I just discovered the following. But since half the things I find in topology turn out to be wrong, I feel I better check with you guys.(adsbygoogle = window.adsbygoogle || []).push({});

What I convinced myself of this time is that if you have a function f:(X,T)-->(Y,S) btw topological spaces, and S' is a basis for S, then to show f is continuous, is suffices to show that [itex]f^{-1}(S') \subset T[/itex].

In words, that is because every open set of S can be written as a union of sets of S', and the operations of f^{-1} and union commute. (and that a union of open sets is open)

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# A result in topology

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