What does this mean: there exist an integer N such that $n\ge N$? I'm reading Rudin's, Principles of Mathematical Analysis, and I keep tripping over this phrase. Usually the phrase by that it implies a some equation with n being the index, subscript, of a point. My guess is that the phrase means that there is two points and that 'n' is the limit. Not sure, I'll greatly appreciate any help with this.
(edit): I admit the question is vague without its implication but I wanted to clear with the premise first. Thanks for the fast and helpful responses.
 A: The phrase "there exist an integer N such that [for all] $n\ge N$, ..."
is quantified prepositional clause, (is not an assertion), itself, after which something is asserted about $n$ in relation to $N$.
That is, "there exist an integer N such that [for all] $n\ge N$ ... " [something is claimed to be  true about $n$ when for all $n \ge N$].
So, we say something like 

"Let $\{s_n\}$ be a sequence of real numbers with the following property: For every real M, there exists an integer $N$ such that $n\geq N$ implies $x_n\leq M,$ we then write $s_n \to +\infty$."  (p. 55 Rudin PMA, Definition 3.15)

Here, indeed, $n$ refers to the index (position) of some point in the sequence, $n\geq N$ meaning $s_n$ appears later in the sequence than does $s_N$, after which the sequence diverges. 
But $n$ can be a real number, as well, when the clause you are asking about is used in a different context. 
A: Let's consider a simple example.  Using the definition of the limit of a sequence, show that the sequence defined by
$$a_n=\frac{1}{n}$$
converges to $0$.  The definition says that no matter how small we choose a positive number $\epsilon,$ we should be able to find a big enough number $N$ so that as long as $n\ge N$, the terms $a_n$ are really close to $0$, that is, they are within a distance $\epsilon$ of $0$.
Let's start concretely and then move to the abstract.  Let's first choose $\epsilon=.1$.  Then, notice that $a_{10}=.1$, and that any term beyond $a_{10}$ is even closer to $0$.  This means that if we let $N=10$, we know that for any $n\ge 10$, the value of $a_n$ is within a distance $.1$ of $0$.  Notice we also could have chosen $N=11,12,\ldots$ because we would still have $a_n$ within $.1$ of $0$ so long as $n\ge 11,12,\ldots$.  Notice also that if $n<10$, then $a_n$ is not close enough to $0$.
Similarly, if we chose $\epsilon=.01$, you will see we could choose $N=100,101,102,\ldots$, and that for all $n\ge N$, $a_n$ is small enough:
$$|a_n-0|\le.01$$
Now, we can't continue to just choose different values for $\epsilon$ and prove it for those cases, because there are infinitely many possible values, and we don't have enough time to check all of them.  So let's just assume $\epsilon$ is any positive number.  For large enough values of $n$, we want the following inequality to hold:
$$|a_n|=\left|\frac{1}{n}\right|=\frac{1}{n}\le\epsilon$$
Rearranging the inequality, we see that we need $n\ge\frac{1}{\epsilon}$, hence, if we choose $N$ to be anything greater than $\frac{1}{\epsilon}$, then for all $n\ge N$, our desired inequality will hold, because then
$$\frac{1}{\epsilon}\le N\le n$$
Notice that as $\epsilon$ gets smaller, we must choose our $N$ to be larger and larger to obtain the desired closeness of $a_n$ to $0$, but it is always possible.
In this example, as in most, $N$ is a fixed large number such that a certain property holds for all $n\ge N$.  Here, the property is that the terms $a_n$ are all very close to the limit $0$ for all $n\ge N$.
A: an example that might have nothing to do with what is in rudin but wich could make you understand the concept a bit is this question : does there exist a number $N$ such that for any number of person , call it $n$ with $n\geq N$ there are at least two person of the same sex among those people? 
This number $N$ infact exists and is equal to $3$. So in this case : $\exists N : \forall n\geq N$ there are at least two person of the same sex among $n$ person.
