# limit superior of a sequence - showing an alternate definition

I am wondering if anybody can help me with a problem regarding the definition of the limit superior of a sequence - or rather showing an alternate but equivalent defintion holds.

The question is: The limit superior of a numerical sequence $\{x_{k}\}$ (presumably this means the sequence is real valued) can be defined as the supremum of the set of limit points of the sequence. Show that this is the same thing as defining

$$\limsup _{n \to \infty} ~ x_n = \bigwedge_{n=1}^{\infty} \bigvee_{k=n}^{\infty} x_k .$$

This question comes from Chapter 4 of "Probability and Measure" by Patrick Billingsley. The problem is that Billingsley assumes the reader knows what the symbols $\bigwedge$ and $\bigvee$ are - but I do not! The best I have come up with is that they are the "meet" and the "join" symbols used in a lattice? Could anybody shed some light on how this problem might be attacked?

Billingsley does give some hints to the problem. He says that the following are all equivalent: $x<x_k$ i.o. OR $x<x_k$ for some $k\geq n$, for all $n\geq 1$ OR $x<\bigvee_{k=n}^{\infty} x_k$ for all $n\geq 1$. Using this kind of logic he shows that

$$\bigwedge_{n=1}^{\infty}\bigvee_{k=n}^{\infty}x_{k} = \sup\{ x: x<x_n \text{ i.o.}\} .$$

Apparently the supremum of the set above can be seen to be the supremum of the limit points of the sequence - this would prove the result. I think I follow this derivation but I was taking the $\bigvee$ and $\bigwedge$ symbols to simply mean $\bigcup$ and $\bigcap$ for singleton sets $\{x_{k}\}$. I think this is the wrong assumption. I also cannot see the last assertion about the limit points of the sequence.

Any help would be much appreciated.

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Maybe $\bigvee$ means sup and $\bigwedge$ means inf? What I have seen is that $\vee$ was used for max (between two numbers) and $\wedge$ for min; and in your case it would yield the right definition. – Florian Nov 3 '11 at 18:55
Many thanks Florian - the problem now makes more sense. I will see if I can complete it! – dandar Nov 3 '11 at 23:33

Aliprantis and Burkinshaw, in Principles of Real Analysis (3rd ed.), make the following definitions on page 24:

$$\bigvee_{k=n}^{\infty} x_k = \sup_{k \geq n} \; x_k$$

$$\bigwedge_{k=n}^{\infty} x_k = \inf_{k \geq n} \; x_k$$

I assume that your text is using the same notational convention.

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Many thanks 3Sphere - these definitions now mean the problem makes more sense. – dandar Nov 3 '11 at 23:44

The correct statement is that $\displaystyle \overline{\lim}x_n=\inf_{n\in\mathbb{N}}\sup_{m\geqslant n}x_m$. So, what you want to show is that this right hand expression gives you the supremum over all the subsequential limits of your sequence (i.e. limit points).

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Thanks Alex for this. From what Florian has said then I get just what you have stated here, that is $\bigwedge_{n=1}^{\infty}\bigvee_{k=n}^{\infty}x_{k}=inf_{n\geq 1}sup_{k\geq n}x_{k}$ – dandar Nov 3 '11 at 23:34
As you state, proving the RHS of the above equation is the supremum over the set of all limit points of the sequence would give the result. It would also be good to show $inf_{n\geq 1}sup_{k\geq n}x_{k}=\{x:x<x_{n}~i.o.\}$, where the RHS is supremum over the set of all limit points of the sequence since this would be the way Billingsley solves it. However I will have a look at both. – dandar Nov 3 '11 at 23:49
Hi Alex and thanks again for your response. Where am I going wrong here? Define the sequence $x_{n}=sin(n)$. We have $\underset{k\geq n}{sup}~x_{k}=1$ for any subsequence. Thus $\underset{n\geq 1}{inf}\underset{k\geq n}{sup}~x_{k}=1$. Now the sequence has no limit points since it does not converge to anything so the supremum of all the subsequence limits is the supremum of the empty set which presumably is not equal to 1 (I actually do not know what it is). So we have a sequence where the supremum of limit points does not equal $\underset{n\geq 1}{inf}\underset{k\geq n}{sup}~x_{k}$? – dandar Nov 5 '11 at 9:52
For info I subsequently asked this question in another post link and it was answered. – dandar Nov 19 '11 at 14:28

Thanks again for everyody's help with his one. Here is the proof of the answer to my question. If there are any errors here then please feel free to point them out.

