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Open sets and cartesian products

  1. Mar 29, 2013 #1
    Let f be a continuous function from R to R and let A be a subset of R^2. Define A={(x,y): y<f(x)}.

    Can you express A as a cartesian product of two open sets? I tried RxU alpha_x where alpha_x = {y:y<f(x)}. But that didn't work, i need to change something about R.
     
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  3. Mar 29, 2013 #2

    mfb

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    If those open sets should be subsets of R, I don't see how this can be possible.
    Two open sets B,C of R or R^2, where A={b+c|B in B, c in C} are possible, but that is not B x C.
     
  4. Mar 29, 2013 #3
    Let A=U_y {{f^-1(x):f(x)>y}x{y':y<maxf(R) is there is such a max and y'<y}}. Does that work?
     
  5. Mar 29, 2013 #4
    But this doesn't contain those y's. If we add them manually to A will that change something?

    Edit:
    If f has a max, say f(D), then add (f^-1(D), y) to A. If not then there exists an f(x') such that f(x')>y so add (f^-1(x'), y) to A. What do you think?
     
    Last edited: Mar 29, 2013
  6. Mar 29, 2013 #5

    mfb

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    What does "{f^-1(x):f(x)>y}" mean? You use "x" to denote values otherwise written as y, and get a set of values usually described as x. In addition, it is meaningless to use the same x in f and f^(-1). And how does y get defined here?

    What is "{y':y<maxf(R) is there is such a max and y'<y}"? What happens if there is no such maximum? And if there is one, what does that mean?
     
  7. Mar 29, 2013 #6
    Actually you can ignore my last reply above.
    Alright so write {x:f(x)>y} instead. We know this is open because f is continuous. y is defined as an arbitrary real number such that y<sup f so that y<f(x) could make sense. And if we put y'<y we get an open subset of real numbers consisting of y' s such that y'<y<f(x).
     
  8. Mar 29, 2013 #7

    mfb

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    I don't see what you are doing.

    Let's consider two examples:
    (1) f(x)=-y^2
    (2) f(x)=y

    How do your sets look like?
     
  9. Mar 29, 2013 #8
    (1)
    A={{x: f(x)=-y^2>-1}x{y'<-1: y' is a real}}U{{x: f(x)=-y^2>-2}x{y'<-2: y' is a real}}U{{x: f(x)=-y^2>-sqrt2}x{y'<-sqrt2: y' is a real}}U....

    (2) is similar, I just substitute random real values for y.
     
  10. Mar 29, 2013 #9

    WannabeNewton

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    The assumption isn't even true in general so I'm not sure what you're trying to prove with all that work. Given ##f:\mathbb{R}\rightarrow \mathbb{R}##, assume that ##A = \mathbb{R}\times U## is in fact true, where ##U\subseteq \mathbb{R}## is open. If ##f## isn't constant then there exist ##x,y\in \mathbb{R}## such that ##f(x) < f(y)##. Therefore ##(y,f(x))\in A## but then ##f(x)\in U## implying ##(x,f(x))\in A## since ##A = \mathbb{R}\times U## by hypothesis, which is a contradiction.
     
  11. Mar 29, 2013 #10
    Yes, this is why I said R x {something} doesn't work. I think I should write these in latex, it may be the case that you think I wrote U as a set but it is the union operation.
     
    Last edited: Mar 29, 2013
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