Meaning of "occurs for infinitely many" and "all but finitely many" Suppose $P(n)$ is certain statement. I hear all the time that my teachers say 
$$ P(n) \; \; \text{occurs for infinitely many} \; \; \;n $$
$$ P(n) \; \; \text{for all but finitely many} \; \; n $$
MY question: What are the precise definitions of this statements ? 
 A: I am assuming that $n \in \mathbb{N}$ in the following:
The first statement corresponds to $\forall n$ $\exists k\ge n$ such that $p(k)$ is true.
The second statement corresponds to $\exists n$ such that $\forall k \ge n$ $p(k)$ is true.
A: The first statement ($P(n) \; \; \text{occurs for infinitely many} \; \; \;n$) means that there is an infinite number of ways to satisfy the proposition $P(n)$. Just like Michael says, in $\mathbb(n)$ let $P(n)$ : n is even, then there is an infinite number of $n \in\mathbb N$ that satisfies $P(n)$ (because there are infinitely many even numbers).
The second statement ($P(n) \; \; \text{for all but finitely many} \; \; n$) means that, although there is an infinite number of ways of satisfying a condition $P(n)$, there is a finite number of exceptions. For example, in $\mathbb N$ let $P(n)$ : n > 5. There is a inifinite number of $n \in\mathbb N$ that satisfies $P(n)$ (there are infinitely many natural numbers greater than 5), but there is a finite number of exceptions, in our case, the numbers: 0, 1, 2, 3, 4 and 5 are those exceptions.
A: We say $P(n)$ occurs for infinitely many $n$ if $\{n \mid P(n)\ \text{is true}\}$ contains infinitely many elements.
We say $P(n)$ occurs for all but finitely many $n$ if $\{n \mid P(n)\ \text{is not true}\}$ contains only finitely many elements.
Example: For the set $\mathbb{N}$, consider $P(n):$ $n$ is even. We have
$$\{n \mid P(n)\ \text{is true}\} = \{2, 4, 6, \dots \}$$
and
$$\{n \mid P(n)\ \text{is not true}\} = \{1, 3, 5, \dots \}.$$
As $\{n \mid P(n)\ \text{is true}\}$ contains infinitely many elements, we can say that $P(n)$ occurs for infinitely many $n$.
As $\{n \mid P(n)\ \text{is not true}\}$ contains infinitely many elements, we cannot say that $P(n)$ occurs for all but finitely many $n$.
A: Some concrete examples.
For $P(n)$ = "$P$ is prime", $P(n)$ and $\neg P(n)$ both occur for infinitely many $n$: there is no last prime, nor a last non-prime (composite).
For $P(n)$ = "$P$ is a McNugget Number", P(n) is true for all except finitely many $n$: there's only 22 different McNugget counts that cannot be created (using boxes of 6, 9, and 20).
