Meaning of "occurs for infinitely many" and "all but finitely many"

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 $n$.

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.

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.