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By Wolfgang Lück

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Proof. 9IJ = {Un} be a countable base of'iJ. 9IJ [where n(x) depends on x] such that x E Un(x) C P. 9IJ, {Un(x)} is a countable covering of X. Hence (X, 'iJ) is a Lindelof space. §12 • 33 Metric Spaces Proposition 15. In a second countable topological space (X uncountable subset A has a limit point in A. ~), every Proof. Suppose 93' = {Un} is a countable base of~. Suppose A does not have a limit point in A. Then for each a E A there exists Un(a) in {Un} such that a E Un(a) and Un(a) does not contain any other element of A.

Al > a 2 and UI C U2 ), we have p(al , UI ) > p(a2 , U 2 ). The range of pis cofinal in r because {x"" a E r} is frequently in each U. , {xp, {J E B} is eventually in U. Therefore {xp, {J E B} converges to x. Conversely, suppose a sub net of {x"" a E r} converges to x. If x is not a limit point of {x"" a E r} then there is a neighborhood U of x such that {x"" a E r} is not frequently in U. That means it is eventually in ~U. Hence a subnet of {x"" a E r} is in ~U and therefore it cannot converge to x.

Therefore at least one of r A , r E is cofinal with r. Hence at least one of the subsets {x", a E r A } and {x", a E r B } of {x", a E r} has the property that {(x"' a ErA), x} Ef(X) or ({x", a ErE), x) Ef(X) by condition (b). Hence x E q;(A) u q;(B) and we have shown that q;(A) u q;(B) = q;(A u B). Thus all the conditions of Theorem I are satisfied. Therefore there exists a unique topology Wi on X such that q;(A) = ClwA for each A E g(X). Finally, to show that lim"x" = x iff ({x,,}, x) E f(X), let ({x,,}, x) E f(X), and assume lim"x~ =I=- x.

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