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Isolated points and continuity

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Let f : A --> R be a function, and let c in A be an isolated point of A. Prove that f
    is continuous at c


    2. Relevant equations



    3. The attempt at a solution

    I'm kind of confused by this problem.... if c is an isolated point, then the limit doesn't exist. So I can't really use the fact that a function is continuous at c if for all epsilon>0 there exists a delta>0 such that whenever |x-c|<delta, it follows that |f(x)-f(c)|<epsilon.

    Any hints would be great!
     
  2. jcsd
  3. Nov 1, 2009 #2

    tiny-tim

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    Hi dancergirlie! :smile:

    (have a delta: δ and an epsilon: ε :wink:)
    How can that statement not be true? :wink:
     
  4. Nov 1, 2009 #3
    Well it wouldn't be true for x=c. Because |x-c| would be equal to zero, and that contradicts the statement that delta is greater than zero.
     
  5. Nov 1, 2009 #4
    I am assuming that A is a subset of the real numbers. Since c is an isolated point of A, there exists a [itex]\delta[/itex] such that [itex](c-\delta,c+\delta)[/itex] contains no other points of A. Hint: that is your [itex]\delta[/itex] to show that f is continuous at c. So now let x be in A and [itex]\varepsilon>0[/itex]. Then...
     
  6. Nov 1, 2009 #5
    I am not sure how you are choosing delta.... are you saying that you choose delta to be equal to the delta satisfying (c-delta, c+delta) containing no other points of A but c?
     
  7. Nov 1, 2009 #6
    Yes. That delta exists by the fact that c is an isolated point. Now just go through the epsilon-delta steps of proving f is continuous at c. Using that delta, what does [itex]|x-c|<\delta[/itex] imply if x is in A?
     
  8. Nov 1, 2009 #7
    Would it just mean that x-c is in a as well?
    Because I'm trying to show that
    |f(x)-L|< epsilon, but there is no limit so that is why I'm getting confused....
     
  9. Nov 1, 2009 #8
    Well your L is f(c). So you are trying to show that for x in A and any [itex]\varepsilon>0[/itex], there is a [itex]\delta>0[/itex] such that [itex]|x-c|<\delta[/itex] implies that [itex]|f(x)-f(c)|<\varepsilon[/itex]. But if x is in A and [itex]|x-c|<\delta[/itex], where the delta is the one described above, then x can only be one point! And that point is ...?
     
  10. Nov 1, 2009 #9

    tiny-tim

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    Sorry, but that doesn't make sense. :redface:

    How does any of that answer the question, can the statement "a function is continuous at c if for all epsilon>0 there exists a delta>0 such that whenever |x-c|<delta, it follows that |f(x)-f(c)|<epsilon" be untrue?

    Choose any ε … for example choose ε = 2009 …

    can you find a δ that works for that ε? :smile:
     
  11. Nov 1, 2009 #10
    Wouldn't that mean that x has to be c??? If it was then that contradicts the fact that it is continuous since that would mean that epsilon would=0 and delta=0, so the function wouldn't be continuous.... but I'm trying to prove that it is continuous. I don't know if I m completely off here... but that is what I'm getting from what you're saying..
     
  12. Nov 1, 2009 #11
    [How does any of that answer the question, can the statement "a function is continuous at c if for all epsilon>0 there exists a delta>0 such that whenever |x-c|<delta, it follows that |f(x)-f(c)|<epsilon" be untrue?

    Choose any ε … for example choose ε = 2009 …

    I understand that I can find a delta for any specific epsilon, but how am I supposed to phrase that in my proof for an arbitrary epsilon?
     
  13. Nov 1, 2009 #12
    Yes, it would mean that x=c, but that does in no way imply that [itex]\varepsilon=0[/itex]. Look. I'll just rewrite everything so hopefully it makes sense now.

    Let c be an isolated point of A. Then there exists a [itex]\delta>0[/itex] such that the interval [itex](c-\delta,c+\delta)[/itex] contains no other points of A besides c.

    Let x be in A and [itex]\varepsilon>0[/itex]. Then [itex]|x-c|<\delta[/itex] implies that x is in the interval [itex](c-\delta,c+\delta)[/itex]. This means x=c because there are no others points of A in that interval. This implies that [itex]|f(x)-f(c)|=|f(c)-f(c)|=0<\varepsilon[/itex] for absolutely any [itex]\varepsilon>0[/itex] you choose. Therefore, f is continuous at c.
     
  14. Nov 1, 2009 #13
    Oh it makes sense, I was thinking that implied that epsilon equals zero, when it was just that |f(x)-f(c)|=0 which would be less than any epsilon greater than zero.... thank you so much for your help, i really appreciate it!!
     
  15. Nov 1, 2009 #14
    Good! No problem for the help. This problem is subtle, so I'm glad it makes sense now.
     
  16. Nov 1, 2009 #15

    tiny-tim

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    ok, then find it for ε = 2009 :smile:
    well, for example, if you can use the same δ for every ε, that would do it, wouldn't it? :wink:
     
  17. Nov 2, 2009 #16

    HallsofIvy

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    Take [itex]\delta[/itex] to be less than the distance from closest point in the domain of f other than c. There are no points in the domain of f such that [itex]0< |x-c|< \delta[/itex] so the hypothesis, "if [itex]0< |x-c< \delta[/itex]" is always false. If the hypothesis of a statement is false then the statement is _____.
     
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