epsilon-delta proofs 8

In a metric space, we are able to estimate the distance between elements with the distance function. Therefore, without a real numerical sequence explicitly we can define the "limit".

We write $\lim_{x\rightarrow c}f(x)=y$  when $f(x)$  approaches $y$ as $x$ approaches $c$. Suppose that the set $A$ including $x$ and the set $B$ onto which $f(x)$ maps are both metric spaces. Hence, there are the distance function $d_A(p,q)$  of $A$  and $d_B(p,q)$  of $B$.

$\lim_{x\rightarrow c}f(x)=y$  means that for any $\epsilon>0$ , there exists a $\delta>0$  such that if
$$d_A(x,c)<\delta$$  
then 
$$d_B(f(x),y)<\epsilon$$

Preceding definition was as follow.

$a_n$ approches $a$, when for any $\epsilon >0$, there is a $N>0$ such that if $n\geq N$,
$$|a_n-a|< \epsilon$$

Let us equate $d_A(x,c)$  with $n\geq N$  and $d_B(f(x),y)$  with $|a_n-a|$ . Only $n\geq N$ is not a distance function. Please understand carefully that $d_B(f(x),y)<\epsilon$ is true for "every" $x$ which satisfies $d_A(x,c)<\delta$ .