Limit of function from one metric space to another metric space

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September 17th 2010, 08:53 AM
tarheelborn

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!
September 17th 2010, 01:17 PM
Opalg

Quote:

Originally Posted by tarheelborn

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

... 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

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 $f(a)$ .
September 17th 2010, 01:32 PM
tarheelborn

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!
September 17th 2010, 02:04 PM
bubble86

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 $V_\epsilon (f(a))$