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Continuous, Onto Function: (0,1)x(0,1)->R^2

  1. Aug 6, 2011 #1
    Continuous, Onto Function: (0,1)x(0,1)-->R^2

    Hi, All:

    Just curious about finding a continuous onto function from the open unit square (0,1)x(0,1)
    into R^2. All I can think is that the function must go to infinity towards the edges, because
    if it could be continued into the whole square, so that we would get a continuous bijection between compact [0,1]x[0,1] and Hausdorff R^2, which would then necessarily be a homeomorphism, which cannot happen for many reasons (e.g., square has boundary, R^2 does not; square has a cutset of two points and R^2 does not).

    It seems at first that pairing up two bijections f,g:(0,1)-->R^2 into h=fxg would work, but it does not, I don't think; the technique does not work in the general case, e.g., f(x)=x and g(x)=2x, both from R-->R.

    Any Ideas?

  2. jcsd
  3. Aug 7, 2011 #2
    Re: Continuous, Onto Function: (0,1)x(0,1)-->R^2

    where x=Re{z} and y=Im{z}.
    Does it work?
    Last edited by a moderator: Apr 26, 2017
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