# Limit Inf/Sup of Sequence of Set Example

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?

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@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 $\displaystyle\bigcup_{n\ge 1}I_n=[0,1)$.

Now

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

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Thanks for the explanation. It was very clear. – user40200 Sep 13 '12 at 14:08