# Set of finite and cofinite subsets of the integers

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 $\mathbb Z_2$?

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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
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

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.