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Thread: Pretty confusing proof: I don't even know what it is I am supposed to prove...

  1. #1
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    Pretty confusing proof: I don't even know what it is I am supposed to prove...

    Let f-->R and xo is a cluster point of D. Prove that f has a limit at xo if for each epsilon>0 there is a neighborhood Q of xo such that for all x,y belonging to QintersectD x=/=xo, y=/=xo, we have |f(x)-f(y)|<epsilon.

    I have an exam coming up in a week and I don't even know where to start with such weird problems.
    I found a proof of this statement on the web (Since we don't have a solution manual), but it is very long and complicated..using many things that we have not used in this course. I was hoping someone could help me get an easier proof so that if such a problem appeared on my exam, I might actually have a chance because there is no way I could come up with the proof I found. Thanks
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by zhupolongjoe View Post
    Let f-->R and xo is a cluster point of D. Prove that f has a limit at xo if for each epsilon>0 there is a neighborhood Q of xo such that for all x,y belonging to QintersectD x=/=xo, y=/=xo, we have |f(x)-f(y)|<epsilon.

    I have an exam coming up in a week and I don't even know where to start with such weird problems.
    I found a proof of this statement on the web (Since we don't have a solution manual), but it is very long and complicated..using many things that we have not used in this course. I was hoping someone could help me get an easier proof so that if such a problem appeared on my exam, I might actually have a chance because there is no way I could come up with the proof I found. Thanks
    Couple of clarification questions:

    What do you mean by:

    f-->R

    (Last I checked, "Big Grin" was not a set...)

    And what set are you denoting as D?
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  3. #3
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    Oops lol, didn't notice that. It's just any set D
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by zhupolongjoe View Post
    Let f: D-->R and xo is a cluster point of D. Prove that f has a limit at xo if for each epsilon>0 there is a neighborhood Q of xo such that for all x,y belonging to QintersectD x=/=xo, y=/=xo, we have |f(x)-f(y)|<epsilon.

    I have an exam coming up in a week and I don't even know where to start with such weird problems.
    I found a proof of this statement on the web (Since we don't have a solution manual), but it is very long and complicated..using many things that we have not used in this course. I was hoping someone could help me get an easier proof so that if such a problem appeared on my exam, I might actually have a chance because there is no way I could come up with the proof I found. Thanks
    Let me clean this up a bit:

    Let $\displaystyle f: D\longrightarrow\mathbb{R}$ be a function and let $\displaystyle x_0$ be a cluster point of $\displaystyle D$. Prove that $\displaystyle \lim_{x_n\to x_0}f(x_n)$ exists if $\displaystyle \forall~\epsilon>0$, $\displaystyle \exists~\delta$ such that $\displaystyle x,y\in Q_{\delta}(x_0)$ implies that $\displaystyle |f(x)-f(y)|<\epsilon$.

    ----------

    Proof: Let $\displaystyle \{x_n\}\subset D$ be a sequence that converges to $\displaystyle x_0$. (We know such a sequence exists because $\displaystyle x_0$ is a cluster point.) We want to prove that the sequence $\displaystyle \{f(x_n)\}\subset\mathbb{R}$ converges.

    We know that given a $\displaystyle \delta$, all but finitely many $\displaystyle x_n$ are contained in $\displaystyle Q_{\delta}(x_0)$, that is, $\displaystyle n>N$ implies $\displaystyle x_n\in Q_{\delta}(x_0)$. So for $\displaystyle n,m>N$ we know that $\displaystyle x_m,x_n\in Q_{\delta}(x_0)$ and therefore $\displaystyle |f(x_n)-f(x_m)|<\epsilon$.

    But this means that $\displaystyle \{f(x_n)\}$ is Cauchy, and therefore converges because $\displaystyle \mathbb{R}$ is complete. In other words, $\displaystyle \lim_{x_n\to x_0}f(x_n)$ exists. $\displaystyle \square$
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  5. #5
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    Thanks a bunch! I understood that
    a lot more than the other proof I found. BTW: I love your David Hilbert quote...
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