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Inspiration for the following question comes from an exercise in Spivak's Calculus, there too are considered finite sets of real numbers in interval $[0,1]$ but in completely different setting. I will state formulation of the question and my attempt to solve it. I should note that all my knowledge of set theory mostly comes from reading wikipedia, math.SE, and some abstract algebra textbook introductions, so I have no "real" knowledge of it.


Consider collection of sets $A_i$ for every natural number $i$ such that every set in that collection contains a finite amount of real numbers in $[0,1]$. Then the question is: what is the cardinality of set $C = \bigcup_{i=1}^{\infty} A_i$


My intuition says that it is equal to cardinality of integers. My first thoughts of bijection were bijecting all the elements in sets with particular subsets of rational numbers, but then I thought of way that almost seems too easy: Biject first $|A_1|$ natural numbers with elements of $A_1$, biject elements of $A_2$ with next $|A_2|$ numbers, in general, biject elements of $A_n$ to natural numbers from $k = \sum\limits_{i=1}^{n-1}|A_i| + 1$ to $k + |A_n|$. Does this resolve the issue?

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Yes, it really is that easy. –  Qiaochu Yuan May 29 '11 at 15:23
3  
Yes and no. It solves the problem only if $A_i$ are pairwise disjoint. What happens if some elements repeat? What if all $A_i$ are the same? ;) –  N. S. May 29 '11 at 15:24
    
@user9176: Yeah, I was mostly interested in proving it for case when all sets are disjoint. If all $A_i$ are same, then it is obviously finite number $|A_n|$, for any $n$. If some elements repeat, then if they repeat finitely often and we require that every set has at least one member (because finitely many could mean 0), then it also has the cardinality of natural numbers. If they repeat infinitely.. then both options are possible, right? –  user5501 May 29 '11 at 15:40
    
Right. Why do Wikipedia, math.SE and abstract algebra textbook introductions not impart "real" knowledge of set theory? :-) –  joriki May 29 '11 at 16:35
    
@joriki: Well, I used the word 'real' in quotation marks in lieu of a better word, but what I was trying to convey was that my knowledge isn't complete even at elementary level. If it was, I guess I would have had more confidence in my result. –  user5501 May 29 '11 at 16:43

1 Answer 1

up vote 2 down vote accepted

The fact that the elements of $A_i$ are real numbers in $[0,1]$ is irrelevant. You only need to know that each $A_i$ is finite.
Also, repetitions are easy to handle: if an element of $A_i$ has already occurred in $A_j$ for some $j < i$, just use the value that you already gave it.
Finally, the union of any countable collection of countable (not just finite) sets is also countable. But this requires the axiom of (countable) choice.

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Yeah, I know it is irrelevant, I just used it as it was used in problem that inspired me. Thanks for the answer, I didn't know that last fact. –  user5501 May 29 '11 at 20:25
    
I must complain, you do not need countable choice to prove that every countable union of countable sets is countable. Every step in such a proof can be made explicit. –  Alexander Thumm May 30 '11 at 6:09
    
@Alexander: Yes you do. All you know about each $A_i$ is that there exists at least one bijection from $A_i$ to $\mathbb{N}$. To prove that the union of all the $A_i$ is countable, you need to explicitly choose such a bijection for each $A_i$. –  TonyK May 30 '11 at 19:59
    
Well, I must admit, you are right. Thank you for clarifying this. –  Alexander Thumm May 30 '11 at 21:24
3  
You need choice in general to show that a countable union of finite sets is countable. This is not necessary here, though, since the ordering of $\mathbb{R}$ gives explicit well-orderings on each finite set. –  Chris Eagle May 30 '11 at 21:49

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