A question about the formulation of the definition of a limit for sequences

Question

So I know the definition of a limit of a the sequence is:

$a$ is a limit of a sequence $\{x_n\}$ if given $\epsilon>0$ there exists a positive integer $N$ such that $|x_n-a|<\epsilon$ for all $n \geq N$.

My question is, why does the order of stating given $\epsilon>0$ there exists... matter. In other words, why can we not say there a exists a positive integer $N$ such that given $\epsilon>0$...

In other words, one $N$ value can work for all values of $\epsilon$ because if it works for an arbitrarily small real number then surely it works for all real numbers arbitrarily bigger than it? Thus making defining a different $N$ value for each $\epsilon$ unnecessary.

Answer 1

The $N$ may tend to infinity, which is the problem. Here's my example:

Let $a_n = \frac{1}{n}$. Then given $\epsilon$, we may let $N = \lceil \frac{1}{\epsilon} \rceil$. However, no single $N$ will work for all $\epsilon$: such an $N$ would need to be $\geq \frac{1}{\epsilon}$ for all $\epsilon$. Yet $a_n$ clearly tends to 0.

Answer 2

Let me write it a bit more technically:

$$ \forall \varepsilon >0 \colon~~ \exists N\in\mathbb{N}:~~ \forall n\ge N: \left|a_n-a\right|<\varepsilon $$ Then $a_n \to a$

Or even more detailed:

$$ \forall \varepsilon >0 \colon~~ \exists N=N(\varepsilon)\in\mathbb{N}:~~ \forall n\ge N: \left|a_n-a\right|<\varepsilon $$ Then $a_n \to a$

Now the explanation:

You can view such definitions always as a kind of game, that works as follows: At $\forall$ statements it is your opponents turn, at $\exists$ statements it is your turn. The goal is to satisfy the (in)equality or something like that.

If the claim is true, you can always win no matter what your oponent does (if you are clever). If the claim is false, you will always loose (if your opponent is clever).

So for the example above: Your "enemy" picks some $\varepsilon >0$ and tells it to you, then it is your turn. You now have to choose some $N\in\mathbb{N}$ and tell it your opponent. Then again your opponent chooses some $n\ge N$ and now you check if the statement $\left|a_n-a\right|<\varepsilon$ is correct. If so - you win - if not - your opponent wins -

Answer 3

That is not necessarily the case. Stating the epsilon first is important to start having a feel for what N actually is. If you were to pick an N first, then it must have the property of satisfying the condition for every possible epsilon taken.

Take $x_n = \{\frac 1 n\}_{n \in \mathbb{N}} $ and $a=0$

Take $N=1$, is it true to say that $|\frac 1 n|< \epsilon$ after $n \ge 1$, $\forall \epsilon >0$ is true?

Does this fact change for different $N$'s?