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I was looking at the proof on proofwiki for showing that the finite complement topology is separable (http://www.proofwiki.org/wiki/Finite_Complement_Topology_is_Separable) and came across something I don't agree with. I am specifically interested in the finite complement topology on the real line. It says "Let H be a countably infinite subset of S" or, in my case, let H = rationals. Then, it goes on to say that H separates S. I agree that the rationals are a countable dense subset of the reals under this topology, but the rationals are not in the finite complement topology on the reals. Does this mean that the countably dense subset need not be an open set in the topology? In Lee's Introduction to Topological Manifolds the definition of separability is: "A topological space X is said to be separable if it contains a countable dense subset." To me the reals with the finite complement topology does not contain the rationals.

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The definitions says that a space is separable if the space, not the topology, contains a countable dense subset. The subset isn't required to be open.

This could be confusing since a separation $A \cup B$ of a space (an entirely different concept) does require that $A$ and $B$ are open.

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  • $\begingroup$ Thanks, this is what I thought. $\endgroup$ – Jeremy Upsal Sep 22 '13 at 1:19

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