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Point-wise continuity on all of R using compact sets

  1. Nov 1, 2011 #1
    Ok, so basically I am trying to decide whether my mathematics is valid or if there is some subtly which I am missing:

    Lets say I have a 1-1 strictly increasing point-wise continuous function f: R -> R, and I want to show that the inverse function g: f(R) -> R is also point-wise continuous.

    Now from Rudin's PMA Theorem 4.17 it says that if I have a continuous 1-1 mapping of a compact metric space X to a metric space Y. Then the inverse mapping g defined on Y by g(f(x)) = x is a continuous mapping of Y onto X.

    Ok so basically I have everything I need accept for one thing, that R is not compact. Ok so first my reasoning goes that given any f(x) ∈ f(R) I simply restrict the domain of f to the mapping f: [x-1 , x+1] -> R and now since f is strictly increasing and by theorem 4.17, since f(x) ∈ f([x-1 , x+1]), I can conclude that g: f(R) -> R is continuous at f(x). And since I can do this for any point in the range of f, I can conclude that g is continuous.

    I suppose my main concern is that while I can certainly conclude that g: f([x-1 , x+1]) -> [x-1 , x+1] is continuous and thus continuous at f(x), I'm not entirely convinced that just because I can draw a compact set around any point in R that still doesn't allow me to generalize to saying that g is continuous on all of f(R) although I am inclined to say it is ok since I'm only looking for point-wise continuity. Hope this made since, thanks
     
  2. jcsd
  3. Nov 1, 2011 #2
    Your proof is almost correct, there's just one thing: In order to make the inference "f-1 is continuous on a set U containing y ⇒ f-1 is continuous at y", you need the set U to be a neighborhood of y (that is, U must contain an open set containing y). Now, in this case, there's no problem, since f([x-1, x+1]) will be an interval of the form [a, b], where a<f(x)<b, and that contains the open set (a, b) which contains f(x). But there are some cases where it would present a problem -- for instance, let f:(0, 1)⇒R2 be given by:

    [tex]f(x)=\begin{cases} (4x, 0) & \text{if } x<\frac{1}{4} \\ (1, 4x-1) & \text{if } \frac{1}{4} \leq x < \frac{1}{2} \\ (\frac{3}{2} - x, 1) & \text{if } \frac{1}{2} \leq x \leq \frac{3}{4} \\ (\frac{3}{4}, 4-4x) & \text{if} \frac{3}{4} \leq x \end{cases}[/tex]

    This function is continuous, and at each point x, there's a compact interval [x-ε, x+ε] such that f-1 is continuous on f([x-ε, x+ε]), but f-1 is not continuous at (3/4, 0).
     
  4. Nov 1, 2011 #3
    A yes good show, it appears you are correct.

    Although I'm not sure about your implication: "f-1 is continuous on a set U containing y ⇒ f-1 is continuous at y" It seems that if a function is continuous on a set, then by definition it is continuous on every point of that set. I think in our case we either take the domain of f-1 ( f inverse ) to be just say U=f([1/100,1/5]), in which case by definition it's continuous at y, or we say the domain of f-1 is all of f((0,1)) and in that case f-1 was never continuous on U to begin with; which is fine and does not contradict Rudin's Theorem since (0,1) is not compact. I do see what is going wrong however, that balls on R^2 are two dimensional and thus the one centered at (3/4,0) ends up catching the tail end of curve which has wrapped itself around.

    It seems best to say that there do exist some restrictions on when we can piece together individual domains on which a function f is continuous and obtain a continuous function. I understand that in my case things are very well behaved and my open balls, in this case intervals, aren't going to be reaching out into another part of the curve through an extra dimension in order muck things up. The question is how to show that this won't happen.
     
    Last edited: Nov 1, 2011
  5. Nov 1, 2011 #4
    There's a bit of ambiguity in saying that a function f is continuous on U. It could mean either:
    1. the restriction of f to U is a continuous function, or:
    2. for every point x∈U, f, considered on its original domain, is continuous at x

    The second condition is stronger than the first, and the two conditions are equivalent iff U is an open set. Now, when we apply the theorem from Rudin to f-1:f([x-1, x+1]) → [x-1, x+1], we are proving that f-1 satisfies the first condition for [x-1, x+1]. But what we need is the second condition to hold.
    The good news is that if the restriction of f to some set U is continuous, then for every point x in the interior of U, f is continuous at x. So as long as every point in the set lies in the interior of at least one of the domains you're piecing together, the resulting function will be continuous. This is certainly true in your case, since if f([x-1, x+1]) = [a, b], then x∈(a, b)=int([a, b]).
     
  6. Nov 1, 2011 #5
    Excellent thank you, that all makes sense. One curiosity (my lack of a formal course in topology is really showing) is how do I interpret say the interval [0,1] in R^2? In R, (0,1) would be the interior of this set, but in R^2 there's no longer any interior is there?
     
  7. Nov 1, 2011 #6
    Yes, that is correct.
     
  8. Nov 1, 2011 #7

    mathwonk

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    i recommend you prove this result without appealing to rudin's compactness theorem, since a direct proof is even easier and more instructive for you.
     
  9. Nov 2, 2011 #8
    ya, I tried that for a while and was never confident enough to take it all the way to completion, plus I imagine I wouldn't have to justify that extra step of having to stitch all the separate domains back together.
     
  10. Nov 2, 2011 #9

    mathwonk

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    this should be "trivial", look:

    given f:R-->R continuous and increasing consider g = f^-1. If f(a) = b, then g(b) = a. claim g is continuous at b. let e>0 be given. we must find d>0 so that b-d < y < b+d implies a-e < g(y) < a+e.

    Note that by the intermediate value theorem, f maps R onto an interval in R. I.e. g is defined on the interval of values assumed by f.

    just look at f(a-e) and f(a+e). By increasingness of f, we have

    f(a-e) < f(a) = b < f(a+e).

    Now just choose d>0 so that f(a-e) < b-d < b < b+d < f(a+e).

    claim this d works.

    I.e. g is defined on a interval so is certainly defined on all points in this sequence of

    inequalities.

    And f increasing implies also g increasing, so we have

    a-e < g(b-d) < g(b) < g(b+d) < f(a+e).

    since g is increasing, if b-d < x < b+d, then a-e < g(b-d)< g(x) < g(b+d) < a+e.


    ok, i agree this is tedious and complicated, but the complication is inherent in the meaning of the concepts: continuity, increasing, inverse, ...


    you need to understand these terms thoroughly. trying to be clever and avoid dealing with them by applying some slick theorem out of rudin is dangerous laziness. it will let you down sometime when you need to know the stuff yourself, not piggy back on some crutch provided by others. this is the time to build your own muscles. do the work. you'll be glad you did.


    by the way, to see why this is trivial, draw a picture, don't try to write it out cold and blind as I did. the picture makes it obvious, and obvious how to write it up.
     
  11. Nov 3, 2011 #10
    You're right, you make good points, although I do think there is something to be said for also sometimes practicing the proper application of important theorems, as that too has its own set of nuances.
     
  12. Nov 3, 2011 #11

    disregardthat

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    Theorems should only be used when needed. Realizing a theorem is unecessary is much better than applying it needlessly.
     
  13. Nov 3, 2011 #12

    mathwonk

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    before using rudin's theorem you need to prove that [a,b] is compact. can you do that?
     
  14. Nov 5, 2011 #13
    Yes.
     
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