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Well, it follows immediately that A is countable, not? Since both B and B' are countable basis?
The discussion revolves around the countable basis problem in topology, specifically addressing the implications of having a countable basis for a topological space X and the properties of other bases for X. Participants explore the relationships between different bases and their intersections.
The discussion is active, with participants providing insights and questioning each other's reasoning. Some have identified errors in reasoning and are reconsidering their approaches, while others suggest alternative methods and reference hints from the textbook.
There is an ongoing examination of the definitions and properties of bases in topology, particularly in the context of second-countable spaces. Participants note the importance of ensuring that any proposed basis generates the correct topology.
radou said:Hence, the collestion C is countable. Now, if we apply our lemma here, let U be any open set and x some element in U, then ma post #28 implies that there is a C from the collection A such that x is in C and C is in U, hence C is a countable basis?
radou said:Ah, OK, thanks a lot!
The real solutions to such problems are actually always quite simple, but require a certain amount of creativity. The proof I tried is more "definition-based", and such proofs ofteh lead to nothing.![]()
micromass said:I suppose that argument would work out to...
I'm still wondering if you can't prove the problem directly from the definition of a basis. I would deem it possible, but it would be more difficult.