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September 4th, 2010, 10:34 AM   #1
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Topology generated from subbasis

I'm teaching myself topology from Munkres' book, and I ran across this question in the exercises:
Quote:
 Show that if A is a basis for a topology on X, then the topology generated by A equals the intersection of all topologies on X that contain A. Prove the same if A is a subbasis.
Proving it isn't the problem - I can do that - but what's troubling me is the implication; is it saying that the topology generated by a basis A is equal to the topology generated by A as a subbasis? Because it seems like you're trying to prove that both topologies equal the intersection of all topologies on X that contain A [and are consequently equal themselves]. If so, then what's the use of using a subbasis if it's just going to end up working identically to a basis?

 September 4th, 2010, 11:48 AM #2 Senior Member   Joined: Jun 2010 Posts: 618 Thanks: 0 Re: Topology generated from subbasis Hello martexel. I think the question is asking you to prove that if A is a basis for a topology on X, then the topology generated by A equals the intersection of all the topologies on X containing A, and if alternatively A is a subbasis for a topology on X, then the topology generated by A equals the intersection of all the topologies on X containing A. Therefore, it is not implicit in the question that the two topologies so generated be equal, since A as a basis in the first case and as a subbasis in the second are in no way related. It may help to subscript the two differently. The crux of the matter is how we define "the topology generated by a basis" versus "the topology generated by a subbasis", as well as the difference in the definition of "basis" and "subbasis". (Keep in mind that a basis is automatically a subbasis, so a subbasis is "easier" to produce.) Look over these differences again and your worries will disappear. Regards, Ormkärr

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