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Math Help - Uniform Continuity

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    Uniform Continuity

    Here is a proof that I am having trouble getting started with. I'd appreciate any help.

    Let D be a subset of \mathbb{R}. Let \rightarrow \mathbb{R}" alt="f\rightarrow \mathbb{R}" /> be uniformly continuous. Let x_0 be a limit point of D. Suppose x_{0}\notin D. Prove there is a continuous function \bigcup \{x_0\}\rightarrow \mathbb{R}" alt="g\bigcup \{x_0\}\rightarrow \mathbb{R}" />, such that g(x)=f(x) for all x\in D.

    I'll be happy to attempt the proof. If anyone could maybe just tell me exactly what direction I should be looking to go I would would appreciate it.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zebra2147 View Post
    Here is a proof that I am having trouble getting started with. I'd appreciate any help.

    Let D be a subset of \mathbb{R}. Let \rightarrow \mathbb{R}" alt="f\rightarrow \mathbb{R}" /> be uniformly continuous. Let x_0 be a limit point of D. Suppose x_{0}\notin D. Prove there is a continuous function \bigcup \{x_0\}\rightarrow \mathbb{R}" alt="g\bigcup \{x_0\}\rightarrow \mathbb{R}" />, such that g(x)=f(x) for all x\in D.

    I'll be happy to attempt the proof. If anyone could maybe just tell me exactly what direction I should be looking to go I would would appreciate it.
    I'll give you three things to think about

    1) If x_0\in\overline{D} then there is a sequence of points in D converging to it.

    2) The image of Cauchy sequences under unif. cont. maps is unif. cont.

    3) \mathbb{R} is complete.


    Try to connect these three.
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    To be honest we haven't talked about Cauchy sequences so I'm not sure how to connect all these.

    However, I will do my best with what I have gathered form Wikipedia...
    Since x_0 is a limit point in D then for x_0\in\overline{D} there must be a sequence of points that are contained in D that converge to x_0.

    Therefore, we have a sequence of points that are converging to a point x_0. Thus,(from what I gather about Cauchy Sequences), this sequence is a Cauchy sequence since the points are converging to x_0.

    Then, (I got the following from Wikipedia) M is said to be complete (or Cauchy) if every Cauchy sequence of points in M has a limit that is also in M or alternatively if every Cauchy sequence in M converges in M.
    Here, our Cauchy sequences is converging to x_0 where x_0\in \overline{D}. So, \overline{D} is complete?

    Sorry this is might be kinda weak but I did what I could for never have learning about Cauchy Sequences or completeness....
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zebra2147 View Post
    To be honest we haven't talked about Cauchy sequences so I'm not sure how to connect all these.

    However, I will do my best with what I have gathered form Wikipedia...
    Since x_0 is a limit point in D then for x_0\in\overline{D} there must be a sequence of points that are contained in D that converge to x_0.

    Therefore, we have a sequence of points that are converging to a point x_0. Thus,(from what I gather about Cauchy Sequences), this sequence is a Cauchy sequence since the points are converging to x_0.

    Then, (I got the following from Wikipedia) M is said to be complete (or Cauchy) if every Cauchy sequence of points in M has a limit that is also in M or alternatively if every Cauchy sequence in M converges in M.
    Here, our Cauchy sequences is converging to x_0 where x_0\in \overline{D}. So, \overline{D} is complete?

    Sorry this is might be kinda weak but I did what I could for never have learning about Cauchy Sequences or completeness....
    Hmm, well \overline{D} is complete (any closed subspace of a complete space is complete), but the point is this.

    Picture this. Since x_0\in\overline{D} there is a sequence of points \{x_n\} in D such that x_n\to x_0. Now, these points converge to x_0 but all we really care about is that they get really close to one another as n gets big (they're Cauchy). But, it's a fact that if the x_n's get really close to one another (Cauchy again) and f is unif. cont then the values of the sequence \{f(x_n)\} get really close to one another (Cauchy). But, it's a fact that says that all sequences of real numbers which get really close to one another (Cauchy) converge, and so in particular f(x_n) converges to some y. So, what if we said that f(x_0)=\lim_{n\to\infty}f(x_n)=y. I mean, it's not obvious why this construction is independent of the choice of the sequence \{x_n\}, but it is. From there continuity (and in fact uniform continuity, although you don't need to prove that) are clear since if \{d_n\} is any sequence in D\cup\{x_0\} we have that (combining continuity on D with the construction of f(x_0)) that \displaystyle f\left(\lim_{n\to\infty}d_n\right)=\lim_{n\to\inft  y}f(d_n) for any \{d_n\} in D\cup\{x_0\}.
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