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Math Help - Continuity practice

  1. #1
    Senior Member sfspitfire23's Avatar
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    Continuity practice

    thanks all
    Last edited by sfspitfire23; April 13th 2010 at 05:10 PM.
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  2. #2
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    We know that \left| {f(x) - f(y)} \right| \leqslant \left| {f(x) - f(c)} \right| + \left| {f(x) - f(c)} \right|.
    Using continuity make each of the two on the right  < \frac{\varepsilon }{2}.
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  3. #3
    Senior Member sfspitfire23's Avatar
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    Hm, so \left| {f(x) - f(y)} \right| \leqslant \left| {f(x) - f(c)} \right| + \left| {f(x) - f(c)} \right| \leqslant 2\epsilon.

    So, \left| {f(x) - f(c)} \right|\leqslant \frac{\epsilon}{2}

    So \delta= \frac{\epsilon}{2}

    Then 2|f(x)-f(y)|<2\frac{\epsilon}{2}

    How is this?

    Also,
    Quote Originally Posted by Plato View Post
    We know that \left| {f(x) - f(y)} \right| \leqslant \left| {f(x) - f(c)} \right| + \left| {f(x) - f(c)} \right|.
    Is that triangle inequality? How did you recognize that you could do this? What happens when you substitute c for y?

    Thanks
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    Hm, so \left| {f(x) - f(y)} \right| \leqslant \left| {f(x) - f(c)} \right| + \left| {f(x) - f(c)} \right| \leqslant 2\epsilon.

    So, \left| {f(x) - f(c)} \right|\leqslant \frac{\epsilon}{2}

    So \delta= \frac{\epsilon}{2}

    Then 2|f(x)-f(y)|<2\frac{\epsilon}{2}

    How is this?
    I don't get the reference to \delta=\frac{\varepsilon}{2}. I think you mean choose \delta so that d(f(c),f(x))<\frac{\varepsilon}{2}. Otherwise looks fine.

    Also,

    Is that triangle inequality? How did you recognize that you could do this? What happens when you substitute c for y?

    Thanks
    Yes this is the triangle inequlaity and Plato made a slight typo, it should be |f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)|
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  5. #5
    Senior Member sfspitfire23's Avatar
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    So heres what I have, I'm not sure about my logic at one step (I point it out where it is)

    We know that |f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)| \leqslant 2\epsilon. I feel like choosing  2\epsilon because I feel like it will be east to pick my delta to cancel the 2 (is this logic correct?). Now, choosing a delta to get rid of the 2 in the 2\epsilon, I chose delta so that |f(x)-f(c)|<\frac{\epsilon}{2} and |f(y)-f(c)|< \frac{\epsilon}{2}. Then I can take 2|f(x)-f(y)| because this will still be \leqslant |f(x)-f(c)|+|f(y)-f(c)|. But then we can say that |f(x)-f(y)|\leqslant \frac{\epsilon}{2} because really, y is just some c. (THIS is the part that I'm not sure on. Is y just some c?) Then we get 2|f(x)-f(y)|<2\frac{\epsilon}{2}
    Last edited by sfspitfire23; April 12th 2010 at 05:38 PM.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    So heres what I have, I'm not sure about my logic at one step (I point it out where it is)

    We know that |f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)| \leqslant 2\epsilon. I feel like choosing  2\epsilon because I feel like it will be east to pick my delta to cancel the 2 (is this logic correct?). Now, choosing a delta to get rid of the 2 in the 2\epsilon, I chose delta so that |f(x)-f(c)|<\frac{\epsilon}{2} and |f(y)-f(c)|< \frac{\epsilon}{2}. Then I can take 2|f(x)-f(y)| because this will still be \leqslant |f(x)-f(c)|+|f(y)-f(c)|. But then we can say that |f(x)-f(y)|\leqslant \frac{\epsilon}{2} because really, y is just some c. (THIS is the part that I'm not sure on. Is y just some c?) Then we get 2|f(x)-f(y)|<2\frac{\epsilon}{2}
    I can't really follow your logic.

    Try rewriting it with this in mind.

    |f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)|. But, since f is continuous at c you can make those last two terms arbitrarily small by choosing x,y arbitrarily close to c...
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  7. #7
    Senior Member sfspitfire23's Avatar
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    Ok, trying again, please tell me what you think

    We know that |f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)| \leqslant 2\epsilon. Because f is continuous at c, I can make x, y arbitrarily small so that |f(x)-f(c)|<\epsilon and |f(y)-f(c)|<\epsilon
    Now, choosing a delta to get rid of the 2 in the 2\epsilon, I chose delta so that |f(x)-f(c)|<\frac{\epsilon}{2} and |f(y)-f(c)|< \frac{\epsilon}{2}. Then I can take 2|f(x)-f(y)| because this will still be \leqslant |f(x)-f(c)|+|f(y)-f(c)|. But then we can say that |f(x)-f(y)|\leqslant \frac{\epsilon}{2} because really, y is just some c. (THIS is the part that I'm not sure on. Is y just some c?) Then we get 2|f(x)-f(y)|<2\frac{\epsilon}{2}[/quote]
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    Ok, trying again, please tell me what you think

    We know that |f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)| \leqslant 2\epsilon. Because f is continuous at c, I can make x, y arbitrarily small so that |f(x)-f(c)|<\epsilon and |f(y)-f(c)|<\epsilon
    Now, choosing a delta to get rid of the 2 in the 2\epsilon, I chose delta so that |f(x)-f(c)|<\frac{\epsilon}{2} and |f(y)-f(c)|< \frac{\epsilon}{2}. Then I can take 2|f(x)-f(y)| because this will still be \leqslant |f(x)-f(c)|+|f(y)-f(c)|. But then we can say that |f(x)-f(y)|\leqslant \frac{\epsilon}{2} because really, y is just some c. (THIS is the part that I'm not sure on. Is y just some c?) Then we get 2|f(x)-f(y)|<2\frac{\epsilon}{2}
    [/QUOTE]
    You are messing up the order in which you say things.

    The fact that |f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)|<<br />
\varepsilon is the solution to the problem. You have to note that you can do this since you can make |f(x)-f(c)| and |f(y)-f(c)| by taking \delta=\min\{\delta_1,\delta_2\} where |x-c|<\delta_1\implies |f(x)-f(c)|<\varepsilon and |y-c|<\delta_2\implies |f(y)-f(c)|<\varepsilon
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  9. #9
    Senior Member sfspitfire23's Avatar
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    So, we have |f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)| \leqslant 2\epsilon because we can find some delta1/delta2 such that

    |f(x)-f(c)|<\epsilon and |f(y)-f(c)|<\epsilon

    Choose \delta = \epsilon /2

    But then 2|f(x)-f(y)|<|f(x)-f(c)|+|f(y)-f(c)|

    So 2|f(x)-f(y)|\Rightarrow 2|f(x)-f(c)| because y is really just some c and we then have 2|f(x)-f(c)|<2(\epsilon / 2)
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