'm probably wrong. I will post it though and you guys can pick it apart. 
April 15th, 2017, 03:52 PM  #22 
Senior Member Joined: Aug 2012 Posts: 1,887 Thanks: 522  
April 15th, 2017, 05:29 PM  #23  
Senior Member Joined: Aug 2012 Posts: 1,887 Thanks: 522  Quote:
A countably additive, uniform probability measure must assign zero or infinity to countable sets. If you just tell me whether you understand that or not, it would be helpful to know. It's just a little logic puzzle. You assume a couple of things and then logically deduce another thing. So that as a matter of logic, you simply can't have a uniform probability measure on a countable set, because probability measures have to add up to 1; and any uniform measure on the naturals would have to add up to zero or infinity. Once you understand that, you're in a better position to analyze your own argument; and see where you might be assuming something that's not justified. That doesn't mean your idea is wrong. It DOES mean that either: a) You are wrong; or b) Math is busted. You discovered an inconsistency. You will be famous. Now please note, I myself am a very openminded person. Option (b) is perfectly possible. It's very unlikely, but it's possible. After all, we don't even know for sure if set theory is consistent. If the whole thing blows up tomorrow morning, people will be shocked but not entirely surprised. If set theory doesn't work out they'll find another foundation and most mainstream math won't even be affected. So I don't judge. I do say that (a) is a lot more likely; and that in any event, that's where we should focus our efforts. To that end, I propose focussing on the nature and definition of probability measures. This is not a bad place to start ... https://en.wikipedia.org/wiki/Probability_measure Last edited by Maschke; April 15th, 2017 at 05:35 PM.  
April 15th, 2017, 06:38 PM  #24 
Senior Member Joined: Aug 2012 Posts: 1,887 Thanks: 522 
ps  You know what? I just thought about this and realized where your error is. I don't think I realized this last time we talked about it. We take $\mathbb R / \mathbb Q$. Actually we don't start with all of $\mathbb R$, just the unit interval (open or closed doesn't matter). [To do that we have to define "addition mod 1" or else translate the problem to the unit circle. Let's ignore that for now because it's not relevant.] Another way to express this is that we define an equivalence relation on the reals where we say that $x \equiv y$ if $x  y \in \mathbb Q$. This is an equivalence relation. Each equivalence class is countable. For example the rationals are one equivalence class and the rationals are countable. Other equivalence classes are $\mathbb Q + \pi$, $\mathbb Q + \sqrt{2}$, and so forth. But note that there is some collapsing, for example if $x \equiv y$ then $\mathbb Q + x = \mathbb Q + y$. So we've partitioned the reals (or the unit interval) into an uncountable collection of equivalence classes, each one countable. Now this is no good for your idea of picking a random real and mapping it back to some natural number, because there are uncountably many equivalence classes. Now in the construction of the Vitali set, what we then do is use the Axiom of Choice to form a set $V$ that contains exactly one element of each equivalence class. This is an uncountable set. NOW  this is the part you're leaving out, and I'm sorry I didn't realize this was going on before  we can show that the collection of RATIONAL TRANSLATES of $V$ also covers the reals, but this time as a countable collection of uncountable sets. But now, the set $V$ must either have measure $0$ or infinity, yet the countable union of its translates is the unit interval, which has measure $1$. In other words we have just proved that $V$ is not measurable and neither are any of its translates. So when you pick a random real and then see which translate it's in, you can not possibly assign a sensible probability of landing in that translate. You have actually done something very clever. You actually proved the existence of nonmeasurable sets. That's really good. The bad news is that it falsifies your idea. But you should be encouraged, you have great insight into a piece of advanced math that's usually shown to beginning grad students. I apologize if I got about three steps ahead of our conversation, but I've been reviewing this lately in connection with the BanachTarski paradox, and now it all came back to me. By the way this is really a continuation of a conversation we had in another thread, so if this is too far offtopic for the present thread, we should probably take this to a different thread. On the other hand the general topic is the paradoxes of measure, and this is certainly in that ballpark. Also by the way I walked through that proof really fast and skipped a lot of details and used some fancy notation, because I wanted to get the key idea out. But if anyone is interested, I'd be glad to walk through any part of this in detail. There are nonmeasurable sets. That's at the heart of BanachTarski and a lot of other measurerelated anomolies. Historical note, Giuseppe Vitali was the first person to think these thoughts and write them down, in 1905. Here's Wiki's version of the proof, which adds some detail I left out. https://en.wikipedia.org/wiki/Vitali_set Last edited by Maschke; April 15th, 2017 at 07:04 PM. 
April 15th, 2017, 07:32 PM  #25 
Senior Member Joined: Sep 2015 From: USA Posts: 1,931 Thanks: 999  sorry for the obscure reference. It's what the cartoon character Wimpy used to say when trying to borrow money for a hamburger and refers to something we'll never see. 
April 16th, 2017, 05:39 AM  #26  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,305 Thanks: 2443 Math Focus: Mainly analysis and algebra  Quote:
Your sketch of a proof, even without your typing errors, doesn't make sense. The axiom of choice says nothing about probability or probability distributions so your argument is circular.  
April 16th, 2017, 06:22 AM  #27 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,591 Thanks: 546 Math Focus: Yet to find out.  If i had a dollar for every time this is stated...

April 16th, 2017, 06:53 AM  #28  
Senior Member Joined: Jun 2014 From: USA Posts: 363 Thanks: 26  Quote:
 
April 16th, 2017, 10:16 AM  #29 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  
April 16th, 2017, 03:05 PM  #30 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,305 Thanks: 2443 Math Focus: Mainly analysis and algebra  I did later realise that you were highlighting that it makes no sense to say that $\frac1\infty=0$, but my internet connection has been too poor to do anything about it. Your method of demonstrating the fact and/or your reason why it makes no sense couldn't be worse in my opinion because it perpetuates the false idea that algebraic manipulation with the symbol $\infty$ has some validity. In fact it makes as much sense as trying to correctly decline the verb "gluft"  it uses rules that don't apply because because the object they should operate on doesn't exist.


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