# Thread: Limit of function from one metric space to another metric space

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

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

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

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

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

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

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

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

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

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

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

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

Originally Posted by tarheelborn
2) I need to prove that, in fact, $\displaystyle \displaystyle \lim_{n \to \infty} \left(f\left(x_n\right)\right) \neq f\left(a\right)$. This means that it is not true that $\displaystyle \forall \ \epsilon > 0 \ \exists \ N \in \mathbb{N}$ so that, for $\displaystyle n \geq N$, then $\displaystyle \rho_2\left(f\left(x_n\right), f\left(a\right)\right)<\epsilon$. Since when $\displaystyle \delta=\frac{1}{n}$, $\displaystyle f\left(B\left[a;\frac{1}{n}\right]\right) \notin B\left[f\left(a\right);\epsilon\right]$ 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 $\displaystyle \displaystyle \lim_{n \to \infty} \left(f\left(x_n\right)\right) = f\left(a\right)$ then there exists an $\displaystyle N$ such that $\displaystyle \rho_2\left(f\left(x_n\right), f\left(a\right)\right)<\epsilon$ for all $\displaystyle n\geqslant N$. But since $\displaystyle \rho_2\left(f\left(x_n\right), f\left(a\right)\right)\geqslant\epsilon$ for all $\displaystyle n$, that is a contradiction. Therefore $\displaystyle f(x_n)$ does not converge to $\displaystyle f(a)$.

3. Sorry about the typos! So the way to avoid dealing with $\displaystyle \frac{1}{\epsilon}$ not being an integer is to choose N so that it is an integer ... ? Thank you so much!

4. If you want to prove 2 $\displaystyle \Rightarrow$ 1 , then consider this
$\displaystyle \ f: X \rightarrow Y$
$\displaystyle \ f^{-1} V_\epsilon (y)$ is an open set in X for every open set $\displaystyle \ V_\epsilon (y)$ in Y because f is cont
$\displaystyle \ f^{-1} V_\epsilon (f(a))$ contains a and is an open set of X
Lets call $\displaystyle \ f^{-1} V_\epsilon (f(a)) in X, V_\delta (a)$
since $\displaystyle \lim_{n\to\infty} x_n = a$
$\displaystyle \forall$ open neighbourhoods of a, i.e r >0, $\displaystyle \ V_r (a) \bigcap S \not= \emptyset$ , where S is the set containing $\displaystyle \{ x_n \}$; actually $\displaystyle \{ x_n \}$ is eventually in $\displaystyle \ V_r (a) \forall r > 0,$
so $\displaystyle \exists N_0 : \forall n \ge N_0 x_n \in V_\delta (a)$
$\displaystyle \ x_n \in V_\delta (a) \rightarrow f(x_n) \in V_\epsilon (f(a))$
so $\displaystyle \forall \epsilon > 0, \{f(x_n)\}$ is eventually in $\displaystyle V_\epsilon (f(a))$