# Understanding the Lim infimum and Lim supremum

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?

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

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 $$A_{1} \not\subseteq A_{2}$$ 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)