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.

Let $a_i$ be a cardinal number for every $i \in\ I$.

Let $\{A_i\}$ and $\{A'_i\}$ be families of sets and let $A_i$, $A'_i$ and $a_i$ be equipotent for every $i \in I$.

Then show that $\prod_i\ A_i $ is equipotent with $\prod_i\ A'_i $.

This seems obviously true but I don't know how to actually show the bijection between them..

share|improve this question
1  
Hint: Since you are given that $A_i$ and $A_i\!'$ are equipotent for each $i$, you can assume that you have a collection of bijections $f_i$, where $f_i$ is a bijection between $A_i$ and $A_i\!'$. These are your raw materials. Your job is to figure out how to build the bijection for the product out of these raw materials. –  MJD May 14 '12 at 18:59
1  
It really feels that you are afraid to try things on your own. You should know that without chewing a lot on problems and trying hard you will not learn how to do this on your own. I say that because you ask questions which often appear in a rather similar form on this site before, and you often accept the answer in minutes. This signals me that you haven't really tried to tackle the problem on your own before asking. –  Asaf Karagila May 14 '12 at 19:27
    
Note that this is basically a reformulation of this question –  Martin Sleziak May 15 '12 at 7:22

1 Answer 1

up vote 4 down vote accepted

For each $i\in I$ you have a bijection $\varphi_i:A_i\to A'_i$. Define $$\varphi:\prod_{i\in I}A_i\to\prod_{i\in I}A'_i:\langle a_i:i\in I\rangle\mapsto\Big\langle\varphi_i(a_i):i\in I\Big\rangle$$ and prove that it’s a bijection.

This is what I call a follow-your-nose proof: there really is only one reasonable thing to try. All you’re given is the equipotence of $A_i$ and $A'_i$ for $i\in I$. All that gives you is the existence of the bijections $\varphi_i$, so either the result is very hard (unlikely) or somehow it must be possible to use those bijections to get the one that you want. Since a typical element of $\prod_iA_i$ is just a function $\langle a_i:i\in I\rangle$ from $I$ to $\bigcup_iA_i$, about the only thing to try is to apply the bijections $\varphi_i$ to the components of $\langle a_i:i\in I\rangle$.

share|improve this answer

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.