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Math Help - Limit of function from one metric space to another metric space

  1. #1
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    Limit of function from one metric space to another metric space

    Given the properties (1) \forall \ \delta > 0 \ \exists \ \epsilon > 0 so that f\left(B\left[a;\delta\right]\right) \in B\left[f\left(a\right);\epsilon\right] and (2) f is continuous at a if, for every sequence \{ x_n \} \in M_1 such that \displaystyle\lim_{n \to \infty} \left(x_n\right)=a, the sequence  \{ f \left(x_n\right) \} converges to f\left(a\right).

    I need to prove  2 \Longrightarrow \ 1 . We did the first part in class:

    I employ the contrapositive approach. Suppose f does not have property (1) and suppose that \displaystyle \lim_{n \to \infty} \left(x_n\right)=a, \ a \ \in \ M_1. Now if (1) does not hold, then there is \epsilon > 0 so that, \forall \delta > 0, f\left(B\left[a;\delta\right]\right) \notin B\left[f\left(a\right);\epsilon\left]. When \delta = \frac{1}{n}, \ f\left(B\left[a;\frac{1}{n}\right]\right) \notin B\left[f\left(a\right);\epsilon\right]. So in B\left[a;\frac{1}{n}\right] there is some x_n \in B\left[a;\frac{1}{n}\right] so that f \left(x_n \right) \notin B \left[f \left(a\right);\epsilon\right].

    Now I am to my problems: 1)I need to prove that, in fact, \displaystyle \lim_{n \to \infty}\left(x_n\right) = a. Let \epsilon>0. I need to find N \in \mathbb{N} so that, for n\geq N, \ \rho\left(x_n,a\right)<\epsilon.

    I am inclined to let \epsilon=1/n, but then \epsilon might not be an integer. So I am not sure what to do about that. Once I find my \epsilon, I simply need to finish the epsilon proof of this part.

    2) I need to prove that, in fact, \displaystyle \lim_{n \to \infty} \left(f\left(x_n\right)\right) \neq f\left(a\right). This means that it is not true that \forall \ \epsilon > 0 \ \exists \ N \in \mathbb{N} so that, for n \geq N, then \rho_2\left(f\left(x_n\right), f\left(a\right)\right)<\epsilon. Since when \delta=\frac{1}{n}, f\left(B\left[a;\frac{1}{n}\right]\right) \notin B\left[f\left(a\right);\epsilon\right] it seems like I am done, but I am not sure how to make the conclusion.

    Sorry this post is so long, but this is a long problem. Thank you for your help!
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  2. #2
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    Quote Originally Posted by tarheelborn View Post
    Given the properties (1) \forall \ \delta > 0 \ \exists \ \epsilon > 0 so that f\left(B\left[a;\delta\right]\right) \in B\left[f\left(a\right);\epsilon\right] ...
    Two mistakes in that property (1). The first is that \forall \ \delta > 0 \ \exists \ \epsilon > 0 should be \forall \ \epsilon > 0 \ \exists \ \delta > 0. The second is that f\left(B\left[a;\delta\right]\right) \in B\left[f\left(a\right);\epsilon\right] should be f\left(B\left[a;\delta\right]\right) \subseteq B\left[f\left(a\right);\epsilon\right]. But these may just be typos, because they don't seem to affect the mostly correct argument that follows.
    Quote Originally Posted by tarheelborn View Post
    ... and (2) f is continuous at a if, for every sequence \{ x_n \} \in M_1 such that \displaystyle\lim_{n \to \infty} \left(x_n\right)=a, the sequence  \{ f \left(x_n\right) \} converges to f\left(a\right).

    I need to prove  2 \Longrightarrow \ 1 . We did the first part in class:

    I employ the contrapositive approach. Suppose f does not have property (1) and suppose that \displaystyle \lim_{n \to \infty} \left(x_n\right)=a, \ a \ \in \ M_1. Now if (1) does not hold, then there is \epsilon > 0 so that, \forall \delta > 0, f\left(B\left[a;\delta\right]\right) \notin B\left[f\left(a\right);\epsilon\left]. When \delta = \frac{1}{n}, \ f\left(B\left[a;\frac{1}{n}\right]\right) \notin B\left[f\left(a\right);\epsilon\right]. So in B\left[a;\frac{1}{n}\right] there is some x_n \in B\left[a;\frac{1}{n}\right] so that f \left(x_n \right) \notin B \left[f \left(a\right);\epsilon\right].

    Now I am to my problems: 1)I need to prove that, in fact, \displaystyle \lim_{n \to \infty}\left(x_n\right) = a. Let \epsilon>0. I need to find N \in \mathbb{N} so that, for n\geq N, \ \rho\left(x_n,a\right)<\epsilon.

    I am inclined to let \epsilon=1/n, but then \epsilon might not be an integer. So I am not sure what to do about that. Once I find my \epsilon, I simply need to finish the epsilon proof of this part.
    You don't need to "find \epsilon" at all. In fact, \epsilon is given, and what you have to find is the N that depends on it. Choose N to be an integer greater than 1/\epsilon, and you should be able to complete that section of the proof.

    Quote Originally Posted by tarheelborn View Post
    2) I need to prove that, in fact, \displaystyle \lim_{n \to \infty} \left(f\left(x_n\right)\right) \neq f\left(a\right). This means that it is not true that \forall \ \epsilon > 0 \ \exists \ N \in \mathbb{N} so that, for n \geq N, then \rho_2\left(f\left(x_n\right), f\left(a\right)\right)<\epsilon. Since when \delta=\frac{1}{n}, f\left(B\left[a;\frac{1}{n}\right]\right) \notin B\left[f\left(a\right);\epsilon\right] it seems like I am done, but I am not sure how to make the conclusion.
    That is more or less correct. In fact, if \displaystyle \lim_{n \to \infty} \left(f\left(x_n\right)\right) = f\left(a\right) then there exists an N such that \rho_2\left(f\left(x_n\right), f\left(a\right)\right)<\epsilon for all n\geqslant N. But since \rho_2\left(f\left(x_n\right), f\left(a\right)\right)\geqslant\epsilon for all  n, that is a contradiction. Therefore f(x_n) does not converge to f(a).
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  3. #3
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    Sorry about the typos! So the way to avoid dealing with \frac{1}{\epsilon} not being an integer is to choose N so that it is an integer ... ? Thank you so much!
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  4. #4
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    If you want to prove 2 \Rightarrow 1 , then consider this
    \ f: X \rightarrow  Y
    \ f^{-1} V_\epsilon (y) is an open set in X for every open set \ V_\epsilon (y) in Y because f is cont
    \ f^{-1} V_\epsilon (f(a)) contains a and is an open set of X
    Lets call \ f^{-1} V_\epsilon (f(a))  in X,  V_\delta (a)
    since \lim_{n\to\infty} x_n  =  a
    \forall open neighbourhoods of a, i.e r >0, \ V_r (a) \bigcap S \not= \emptyset , where S is the set containing \{ x_n \} ; actually \{ x_n \} is eventually in \ V_r (a) \forall r > 0,
    so  \exists N_0 : \forall n  \ge N_0  x_n \in V_\delta (a)
    \ x_n \in V_\delta (a) \rightarrow  f(x_n) \in V_\epsilon (f(a))
    so \forall \epsilon  > 0,  \{f(x_n)\} is eventually in  V_\epsilon (f(a))
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