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Let $G$ be the usual: a topological Abelian group with a topology induced by a countable neighbourhood basis $G_n$ of zero such that $G = G_1 \supset G_2 \supset \dots$. Let $\widehat{G}$ denote the completion of $G$.

In a proof I am using $ \widehat{G/G_n} \cong G/G_n$

(Atiyah-MacDonald, page 105, Corollary 10.4.)

Can you tell me if the following way is the correct way of thinking about it:

We know that $\widehat{G} \cong \varprojlim_n G/G_n$

so that $\widehat{G}$ are all sequences $\vec{g}$ in $G$ with $g_n$ in $G_n \subset G_{n-1} \subset \dots \subset G_1 = G$. Then $\widehat{G/G_n}$ are all sequences in $G/G_n$ with $g_n$ in $G_n/G_n = \{0\}$. Hence $\widehat{G/G_n}$ looks like all sequences that are zero after $n$. But this is just what sequences in $G/G_n$ look like.

I guess I should somehow say "in the inverse limit of $G/G_n$".

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    I'm not sure what other tags could fit here. Feel free to add tags.2012-07-29

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The sequence $\vec g=(g_n)$ should have $g_n\in G/G_n,$ rather than $g_n\in G_n.$ Similarly, an element of $\widehat{G/G_m}$ should be given by a sequence $\vec g=(g_n)$ such that $g_n\in (G/G_m)/(G_n/G_m)$ which is either $G/G_n$ for $n or $G/G_m$ for $n\ge m.$ Since the sequences stabilize, by the definition of inverse limit we see that $\widehat{G/G_m}\cong G/G_m.$ In particular, $\widehat{G/G_m}$ looks like all sequences that stabilize to any particular element $g_n$ of $G/G_m$ for $n\ge m$ (not necessarily $g_n=0,$ although this is one possibility).