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Is this true? A refinement of a totally bounded topology and an indiscrete topolgy is an indiscrete topology.

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What is a totally bounded topology? And if by refinement you mean a topology which contains another topology, then every topology refines the indiscrete topology, which makes your question rather trivial. Please clarify. – Pete L. Clark Jul 12 2010 at 7:31
(This seems to be one in a series of bitty questions about a certain paper on abelian topological groups, so presumably the question is related to a more special setting which makes more sense.) – Yemon Choi Jul 12 2010 at 7:41

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