I'm trying to understand what $\limsup A_n= \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N} A_n \right)$ and $\liminf A_n = \bigcup_{N=1}^\infty \left(\bigcap_{n \ge N} A_n\right)$ actually means.
I've already seen this answer, but there is something I don't understand about the lim sup explanation. If $x$ is in lim sup, then x should be in $A_1 \cup A_2 \cup A_3 ...$, and also $A_2 \cup A_3 \cup A_4...$, if we continue like this we see that x shouldn't exist, because $(A_1 \cup A_2 \cup A_3 ...) \cap (A_2 \cup A_3 \cup A_4...) = A_2 \cup A_3 \cup A_4...$ This means that the set is getting "smaller" (I know that it does not get smaller, because $n$ goes to infinity, but I don't know how to explain it better. For whichever n I choose, there will always be a bigger $n+1$ and so we know that $x$ will not be in I know that my logic is wrong, but I can't wrap my head around it.
Lim inf is easier to understand because it does make sense (atleast to me), the one thing I don't understand about it is "it's a member of all except finitely many of the $A$". Maybe it's my English, but what does it mean to be a menber of all except finitely many of the $A$? The way I understand lim inf, is that $x$ could be in the intersection of infinitely many sets, as well as all those sets except one, and so on...
You may be thinking of the case in which the sets $A_n$ are pairwise disjoint; in that case $\liminf_nA_n$ really is empty. But consider the case in which all of the sets $A_n$ are the same, so that $A_1=A_2=A_3=\ldots\;$. Then $$\bigcup_{n\ge N}A_n=\bigcup_{n\ge N}A_1=A_1\,,$$ so $$\bigcap_{N\ge 1}\bigcup_{n\ge N}A_n=\bigcap_{N\ge 1}A_1=A_1$$ as well. That rather extreme example shows clearly that going from $\bigcup_{n\ge 1}A_n$ to $\bigcup_{n\ge 2}A_n$ doesn’t mean that you lose everything in $A_1$: it just means that you lose the things that aren’t also in $A_n$ for some $n>1$.
When we say that $x$ is a member of all but finitely many of the $A_n$, we mean that $\{n\in\Bbb Z^+:x\notin A_n\}$ is finite. Suppose that $A_n=\left\{x\in\Bbb R:x>\frac1n\right\}$. Let $y$ be any positive real; there is a smallest positive integer $m$ such that $\frac1m<y$. If $y=\frac25$, for instance, $m=3$: $\frac13<\frac25$, but $\frac12\not<\frac25$. Then $\frac1n<y$ whenever $n\ge m$, so $\{n\in\Bbb Z^+:y\notin A_n\}=\{n\in\Bbb Z^+:n<m\}$. This set has just $m-1$ members, so it’s finite.