Dtyjlui

There are a bunch of questions I've gone through here that talk about how to solve a specific question. I am more curious on the intuition behind a particular problem. I am self studying probability theory, and am now in the analysis section.

I understand the general concepts of infinum and supremum:

$inf A_k = cap_{k=n}^infty A_k$

The intuition here is that the intersection of all possible of k-length subsequences will be the greatest lower bound.

The same goes for supremum:

$sup A_k = cup_{k=n}^infty A_k$

In this case the union of all possible k-length subsequences will be the least upper bound.

This problem however has me confused:

Check that $liminf_{n to infty} [0, frac{n}{n+1}) = limsup_{n to infty}[0, frac{n}{n+1}) = [0, 1)$

Setting out to try this I attempted to solve one part of it:

$liminf_{n to infty} [0, frac{n}{n+1})$

$lim_{n to infty} [inf_{k le n} [0, frac{n}{n+1})]$

$cup_{n = 1}^infty cap_{k=n}^infty [0, frac{n}{n+1})$

So I need to try to find the union of all intersections of the subsequences. This is where my intuition falls apart. Writing down some test sequences:

$k = 1: [0, frac{1}{2}) cap [0, frac{2}{3})cap [0, frac{3}{4}) cap ...$

$k = 2: [0, frac{2}{3}) cap [0, frac{3}{4}) cap [0, frac{4}{5}) cap ...$

So now looking a bit deeper at $k = 1$

$[0, frac{1}{2}) cap [0, frac{2}{3})$

Would give me $[0, frac{1}{2})$ since its the only common subsequence right? Continuing down the line, it would appear eventually this repeated intersectioning would result in $[0, 1)$. This would mean that unioning all of the results of the intersectioning would also be $[0, 1)$ - right? Doing this for $k = 2$ reveals the same thing.

Am I proceeding through this question correctly? I don't have a professor available to ask...so I want to take the time to make sure I am doing this correctly.

Thank you!

EDIT: I've realized part of my interpretation was wrong:

For

$k = 1: [0, frac{1}{2}) cap [0, frac{2}{3})cap [0, frac{3}{4}) cap ...$

This intersection will only ever result in $[0, frac{1}{2})$ because it is all the sequences here have in common. For $k = 2$ this value will be $[0, frac{2}{3})$. This continues on. Then, when you get to the unioning part you end up with:

$[0, frac{1}{2}) cup [0, frac{2}{3}) cup ...$

Which is the union of all of the k intersection results up to infinity.

etc etc. If you do this, it is obvious the union becomes $[0, 1)$ in the limit.

Is this more correct?

First, here is my mental picture:

Imagine you have a set of friends $X$. Every day some of your friends come as guests to your house. In day $n$, the set of guests $A_n$ arrive.

Then:

The $limsup A_n$ are those friends that you, at any given day, are garanteed to see again, at some future day.

The $liminf A_n$ are those friends that, on some day, stop going home, they are present on each following day.

In this case, each set the sets keep "growing", $A_{n} subseteq A_{n+1}$. Since $frac{k}{k+1} to 1 $ We get:

$$limsup A_n = [0,1) cap [0,1) cap [0,1) cap ... = [0,1) $$

and

$$liminf A_n = [0,frac{1}{2}) cup [0,frac{2}{3}) cup [0,frac{3}{4}) cup ... = [0,1) $$.

If, instead $A_n = (-frac{1}{n}, frac{n}{n+1}) $, then $A_{1} notsubseteq A_{2}$ since $-1 < -frac{1}{2}$ and but we have:
$$limsup A_n = (-1, 1) cap (-frac{1}{2}, 1) cap (-frac{1}{3},1) cap ... = [0,1) $$ since all "guests" below $0$ will leave forever at some point ($0$ we are garanteed to see again, on any present $n$, and $1$ will never come). Aslo note that this is an intersection of open sets that is not open (and also not closed), although every finite intersection is open. And

$$liminf A_n = bigcup_{n=1}^{infty} bigcap_{k = n}^{infty} (-frac{1}{k}, frac{k}{k+1}) = [0,frac{1}{2}) cup [0,frac{2}{3}) cup [0,frac{3}{4}) cup ... = [0,1) $$ since all "guests" below $1$ will be forever present after some point. (and $1$ will never come)

Let $A_k = [ 0, frac{k}{k+1})$.

We have $A_k subset A_{k+1}$, for all $k > 0$. Hence, $bigcap_{k=n}^infty A_k = A_n$. Therefore, $lim inf_{n rightarrow infty} A_n = bigcup_{n=1}^infty (bigcap_{k=n}^infty A_k) = bigcup_{n=1}^infty A_n = [0,1)$.

Now, $bigcup_{k=n}^infty A_k = bigcup_{k=n}^infty [0,frac{k}{k+1}) = [0,1)$. Hence, $lim sup_{n rightarrow infty} A_n = bigcap_{n=1}^infty (bigcup_{k=1}^infty A_k)= bigcap_{n=1}^infty [0,1) = [0,1)$.