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Openness of continous real valued functions

  1. Sep 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Let X be the set of continuous real-valued functions on [0,1]. Prove that S={g in X: g(t) > 0 for all t} is an open subset.


    2. Relevant equations



    3. The attempt at a solution
    I was thinking of taking an arbitrary t[itex]_{0}[/itex] in g and another function f that is also continuous on [0,1], then finding a radius r > g(t[itex]_{0}[/itex])-f(t[itex]_{0}[/itex]) and making the open ball or tube or whatever with that radius about the function g. Somehow I feel that this is incorrect though.
     
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  3. Sep 6, 2011 #2

    micromass

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    Hi mohaque! :smile:

    You have the right idea. Take an arbitrary g in S. You must find a radius r such that [itex]B(g,r)\subseteq S[/itex]. So you must find r such that all f with d(f,g)<r, have f>0.

    The thing is, you can't just choose [itex]t_0[/itex] and take [itex]r>f(t_0)-g(t_0)[/itex], because
    1) your r must be independent of f
    2) your r must be good for all t.

    So, how do we choose r? Well, why don't we take the minimal value of g?? Why does a minimal value exist?
     
  4. Sep 7, 2011 #3
    Ahh, I don't know how to choose r. In a continuous function, within a local interval, there will always be some max and min. I don't understand why we would take the min. Shouldn't we take r to be the max of the absolute value of whatever the max and min values may be of g? So that if we chose that as r, it'll contain the entire function? I don't know if you understood that.

    I guess example-wise, if g in [0,1] has a local max of 2 and local min of -1, shouldn't we take r to be 2 because that way it'd include both the local max and min? Forgive any bad wording on my part and feel free to correct me.
     
  5. Sep 7, 2011 #4

    micromass

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    But there is no reason to do that. The funciton will be in our ball automatically. Indeed every g is contained in the ball B(g,r). We only need to choose r small enough so that no functions outside S come in our ball. So, we need to find r such that for all f in B(g,r), we have f>r.

    The maximum value of g doesn't matter here. It is the minimum value that matters.
     
  6. Sep 10, 2011 #5
    I got it, sort of. Was easier to understand it with a picture, has to be the min so that when you make that the r, it doesn't go below the zero line since that'd put it past the boundaries. Is there some graphing posting option here? Would be great if there was, seems kind of necessary as well here.

    I have another question, to test the if something (lets call it T) is closed, should I let some S be the open set that is the complement of T and make an open ball in S and check to see that it is open with another arbitrary point in that ball in S? And because that arbitrary point was not in T, but was in the open ball, the complement of the open set which contained that ball would be the closed set, T.

    If that is the right way, I was wondering if you could perhaps show me a few examples of such things, like how to go about doing the proof for with with specific cases. I can think of how to do it with the d(x,y) stuff but it somehow doesn't feel right because that seems like a general case and doesn't seem to be appropriate when given some specific closed set with some real numbers as boundaries and whatnot. I can't seem to find any good examples in my analysis book that I have, there are only the definitions/theorems and one or two simple non-helpful examples.
     
  7. Sep 10, 2011 #6

    micromass

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    Yes, that's how we would prove closedness.

    You want some nontrivial examples of open sets? Well,

    [tex]\{(x,y)\in \mathbb{R}^2~\vert~x+y>0\}[/tex]
    [tex]\{f\in \mathcal{C}(\mathbb{[0,1]},\mathbb{R})~\vert~\forall x:~x<f(x)<x^2+1\}[/tex]
     
  8. Sep 10, 2011 #7
    I meant how to prove that they're open.

    But I guess for the first one, for some s=(x[itex]_{1}[/itex],y[itex]_{1}[/itex])[itex]\epsilon[/itex]S, you take r = (x[itex]_{1}[/itex]+y[itex]_{1}[/itex])/[itex]\sqrt{2}[/itex]. Then you have the open ball with that center and radius and you pick another point, say q=(a,b) within that open ball and prove that q[itex]\epsilon[/itex]S. Then for L symbolizing the line, we have

    (x[itex]_{1}[/itex]-a)[itex]^{2}[/itex]+(y[itex]_{1}[/itex]-b)[itex]^{2}[/itex] < r[itex]^{r}[/itex]
    [itex]\Rightarrow[/itex] |x[itex]_{1}[/itex]-a| < r, |y[itex]_{1}[/itex]-b| < r
    [itex]\Rightarrow[/itex] -r-x1 < -a < r-x1 or x1-r < a < x1+r and similarly y1-r < b < y1+r
    [itex]\Rightarrow[/itex] x1+y1-2r < a+b, then I get stuck. I don't know what to do after this. I know I'm supposed to get 0 < a+b to prove that it is in S, but I don't see how to get that from x1+y1-2r and r=(x[itex]_{1}[/itex]+y[itex]_{1}[/itex])/[itex]\sqrt{2}[/itex].

    The second problem confuses me because in that interval, I think the radius is supposed to be the min of whatever function you choose to be arbitrarily in the set, say f. But then I don't understand what is supposed to happen next. We pick another arbitrary function, say g, within the ball created by the radius and that has to stay within the set, i.e.
    x < g(t) < x[itex]^{2}[/itex]+1. I don't know how to go about doing this using math speak but I'll try.

    |f(t)-g(t)| < r and |f(t[itex]_{0}[/itex])-g(t[itex]_{0}[/itex])| < r where t[itex]_{0}[/itex] is where f takes its min value.
    f(t)-r < g(t) < f(t)+r and f(t[itex]_{0}[/itex])-r < g(t[itex]_{0}[/itex]) < f(t[itex]_{0}[/itex])+r
    Then I don't know what to do.
     
  9. Sep 10, 2011 #8

    micromass

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    Well, what is the minimum distance between f and [itex]x^2+1[/itex] and [itex]x[/itex]? Is that minimum distance nonzero? Can we take that as radius?
     
  10. Sep 10, 2011 #9
    No, I guess not, I guess it has to be min of whatever f is closest to, either the x or x^2 +1.
     
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