This is from Munkres, 2nd ed., Ch2, supp. exercise 7(a).
Topological Group: is a topological group with identity . This means that the product function is continuous and is a continuous map.
For subsets of . .
Symmetric neighborhood: A neighborhood of such that . Note that, for U open in G and , and are both symmetric.
Claim: Any neighborhood U of e contains a symmetric neighborhood such that .
I can show that if V is symmetric, that is an open set in and trivially. But I'm not seeing how to prove that a construction of a symmetric V from U has ?
Thank you for your help.
Your result looks close, but there is one part that confuses me still; the existence of and . Given open in , since is continuous we know that is open in which can be represented as the union of basis elements from G as . The union doesn't pull through the product, so this set does not factor nicely into separate and .
Example: the open unit ball in is not the product of two open intervals. Am I missing something?