Here is a homework question, TRUE/FALSE:
$$\lim_{n\to\infty}a_n=a\Longleftrightarrow$$
$\forall\varepsilon>0,\ \exists N\in\mathbb{Z^+},\ \text{whenever}\ n>N\Rightarrow|a_n-a|<\varepsilon$. Answer: TRUE
$\exists N\in\mathbb{Z^+},\forall\varepsilon>0,\ \ \text{whenever}\ n>N\Rightarrow|a_n-a|<\varepsilon$. Answer: FALSE
I am confused that what is difference between 1 & 2 (or say why No.2 is wrong)?
Thanks!
The first statement is the definition of convergence of the sequence $\{a_n\}$ to $a$; no matter how close you want the terms in the sequence to get to $a$ (i.e. within $\varepsilon$), after finitely many terms (i.e. $N$ of them), the sequence will stay that close.
The second statement says that after finitely many terms (i.e. $N$ of them) the sequence satisfies $|a_n - a| < \varepsilon$ for every $\varepsilon > 0$. This is equivalent to saying that, after the first $N$ terms, $a_n = a$, i.e. the sequence is eventually constant. This is not equivalent to convergence because not every convergent sequence is eventually constant (but every eventually constant sequence is convergent, so the second statement implies $\lim\limits_{n\to\infty}a_n = a$, but it is not equivalent to it). For example, the sequence given by $a_n = \frac{1}{n}$ converges to $0$ but it is not eventually constant.
Note that $N$ is often dependent on the value of $\varepsilon$, $$ N = N(\varepsilon) $$ not "one size (of $N$) fits all ($\varepsilon$ challenges)".
In the first case you are saying that for every $\epsilon > 0$ there is $N \in \mathbb{Z}^+$ so that the rest holds. In the second case you're saying there is an $N \in \mathbb{Z}^+$ so that $\forall \epsilon > 0$ the following holds.
If this distinction isn't clear think about $a_n = \frac{1}{n}$. For the first definition it's clear that $N = \lfloor \frac{1}{\epsilon} \rfloor +1$ so that the following holds, but if we consider the second definition there is never a single $N$ so that the statement holds for all $\epsilon > 0$. If you want a proof of that then assume there does exist such an $N$ i.e. $$ \forall \epsilon > 0 \; \frac{1}{n} < \epsilon \; \forall n \ge N $$ but what if $\epsilon = \frac{1}{2N}$, then it should hold that $$ \frac{1}{N} < \frac{1}{2N} $$ but this is surely a contradiction.
Note: the order in which we apply quantifiers does matter.
More generally, let $P(x,y)$ be some property of $x,y$: $$\exists x: \forall y \ P(x,y) \nLeftrightarrow \forall y, \exists x: P(x,y).$$
e.g. consider the following two statements, for simplicity, to show that this isn't the case:
$S_1: \forall x \in \mathbb{R}, \exists y \in \mathbb{R}:y>x$ (i.e. for every real number $x$, there is some other real number $y$ which is bigger than $x$). This is true.
Now let's swap the order of the quantifiers:
$S_2: \exists y \in \mathbb{R}: \forall x \in \mathbb{R}, y>x$ (i.e. there exists some real number $y$ which is bigger than all real numbers). This is false (as $\mathbb{R}$ is unbounded above).
This example shows that changing the order in which we apply quantifiers affects the statement.
The same idea applies in your case.
The first is the correct statement. It says: no matter how small $\epsilon$ is, you can always choose a large enough $N$ so that every $a_n$ is within $\epsilon$ of $a$ whenever $n>N$. The second statement says something quite different. It says that if I hand you some large integer $N$, then no matter how small $\epsilon$ is, $a_n$ is within $\epsilon$ of $a$ whenever $n>N$. This second statement is plainly false. For example, after I give you $N$, simply choose $\epsilon$ to be smaller than some difference $|a_n-a|$ for $n>N$.
The second statement says, in effect, that there comes a point beyond which all the terms of the sequence are exactly equal to $a$.
