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Let $ U $ be an arbitrary neighborhood of the identity $ e \in G $. Define $ S = \langle U \rangle $, the subgroup generated by $ U $. Explicitly, $ S $ consists of all finite products of elements from $ U $ and their inverses:
$$S = \{ u_1 u_2 \cdots u_n \mid u_i \in U \cup U^{-1}, \, n \in \mathbb{N} \}.$$
- Show $ S $ is Open
Since $ U $ is open and contains $ e $, for any $ g \in S $, the set $ gU $ is open in $ G $ (as left multiplication by $ g $ is a homeomorphism). Furthermore, $ gU \subseteq S $, because $ g \in S $ and $ U \subseteq S $. Thus, every element of $ S $ has an open neighborhood $ gU $ contained entirely within $ S $, proving $ S $ is open. - Show $ S $ is Closed
Consider the complement $ G \setminus S $. For any $ g \in G \setminus S $, the neighborhood $ gU $ is disjoint from $ S $. If $ gu \in S $ for some $ u \in U $, then $ g = (gu)u^{-1} \in S $, contradicting $ g \notin S $. Hence, $ gU \subseteq G \setminus S $, making $ G \setminus S $ open. Therefore, $ S $ is closed. - Use Connectivity to Conclude $ S = G $
A connected space cannot have non-trivial clopen subsets. Since $ S $ is non-empty (contains $ e $), open, and closed, it must equal $ G $. Thus, $ S = G $, meaning $ U $ generates $ G $.
Conclusion
Any neighborhood of the identity in a connected Lie group $ G $ generates the entire group. This follows from the topological properties of openness, closedness, and connectivity, leveraging the group structure. |
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