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. 
 A: To directly address your question regarding the notation,
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.
A: 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.
A: 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).
