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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

$$\bigwedge_{n=1}^{\infty}\bigvee_{k=n}^{\infty}x_{k} = \sup\{ x: x

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.2011-11-03
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    Many thanks Florian - the problem now makes more sense. I will see if I can complete it!2011-11-03
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    Probably it is written in this way because that way it readily generalizes to more general lattices. For example, it works [for sets](https://en.wikipedia.org/wiki/Set-theoretic_limit) with $\bigwedge\equiv\bigcap$ and $\bigvee\equiv\bigcup$.2017-02-06

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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.

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    Many thanks 3Sphere - these definitions now mean the problem makes more sense.2011-11-03
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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}$2011-11-03
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    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:x2011-11-03
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    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 $\underset{n\geq 1}{inf}\underset{k\geq n}{sup}~x_{k}$?2011-11-05
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    For info I subsequently asked this question in another post [link](http://math.stackexchange.com/questions/80899/limit-superior-of-a-sequence-is-equal-to-the-supremum-of-limit-points-of-the-seq) and it was answered.2011-11-19
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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

Now assume that $x

So we have shown that $x< x_{n}~i.o.~\Rightarrow x\leq r$, and $x

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 $-\epsilon+rr$. By the density of the reals there exists a $\delta>0$ such that $r<-\delta+y