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Extending maps defined on dense sets

  1. Feb 17, 2007 #1


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    Say you have a (continuous) map from a space A into a space Y, where A is dense in some other space X. When is it possible to extend this to a map from X into Y? Intuitively, to find the value of the map at a limit point, just take the limit of the values on a sequence approaching it. However, it's not clear the resulting sequence will always have a limit (eg, 1/x or sin(1/x) on (0,1), which is dense in [0,1)), and even when it does, is the resulting function always continuous?
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  3. Feb 17, 2007 #2


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    The resulting function might not even be well-defined. A=R\{0} is dense in X=R. Consider the map from A to [0,1] which sends positive numbers to 1 and negative numbers to 0. There are sequences in A which converge to 0 that are strictly positive (hence the sequence of their values converges to 1), strictly negative (so the sequence of values converges to 0) and alternating between negative and positive (so the sequence of values doesn't converge).

    But suppose that for all x in X\A there exists a unique y in Y such that if (an) is any sequence in A converging to x, (f(an)) converges to y. Then you can extend f to say f(x) = y. Then it shouldn't be hard to prove that this extend f is continuous, I think.
    Last edited: Feb 17, 2007
  4. Feb 17, 2007 #3
    First of all, what kind of space are we working in? For X and Y metric spaces, here's what I'm thinking:

    If the function f is uniformly continuous, it sends Cauchy sequences to Cauchy sequences, so if we select a point x in X and find a sequence in A converging to x, that sequence is Cauchy so its image is Cauchy. If we further demand Y is complete, then the image sequence converges, to some y in Y. If F is the extension of f to X, let y be F(x). I don't see anything going wrong with the verification that this is well defined and continuous on all of X. Thus f uniformly continuous and Y complete ensures the extension exists. These may not be the weakest conditions that ensure this happens though. Also note that if X is compact, then f is automatically uniformly continuous.
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