## Probability exercise: limit of sequence using definition

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rolo91
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### Probability exercise: limit of sequence using definition

I've got this exercise from my probability classes, and I'm stuck. I could really use some help. I don't want the exercise solved, I just don't know what I'm supposed to do and I would like some guidance; however, I don't have a lot of time since the exercise is supposed to be delivered tomorrow (that is, in about 10 hours ><) so ANY help is going to be appreciated.

I have to study the limit of the following sequence:

{An}n=1, with {An} = { (x,y) € R^2 /[ -1+(1/n)] <= x <=[ 1 + (1/n)] , -1 <= y <= 1 }

Now, if I understand it well, the limit of a sequence of sets is a set which contains the sets of every term in the sequence.

In a non-formal way, I see that the y will always be between -1 and 1, and the x will move between -1 (which is the smallest posible value for [ 1 + (1/n)], and 2 (the largest possible value for [ 1 + (1/n)]).

So, intuitively, I'd say that the solution is -1 <= x <= 2, -1 <= y <= 1.

However, I'm not supposed to do that. Im supposed to find the superior limit, which according to the definition is And that is when my problem comes, because I don't know how I should use that definition.

Any help?

skullturf
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### Re: Probability exercise: limit of sequence using definition

As for the purely "intuitive" part of the question:

When n gets big, -1 + 1/n gets close to -1, but 1 + 1/n gets close to 1.

Speaking purely informally, if you care about large values of n, then 1 + 1/n is close to 1, not 2.

rolo91
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### Re: Probability exercise: limit of sequence using definition

Yes, I know, but if I understood the definition well, then the limit has to contain all "previous" sets, not the ones closer to the limit.

So, for example, for n = 1 there will be a set with x <= 2. And the limit has to contain that, too.

In other words, the way I understand it I have to care for all possible values of n, not only the large ones.

I may be wrong, of course, as I said this is little more than a wild guess...

Qaanol
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### Re: Probability exercise: limit of sequence using definition

The way limsup works here is, you look at the sequence of unions:

B1 = Union of Am for m = 1, 2, 3, …
B2 = Union of Am for m = 2, 3, 4, …
B3 = Union of Am for m = 3, 4, 5, …

Bn = Union of Am for m = n, n+1, n+2, …

Then you take the intersection of all the B’s.
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rolo91
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### Re: Probability exercise: limit of sequence using definition

Oh, ok. So then, what I'm taking as the limit is "what is left" after progressively getting rid of the smaller sets, that is, the sets in the largest numbers of the sequence, right?

If that is correct, then what skullturf said is right, and the limit will be -1=< x =< 1. But the main problem I'm having is still there: How can I show that formally, intuition aside?

I mean, I can't calculate all the unions and then intersect them, so what do I do?

Qaanol
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### Re: Probability exercise: limit of sequence using definition

rolo91 wrote:Oh, ok. So then, what I'm taking as the limit is "what is left" after progressively getting rid of the smaller sets, that is, the sets in the largest numbers of the sequence, right?

If that is correct, then what skullturf said is right, and the limit will be -1=< x =< 1. But the main problem I'm having is still there: How can I show that formally, intuition aside?

It’s actually going to be -1<x<=1, which I’m sure you’ll be able to show formally.

rolo91 wrote:I mean, I can't calculate all the unions and then intersect them, so what do I do?

Since you’re looking at the intersection of all the B’s, you need to find those x-values which are in every single one of the B’s.

What does it mean in terms of the Am to say that a given x-value is in every Bn?

Given an x-value, can you determine if there is some N (which might depend on x) such that BN does not contain that x-value?
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rolo91
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Joined: Mon Aug 06, 2012 10:21 am UTC

### Re: Probability exercise: limit of sequence using definition

Qaanol wrote:What does it mean in terms of the Am to say that a given x-value is in every Bn?

I guess it means that m reaches a point (M) where every Am superior to that M contains x (or at least, there are infinite Am superior to M that contain x). Otherwise, it would be possible to find a Bn large enough so that all the Ams which contain x are "left behind", outside that Bn . I don't know if I'm explaining well what I'm trying to say.

So, I'm trying to show that it is not possible to find a Bn large enough so that all the Ams that contain x are outside the set(?)

Given an x-value, can you determine if there is some N (which might depend on x) such that BN does not contain that x-value?

well, I know that 1/n will always be between 0 and 1, for every possible n. So, because of what I explained before intuitively, I know that if that X is outside (-1,1) I'm supposed to be able to find that x.

Now, I guess I need to find the relation between the x-value and the N from which said x is going to be left outside the set Bn, n>N.

I don't know how, though.

phlip
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### Re: Probability exercise: limit of sequence using definition

Take, say, the point (x,y) where x=1.1 and y=0.5... for which values of n is this an element of An? For which values of n is this an element of Bn? Should this point be in the limit?

Take, say, the point (x,y) where x=0.9 and y=0.5... for which values of n is this an element of An? For which values of n is this an element of Bn? Should this point be in the limit?

Can you generalise this out to an arbitrary (x,y) point?

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