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on page 102 of Atiyah and MacDonald's "Introduction to Commutative Algebra", they state that if $G$ and $H$ are abelian topological groups and $f$ is a continuous homomorphism from $G$ to $H$, then the image under $f$ of a Cauchy sequence in $G$ is a Cauchy sequence in $H$.

But I cannot see how this is true. This is true if we have a uniformly continuous map (between topological spaces where we have notion of uniformity), but what about here?

In fact, taking the multiplicative group of positive reals with the usual topology as $G$ and $H$, $f$ as the inverse map and $\{1/n: n=1,...\}$ as the Cauchy sequence in $G$, then its image in $H$ is the positive integers, which is not a Cauchy sequence. So the claim does not appear to be true, in general for topological groups.

What am I missing?

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Note that topological groups are uniform spaces. We can define $(x,y)\in U\iff y\in U+x=x+U\iff y-x\in U$ where $U$ is a neighborhood of $0$. This way the neighborhoods of $0$ become the entourages of a uniform structure. – Stefan Hamcke Jul 29 '13 at 13:52
How is topological group defined in that text? It is usually defined without a metric, so it seems weird to talk about "Cauchy sequences" in them. You can define Cauchy sequences, either by proving there is a metric, or by using replacing $\epsilon$ with open neighborhoods of $1$ and translating them... – Thomas Andrews Jul 29 '13 at 13:57
@Stefan thanks for that. That was helpful. I saw that on wikipedia, but couldn't find mention of uniform spaces in my copy of Munkres, so lacked a proof. However, the following link has useful references. And the problems with my example were caused by me mixing my group operations. Thanks again! – Pi314 Jul 29 '13 at 20:22

If you have a metric on a topological group, you want uniformity: $$d(x,y)=d(ax,ay)$$ for all $a,x,y\in G$. So your metric on $\mathbb R^+$ is not uniform.

Alternatively, you can define Cauchy sequence in terms of neighborhoods of $1$ rather than $\epsilon$. A sequence $\{x_i\}$ in $G$ is Cauchy if, for any neighborhood $U$ of $1$ there exists $N$ such that if $n,m>N$ then $x_m\in x_nU$.

Under this defintion, again, your sequence is not Cauchy.

Note: The uniform metric on $(\mathbb R^+,\cdot)$ is $d(x,y)=\left|\log x - \log y\right|$. Essentially, it is the standard metric on $(\mathbb R,+)\cong (\mathbb R^+,\cdot)$.

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A sequence $(a_n)$ in a topological group $G$ is Cauchy if it has the following property:

for every neighborhood $U$ of $1$ there exists $k$ such that, for all $m,n\ge k$, it holds $x_m x_n^{-1}\in U$.

A function $f\colon G\to H$ is uniformly continuous if, for every neighborhood $V$ of $1$ in $H$, there exists a neighborhood $U$ of $1$ in $G$ such that, whenever $ab^{-1}\in U$, we have $f(a)f(b)^{-1}\in V$.

Note: there are different notions of “uniform” for non abelian groups, where one has to distinguish between left and right uniform.

Lemma. If $f\colon G\to H$ is uniformly continuous and $(a_n)$ is a Cauchy sequence in $G$, then $(f(a_n))$ is Cauchy in $H$.

Theorem. A continuous homomorphism $f\colon G\to H$ is uniformly continuous.

Conclusion. The sequence $(1/n)$ is not Cauchy in the group of positive reals.

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