How is the distance between the n-th term of a converging sequence and its' limit not zero?

I'm confused as to what $|a_n-a|<\epsilon$ really means. The reason for that is the following proof regarding the assertion that a converging sequence $(a_n)$ has one and only one limit.

This is the proof my professor gave:

Assume the opposite, that the convergent sequence $(a_n)$ has two limits:

$\displaylines{\lim_{n\rightarrow \infty}a_n=a}$ and $\displaylines{\lim_{n\rightarrow \infty}a_n=b}$.

We will now show that $a=b$.

From $\displaylines{\lim_{n\rightarrow \infty}a_n=a}$ we have that $\forall \epsilon > 0$ there exists some $n_1(\epsilon)\in \mathbb{N}$ such that for all $n>n_1(\epsilon)$ the following holds:

$|a_n-a|<\frac{\epsilon}{2}$,

Similarly, from $\displaylines{\lim_{n\rightarrow \infty}a_n=b}$ we have that $\forall \epsilon > 0$ there exists some $n_2(\epsilon)\in \mathbb{N}$ such that for all $n>n_2(\epsilon)$ the following holds:

$|a_n-b|<\frac{\epsilon}{2}$.

Let $n_0=max\{n_1(\epsilon),n_2(\epsilon)\}$. Then, for all $n\geq n_0$ the following inequalities hold:

$|a_n-a|<\frac{\epsilon}{2}$ and $|a_n-b|<\frac{\epsilon}{2}$.

Now, for all $n\geq n_0$ we have:

$|a-b|=|(a-a_n)+(a_n-b)|\leq|a-a_n|+|a_n-b|=|a_n-a|+|a_n-b|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

Therefore, $|a-b|<\epsilon, \forall\epsilon>0\iff|a-b|=0\implies a=b$.

Now, if this is what the line above implies, then doesn't for instance $|a_n-a|<\epsilon, \forall \epsilon > 0$ also imply that $a_n=a$? Am I missing something?

Solution 1:

The definition of limit is: for every $\epsilon > 0$, there exists a positive integer $N$ $\textbf{dependent on $\epsilon$}$ such that $\mid a_n-a\mid < \epsilon$ for every $n \geqslant N$.

The result your professor uses at the very end is: if $a$ and $b$ are $\textbf{fixed}$ real numbers that satisfies $\mid a-b \mid < \epsilon$ for every $\epsilon > 0$, then $a=b$.