Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In "A Probability Path", they have an example that states that the lim inf and lim sup of [0,n/(n+1)) is equal to [0,1). I guess I don't see how [0,1) is in all the sets except a finite number of ties or how it is in an infinite number of sets. Can someone give give a demonstration of why it is so?

share|cite|improve this question
@André: I take it that we’re dealing with this here. – Brian M. Scott Sep 13 '12 at 4:44

Let $$I_n=\left[0,\frac{n}{n+1}\right)=\left[0,1-\frac1{n+1}\right)\;.$$ Note that $I_1\subseteq I_2\subseteq I_3\subseteq\ldots\;$, and that $\displaystyle\bigcup_{n\ge 1}I_n=[0,1)$.


$$\limsup_nI_n=\bigcap_{n\ge 1}\left(\bigcup_{k\ge n}I_k\right)\;,\tag{1}$$ and

$$\liminf_nI_n=\bigcup_{n\ge 1}\left(\bigcap_{k\ge n}I_k\right)\;.\tag{2}$$

Let’s start with $(1)$.

$$\bigcup_{k\ge n}I_k=I_n\cup I_{n+1}\cup I_{n+2}\cup\ldots=\bigcup_{k\ge 1}I_k=[0,1)\;,$$ since the sets $I_k$ are increasing: $I_1$ through $I_{n-1}$ are all contained in $I_n$ anyway. Thus, $$\bigcap_{n\ge 1}\left(\bigcup_{k\ge n}I_k\right)$$ is just the intersection of infinitely many copies of $[0,1)$:

$$\limsup_nI_n=\bigcap_{n\ge 1}\left(\bigcup_{k\ge n}I_k\right)=\bigcap_{n\ge 1}[0,1)=[0,1)\;.$$

If you prefer to think of the $\limsup$ in terms of whether a point is in infinitely many of the sets $I_k$, suppose that $x\in[0,1)$. Then $x<1$, so there is an $n\in\Bbb Z^+$ such that $x<1-\frac1n$. But then $x\in I_n$, and moreover $x\in I_k$ for every $k\ge n$, so $x$ really is in infinitely many of the sets $I_k$; this shows that $[0,1)\subseteq\limsup_nI_n$, and the reverse inclusion is obvious.

Now let’s look at $(2)$. $$\bigcap_{k\ge n}I_k=I_n\cap I_{n+1}\cap I_{n+2}\cap\ldots=I_n\;,$$ since $I_n$ is a subset of all of the later $I_k$’s. Thus,

$$\liminf_nI_n=\bigcup_{n\ge 1}\left(\bigcap_{k\ge n}I_k\right)=\bigcup_{n\ge 1}I_n=[0,1)\;.$$

Again, if you prefer to think in terms of points being in all but finitely many of the sets $I_k$, suppose that $x\in[0,1)$; we just saw that there is an $n\in\Bbb Z^+$ such that $x\in I_k$ for every $k\ge n$, so $x$ is in all but finitely many of the $I_k$, and therefore $[0,1)\subseteq\liminf_nI_n$. Again the reverse inclusion is clear, because if $x$ is in even one of the sets $I_k$, then it’s in $[0,1)$.

share|cite|improve this answer
Thanks for the explanation. It was very clear. – user40200 Sep 13 '12 at 14:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.