Clarification on lim inf and lim sup of a serie of sets

Question

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...

Answer 1

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.