Understanding the supremum limit of a set Given a sequence ${A_n}$, we define the set
lim sup $A_n = \{x : x$ belongs to infinitely many $A_n$'s$\}$
That is -
lim sup $A_n = \bigcap_{m=1}^\infty (\bigcup_{n=m}^\infty A_n)$
I can't see how this makes sense.
When $m = 1$, $(\bigcup_{n=m}^\infty A_n) = A_1 \bigcup A_2 \bigcup A_3 \bigcup ...$
When $m = 2$, $(\bigcup_{n=m}^\infty A_n) = A_2 \bigcup A_3 \bigcup A_4 \bigcup ...$
and so on as $m$ goes to infinity.
So what is the intersection of all these unioned sets supposed to be?...It seems like it 'vanishes' as $m$ goes to infinity.
I can't see how the intersection of these unioned sets correlates with $\{x : x$ belongs to infinitely many $A_n$'s$\}$?
 A: About infinity
Your wording suggests that your concept of infinity is not entirely tuned in. The union
$$
\bigcup_{n=m}^\infty A_n
$$
is infinite for any $m$ so that $1\leq m<\infty$. And though $m$ tends to infinity it NEVER reaches it as infinity is NOT a natural number (in fact not a number at all). It means that $m$ runs through the sequence $1,2,3,...$ that literally goes on forever. So infinity is not a final state or anything like that that $m$ eventually reaches. It is the concept that $m$ keeps counting without an end, without any finite terminating state. So the union given above is never empty (it never 'vanishes'), but instead it produces the union of the infinite tail of sets $A_m,A_{m+1},...$ for any given finite natural number $m$.
The two descriptions of limsup
Let me try to demonstrate how these two descriptions of $\lim\sup$ coincide:
Let $x\in\{x:x\mbox{ belongs to infinitely many }A_n\mbox{'s}\}$ be given. Now consider some $m$. Since $x$ belongs to infinitely many $A_n$'s it must be contained in other $A_n$'s than just $A_1,A_2,...,A_{m-1}$. Hence it follows that $x\in A_{m+k}\subseteq\bigcup_{n=m}^\infty A_n$ for some non-negative integer $k$. Since this holds for any value of $m$ it follows that
$$
x\in\bigcap_{m=1}^\infty\left(\bigcup_{n=m}^\infty A_n\right)
$$
This proves that
$$
\{x:x\mbox{ belongs to infinitely many }A_n\mbox{'s}\}\subseteq\bigcap_{m=1}^\infty\left(\bigcup_{n=m}^\infty A_n\right)
$$
Now let $x\in\bigcap_{m=1}^\infty\left(\bigcup_{n=m}^\infty A_n\right)$ be given. Assume for a contradiction that $x$ belongs to only finitely many $A_n$'s. Let $A_k$ be the last set in the sequence containing $x$. But then we have
$$
\bigcap_{m=1}^\infty\left(\bigcup_{n=m}^\infty A_n\right)\subseteq\bigcap_{m=k+1}^\infty\left(\bigcup_{n=m}^\infty A_n\right)
$$
where $x$ is contained in the subset above but not in the superset which is absurd. This contradicts the possibility of $x$ being contained in only finitely many $A_n$'s. So it follows that $x$ also belongs to $\{x:x\mbox{ belongs to infinitely many }A_n\mbox{'s}\}$. This proves the other inclusion:
$$
\{x:x\mbox{ belongs to infinitely many }A_n\mbox{'s}\}\supseteq\bigcap_{m=1}^\infty\left(\bigcup_{n=m}^\infty A_n\right)
$$
