Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

On Page 60, Set Theory, Jech(2006),

5.9 If $\{X_i : i \in I\}$ and $\{Y_i : i \in I\}$ are two disjoint families such that $|X_i| = |Y_i|$ for each $i \in I$, then $|\cup_{i \in I}X_i| = |\cup_{i \in I}Y_i|$ [Use AC]

Here's how far I goes:

$|X_i| = |Y_i|$ implies there exists a bijective function $f_i: X_i \to Y_i$ for each $i \in I$ ex ante. Since $\{X_i : i \in I\}$ is a disjoint family, for each $x \in \cup_{i \in I}X_i$, there exists exactly one $i \in I$, such that $x \in X_i$. So we could define a bijective function $f:\cup_{i \in I}X_i \to \cup_{i \in I}Y_i$, by $f(x)=f_i(x)$, if $x \in X_i$.

I don't see any usefulness of AC in problem 5.9, as opposed to problem 5.10, in which $|\cup_{i \in I}X_i| = |\cup_{i \in I}Y_i|$ is replaced by $|\prod_{i \in I}X_i| = |\prod_{i \in I}Y_i|$. The reason is that without AC, the cardinality of a cartesan product of non-empty sets is arbitary.

share|improve this question
    
Try to give meaningful titles. 'Why Bother Axiom of Choice' tells me nothing about the question inside... –  Asaf Karagila Dec 1 '12 at 11:30
    
@AsafKaragila: No problem. I should have thought of that. –  Metta World Peace Dec 1 '12 at 11:35
    
The title actually made me think the content is more philosophical in nature (e.g. why do we need the axiom of choice, or why do we bother to mention/verify when it is used). To these questions I have some long answers posted already and I was actually setting my mind to search for them in order to point the duplicate... :-) –  Asaf Karagila Dec 1 '12 at 12:05

2 Answers 2

up vote 3 down vote accepted

You have to choose bijections for every $i$. It is consistent that there is a family $\{P_i\mid i\in\omega\}$ of disjoint pairs which does not have a choice function on any infinite subfamily.

One can show (quite easily too) that $\bigcup P_n$ is uncountable and in fact Dedekind-finite.

However $|P_n|=|\{2n,2n+1\}|$ whereas $|\bigcup P_n|\neq|\bigcup_{n\in\omega}\{2n,2n+1\}|=|\omega|=\aleph_0$.

share|improve this answer
    
Russell sets are so wonderful. –  Asaf Karagila Dec 1 '12 at 11:38
    
Thank you for counterexample and the little excercise which I have to spend some time to figure out why. They are so nice. –  Metta World Peace Dec 1 '12 at 12:00
    
@Metta: I actually wrote something incorrect. It is consistent to have partial choice in the sense that there is an infinite subfamily which do have a choice function, in which case the union of the pairs is Dedekind-infinite, but the corrected version is correct, if no infinite family of pairs have a choice function then the union is Dedekind-finite. –  Asaf Karagila Dec 1 '12 at 12:01

For each $i\in I$ there are in general many bijections from $X_i$ to $Y_i$, so you’re using the axiom of choice when you pick a specific bijection $f_i$ for each $i\in I$.

Specifically, for each $i\in I$ let $B_i$ be the set of bijections from $X_i$ to $Y_i$. Then $\{B_i:i\in I\}$ is a non-empty family of non-empty sets, and you want to pick a set $\{f_i:i\in I\}$ such that $f_i\in B_i$ for each $i\in I$. In order to do that with no further information, you need the axiom of choice.

share|improve this answer
    
Thanks, I got it. –  Metta World Peace Dec 1 '12 at 11:18
2  
@MettaWorldPeace: You’re welcome. (That kind of use of AC is easy to overlook, especially when you’re just learning.) –  Brian M. Scott Dec 1 '12 at 11:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.