# example of a discontinuous group operation in $\mathbb{R}$, under usual metric

someone has an example of that? I know that if G is a topological connectedness group ( operation continuous under the topology) then every neighborhood containing the identity elemental, generates the group, but i could not, find an example that it´s not true, if I don´t have the continuity ( i believe that it´s a necessary condition , but i don´t have an example ) the first example, arose from that, because in $\mathbb{R}$ it´s "familiar" I would be grateful for both examples
Thanks!

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Would you mind if I changed the wording of this post? It's rather difficult to understand. – jspecter Jul 2 '11 at 18:55
-1: please rewrite your question more carefully. For instance I do not understand the title at all. Please begin by carefully asking your question, independently of what you wrote in the title. – Pete L. Clark Jul 2 '11 at 19:04

## 1 Answer

I'll add my interpretation of the question, to at least make clear what I'm attempting to answer.

Question: I know that every connected topological group has the property that every open neighborhood of the identity generates the group. Is there an example of a (discontinuous) group operation on the reals that doesn't have this property? I'm still endowing the reals with the standard metric topology, making it a connected space.

Answer: You can partition the reals into a few intervals (say three, with one centered at $0$), biject each interval with $\mathbb{R}$, and put the structure of $(\mathbb{Z}/3\mathbb{Z}) \times \mathbb{R}$ on the whole space making $0$ the identity element. A sufficiently small interval about $0$ will be contained in the group's version of $\{0\} \times \mathbb{R}$, so it can't generate the whole group.

All you're remembering about the reals is its cardinality, so you have plenty of options here.

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