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According to a modified answer of this question, the direct limit of the sequence $$ \mathbb{Z}\xrightarrow{1}\mathbb{Z}\xrightarrow{2}\mathbb{Z}\xrightarrow{3}\mathbb{Z}\xrightarrow{4}... $$ in the category of abelian groups is $\mathbb{Q}$. What is the inverse limit of the system $$ ...\xrightarrow{4}\mathbb{Z}\xrightarrow{3}\mathbb{Z}\xrightarrow{2}\mathbb{Z}\xrightarrow{1}\mathbb{Z}, $$ i.e. is this an abelian group known under a different name? What about the inverse limit of $$ ...\xrightarrow{5}\mathbb{Z}\xrightarrow{3}\mathbb{Z}\xrightarrow{2}\mathbb{Z}\xrightarrow{1}\mathbb{Z} $$ (i.e. multiplying by the primes)?

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    $\begingroup$ The zero group? If $(a_1,a_2,...)$ is an element then $a_1=2a_2$, $a_2=3a_3$, $a_3=4a_4$, ... If one of the components is zero then all the ones before are zero. If one component is zero, then all the ones before are non-zero. If all components are non-zero, then the first component is $2\times 3\times 4\times ...$. Only the zero sequence satisfies all this. $\endgroup$
    – OR.
    Nov 14, 2013 at 16:22

1 Answer 1

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The problem is somewhat more obvious if you take the diagram

$$ ...\xrightarrow{4}\mathbb{Z}\xrightarrow{3}\mathbb{Z}\xrightarrow{2}\mathbb{Z}\xrightarrow{1}\mathbb{Z}$$

and replace it with the isomorphic diagram

$$ ...\xrightarrow{1}6\mathbb{Z}\xrightarrow{1}2\mathbb{Z}\xrightarrow{1}\mathbb{Z}\xrightarrow{1}\mathbb{Z}$$

and so you're taking the nested intersection of a sequence of subsets of $\mathbb{Z}$.

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