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I was reading chapter 10 of Atiyah where I met the notion of equivalent Cauchy sequences for topological groups.

Atiyah does not explain the reason why equivalent Cauchy sequences indeed give a equivalence relation. I can manage to prove the reflexivity and symmetry of this relation. But I failed to work out the transitivity.

Can anyon help me on this? Thank you.

FYI. A Cauchy sequence of a topological group $G$ is a sequence $(x_n)_{n \in \mathbb{N}}$ such that for any open neighbourhood $U$ of $0$ (the identity of $G$), there is $N \triangleq N(U)$ such that for any $n,m > N, x_n-x_m \in U$. And we say that two Cauchy sequences $(x_n)$ and $(y_n)$ are equivalent if $x_n-y_n$ converges to $0$, that is, for every open neighbourhood $U$ of $0$, there is $N$ such that for all $n > N, x_n-y_n \in U$.

I can see that for any open neighbourhood $U$ 0f $0$, if we can find an open neighbourhood $V$ of $0$ such that $V+V \subset U$, then the transivity will be proved. But, indeed how to explain the existence of such $V$?

Henno Brandsma
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Yu Ning
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2 Answers2

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This is a standard theorem on topological groups, and follows from the continuity of the group operation. I'll write that as addition, and assume the operation is commutative, but the argument works for non-commutative groups too.

As addition is continuous, then if $U$ is an open neighbourhood of $0$ then $W=\{(a,b)\in G\times G:a+b\in U\}$ is an open neighbourhood of $(0,0)$ in $G\times G$. By definition of the product topology, there are open neighbourhoods $V_1$ and $V_2$ of $0$ with $V_1\times V_2\subseteq W$. Let $V=V_1\cap V_2$. It is an open neighbourhood of $0$ and $V\times V\subseteq W$, that is $a+b\in U$ for all $a$, $b\in V$.

Angina Seng
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The existence of such a $V$ follows from the definition of continuity of $+$ at $(0,0)$:

There is a basic open neighbourhood $U_1 \times U_2$ that contains $(0,0)$ such that $+[U_1 \times U_2] = U_1 + U_2 \subseteq U$

Now use $V=U_1 \cap U_2$.

This works in any topological group.

Henno Brandsma
  • 242,131