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:


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

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?


2 Answers 2


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.


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:

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


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

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)


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