
Number Theory Number Theory Math Forum 
 LinkBack  Thread Tools  Display Modes 
April 12th, 2017, 06:15 PM  #1 
Senior Member Joined: Jun 2014 From: USA Posts: 210 Thanks: 6  Countable Union of Countable Sets Question
I am hoping for a better understanding of what a countable union of countable sets means. I have noted that Feferman and Levy famously showed that the reals can be a countable union of countable sets long ago. We can also have an uncountable set that is the countable union of sets of size two (e.g., Russell cardinals). I understand that the axiom of choice plays into this (it fails in Feferman and Levy's model noted above) and in ZFC the well ordering theorem is implied which in turn requires that each cardinal have an initial ordinal. I get that choice may be needed to have an enumeration of each countable set in a countable union. To clarify, if an enumeration is given for each countable set, then an enumeration of their union can be constructed. Alternatively, if no such enumeration is given for each countable set, then choice can be used to select an enumeration instead. What I don't understand is that, where the reals are not countable and neither is $w_1$, how can the reals be a countable union of countable sets? A layman's explanation is the best I'll be able to understand... I'm no math whiz. Thank you! 
April 12th, 2017, 07:42 PM  #2  
Senior Member Joined: Aug 2012 Posts: 1,165 Thanks: 258  Quote:
I Googled FefermanLevy and found this Stackexchange thread. This is advanced set theory. Unless you happen to be knowledgeable in the modern settheoretic technique of forcing, you can't even begin to understand the statement of the theorem. I have just barely enough knowledge to know exactly what I'd need to do in order to grok that Stackexchange thread: Spend a year reading Kunen and Jech, and working at it. In another universe some version of me is doing exactly that. I take comfort in that thought. But what this is all about is basically alternate models of the real numbers. Just as Euclidean and nonEucidean geometry are each models of Euclid's axioms minus the parallel postulate; in the same way, there can be many different models of set theory and the real numbers. You may recall that Gödel showed that the axiom of choice and the Continuum hypothesis were consistent with the rest of set theory back in 1934 or so. Then in 1964 Paul Cohen showed that their negations were consistent as well. In so doing, he cooked up a model of the axioms in which they were false. But that wasn't the most important thing he did. His real achievement was to show how to "cook up" wild models of set theory in order to study independence results. He invented the technique of forcing. Today, forcing is highly abstract and set theorists casually talk about this type of forcing and that type of forcing. This is some of the background material to read that Stackexchange thread. Now the point  and I did promise this would be short  is that they cooked up a crazy model of set theory in which the real numbers, defined in the usual way but with respect to crazy set theory and not usual set theory, happen to be a countable union of countable sets. The purpose of cooking up a model like this is to show that even though the axiom of choice is considered weird and troublesome, without it things are weird and troublesome too. That's why people study this stuff. I hope this is helpful. Tell me if this is the kind of answer you had in mind. Just think nonEuclidean geometry, it's the same thing but for set theory. On the other hand if I got this wrong and you're an expert on this stuff and asking a highly technical question, my apologies, but in that case you should try this out on Stackexchange. They have professional set theorists there, it's incredible what they know. Last edited by Maschke; April 12th, 2017 at 07:57 PM.  
April 12th, 2017, 08:04 PM  #3 
Senior Member Joined: Aug 2012 Posts: 1,165 Thanks: 258 
ps  I put this in a separate post because it's a different answer, shorter and more to the point. Nobody said the reals are countable in this model. In fact the reals are always uncountable, that's a theorem of ZF. It's Cantor's theorem. No Choice needed. The theorem that fails is that a countable union of countable sets is countable. That actually requires Choice. That's not commonly realized. In the usual "snaking diagonals" proof of the countability of the rationals, we don't need choice. That's because we can define an explicit order on the rationals. But in general, here is the problem. Suppose I have a countable collection of countable sets. Since each set is countable, there is a nonempty set of bijections between that set and the positive integers. We know there's at least one, but in fact there may be many. So now ... to get the snaking idea to work, we have to choose a bijection from each of infinitely many nonempty sets! Bam, we just used at least countable choice. That's weaker than full Choice but stronger than ZF. elementary set theory  Prove that the union of countably many countable sets is countable.  Mathematics Stack Exchange So in the absence of Choice, we should expect that there's a countable union of countable sets that's not countable. Because we can see that we made essential use of countable Choice to prove that a countable union is countable. So in this funny model of set theory, the reals are uncountable AND they are a countable union of countable sets. I hope this answer helps too! Last edited by Maschke; April 12th, 2017 at 08:07 PM. 
April 13th, 2017, 05:50 PM  #4  
Senior Member Joined: Jun 2014 From: USA Posts: 210 Thanks: 6  Quote:
Quote:
That helps to put it into perspective, yes. I find it fascinating that something as simple as the axiom of choice can be so powerful. Quote:
Quote:
I think the necessary 'piece' for my understanding is that the reals cannot be expressed as a countable union of countable sets in any model where choice holds true, which I assume to be the obvious case because, if they could be expressed as such in a model where choice holds up, then they would have to be enumerable. Sound correct?  
April 13th, 2017, 06:11 PM  #5  
Senior Member Joined: Aug 2012 Posts: 1,165 Thanks: 258  Quote:
On the other hand, the theorem that a countable union of countable sets is countable depends in a subtle way (usually not mentioned) on countable choice. So in this funny model (funny odd not funny haha), the reals are still uncountable, but they're a countable union of countable sets. It's often claimed that because the axiom of choice leads to weird results, we should reject it. It's not commonly realized that its absence leads to equally weird results. Last edited by Maschke; April 13th, 2017 at 06:16 PM.  

Tags 
countable, question, sets, union 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Countable and uncountable sets  Loren  Number Theory  32  September 12th, 2016 10:09 AM 
Countable Union proof  jstarks4444  Applied Math  1  May 13th, 2011 08:15 AM 
Sets, Countable and Uncountable  Bernie  Applied Math  3  September 9th, 2009 01:28 AM 
I have another one, Countable sets.  Fra  Algebra  5  February 14th, 2008 05:04 PM 
Countable Union proof  jstarks4444  Number Theory  1  January 1st, 1970 12:00 AM 