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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?2012-02-16
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    @BrianM.Scott: Can the direct sum represent both the finite and cofinite subsets?2012-02-16
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    You can safely say that two structures are isomorphic when you have exhibited an isomorphism.2012-02-16
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    The direct product would correspond to the set of _all_ subsets, not just those that are finite or cofinite.2012-02-16
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    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.2012-02-16
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    I wonder whether OP is thinking of "symmetric difference" and intersection. The set is a ring under that pair of operations.2012-02-17

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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.