The limit superior of a sequence $\{x_{n}\}$ in $R$ can be defined as the supremum of the set of limit points of $\{x_{n}\}$. Define $\bigvee_{k=n}^{\infty}x_{k}=\sup_{k\geq n}x_{k}$, $\bigwedge_{k=n}^{\infty}x_{k}=\inf_{k\geq n}x_{k}$ and $r=\bigwedge_{n=1}^{\infty}\bigvee_{k=n}^{\infty}x_{k}=\inf_{n \geq 1}\sup_{k \geq n}x_{k}$. The problem is to show that $r$ is the supremum of the set of limit points of $\{x_{n}\}$ and hence is equal to the limit superior of $\{x_{n}\}$.

Proof. Let $r$ be defined as above and assume $r$ is finite. Now assume $x<r$. Take a subsequence $\{x_{k}\}_{k\geq m}$ for some $m\geq 1$ and let $\text{sup}_{k\geq m}\{x_{k}\}=M$. By assumption and the definition of $r$ we have $x<r\leq M$. Now if $M\in\{x_{k}\}_{k\geq m}$ then for some $x_{m'}\in \{x_{k}\}_{k\geq m}$ we have $x<r\leq x_{m'}$. If $M$ is not equal to any term in the subsequence, then by the definition of $M$, $x<M$ implies there exists a $x_{m'}\in \{x_{k}\}_{k\geq m}$ such that $x<x_{m'}<r\leq M$. So whether $M$ is in the subsequence or not we have that $x<x_{k}$ for some $k\geq m$. Since this holds for all $m\geq 1$ (i.e. for all subsequences), we have that $x<r \Rightarrow x < x_{k}$ for some $k\geq n$ for all $n\geq 1$. Equivalently $x<r \Rightarrow x < x_{k}~i.o.$

Now assume that $x<x_{k}~i.o.$. That is we assume $x<x_{k}$ for some $k\geq n$, for all $n\geq 1$. Again take a subsequence $\{x_{k}\}_{k\geq m}$ for some $m\geq 1$ and let $\text{sup}_{k\geq m}\{x_{k}\}=M$. Then since $x<x_{m'}$ for some $x_{m'}\in \{x_{k}\}_{k\geq m}$ we have $x<M$. This holds for all subsequences and so $x$ is a lower bound for the supremums of all subsequences. Thus $x\leq r$. So $x<x_{k}~i.o. \Rightarrow x\leq r$.

So we have shown that $x< x_{n}~i.o.~\Rightarrow x\leq r$, and $x<r~\Rightarrow x< x_{n}~i.o.$ Letting $A=\{x\in R:x<x_{n}~i.o.\}$ then we have $x\leq r$ for each $x\in A$. So $r$ is an upper bound of $A$. Now if $r'<r$ for some $r'\in R$, then by the density of the reals there exists a $r''\in R$ such that $r'<r''<r$. Since $r''<r \Rightarrow r''<x_{n}~i.o.$, then $r''\in A$. So $r$ is the least upper bound of $A$, i.e. $r=\text{sup}\{x\in R:x<x_{n}~i.o.\}$.

We will now show that $r$ is equal to the supremum of the set of limit points of $\{x_{n}\}$. For any $\epsilon>0$ we have $-\epsilon+r<x_{m}<r$ for some $x_{m}\in A$ (since $r$ is the supremum of A). So any open ball $B_{\epsilon}(r)$ contains a point of $A$ other than $r$, thus $r$ is a limit point of $\{x_{n}\}$. If $A$ has only one limit point then there is nothing to prove. If $A$ contains more than one limit point then let $y$ be one of these and assume $y>r$. By the density of the reals there exists a $\delta>0$ such that $r<-\delta+y<y$. Since $y$ is a limit point there exists a $x_{m'}\in A$ such that $x_{m'}\in B_{\delta}(y)$. So we have either $r<-\delta+y<x_{m'}<y$ or $r<y<x_{m'}<y+\delta$. Either way $r$ is less than a member of $A$ which contradicts $r$ being the supremum of $A$. Thus we must have $y\leq r$. Thus $r$ is the supremum of the set of limit points of $\{x_{n}\}$.Q.E.D.

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