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.

Can I safely say that the set of finite and cofinite subsets of the integers equipped with operations of union and intersection is isomorphic to the direct product of countably infinitely many $\mathbb Z_2$?

share|improve this question
2  
Definitely not: the direct product of countably infinitely many copies of $\mathbb{Z}_2$ is uncountable, but the set of finite and cofinite subsets of the integers is countable. Did you mean the direct sum? –  Brian M. Scott Feb 16 '12 at 23:36
    
@BrianM.Scott: Can the direct sum represent both the finite and cofinite subsets? –  help Feb 16 '12 at 23:41
1  
You can safely say that two structures are isomorphic when you have exhibited an isomorphism. –  André Nicolas Feb 16 '12 at 23:42
    
The direct product would correspond to the set of all subsets, not just those that are finite or cofinite. –  Michael Hardy Feb 16 '12 at 23:50
    
Quite possibly not $-$ I’ve not thought much about it $-$ but at least it has the right cardinality, unlike the direct product. However, it occurs to me that you have another fatal problem: neither union nor intersection is a group operation. You have identities for each, but no inverses. –  Brian M. Scott Feb 16 '12 at 23:51
show 1 more comment

1 Answer

Let $\mathscr{S}$ be the set of subsets of the integers that are either finite or cofinite. $\langle\mathscr{S},\cup,\cap\rangle$ is not a ring: $\varnothing$ is an identity for $\cup$, and $\mathbb{Z}$ is an identity for $\cap$, but neither operation has inverses. Thus, $\langle\mathscr{S},\cup,\cap\rangle$ cannot be isomorphic to any ring; in particular, it cannot be isomorphic to the direct sum $$\bigoplus_{n\in\mathbb{N}}G_n\;,$$ where each $G_n$ is a copy of $\mathbb{Z}_2$. It cannot be isomorphic to the direct product $\mathbb{Z}_2^\omega$ for an even more fundamental reason: it’s countable, and $\mathbb{Z}_2^\omega$ is uncountable.

share|improve this answer
add comment

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.