Examples of taking limits of intervals

[nasshi]

I guess I've fallen through some of the cracks in the plethora of definitions I've learned, or I just never had enough examples of taking limits of intervals. Anyways, which is true, and why?

$\cap^{\infty}_{n=1}(0,\frac{1}{2^{n-1}}]=0?$
$\cap^{\infty}_{n=1}(0,\frac{1}{2^{n-1}}]=\varnothing?$
I think the first.

However, if I were to write the following instead:
$\lim _{n \rightarrow \infty} \cap_{n}(0,\frac{1}{2^{n-1}}]=\varnothing$,
would I be correct?

If not, why?

Thanks.

 

[HallsofIvy]

I guess I've fallen through some of the cracks in the plethora of definitions I've learned, or I just never had enough examples of taking limits of intervals. Anyways, which is true, and why?

$\cap^{\infty}_{n=1}(0,\frac{1}{2^{n-1}}]=0?$
$\cap^{\infty}_{n=1}(0,\frac{1}{2^{n-1}}]=\varnothing?$
I think the first.

No, the first doesn't even make sense- on the left you have a set and on the right you have a number. If you meant
\cap^{\infty}_{n=1}(0,\frac{1}{2^{n-1}}]=\{ 0 \}
That would make sense (both sides of the equation are sets) but would also be incorrect. The notation (0, x] means "all points between 0 and x, including x but not 0". 0 is not in any of those intervals and so cannot be in their intersection. It is the second that is correct.

However, if I were to write the following instead:
$lim _{n \rightarrow \infty} \cap^{n}_{k=1}(0,\frac{1}{2^{n-1}}]=\varnothing$,
would I be correct?

If not, why?

Thanks.

Yes, the second is just definition of
$\cap^{\infty}_{n=1}
and is exactly the same as the second of your first two equations.

 

[nasshi]

HallsofIvy,

I'm trying to show that for F_{n} \subset F_{n+1} for all n (they're sigma-algebras), that \cap ^{\infty}_{i=1}F_{i} may not be a sigma algebra. Counter example is what the exercise says.

I was thinking of a sequence of intervals that never include 0. But I didn't know how to properly express it such that the intersection never included 0. I didn't want to include any epsilons, but I it seems I have to. Here's what I have now.

\cap^{\infty}_{n=1}(0,1-\frac{1}{2^{n-1}}+\varepsilon]

Will that suffice, or is there a better example?

 

[HallsofIvy]

HallsofIvy,

I'm trying to show that for F_{n} \subset F_{n+1} for all n (they're sigma-algebras), that \cap ^{\infty}_{i=1}F_{i} may not be a sigma algebra. Counter example is what the exercise says.

I was thinking of a sequence of intervals that never include 0. But I didn't know how to properly express it such that the intersection never included 0. I didn't want to include any epsilons, but I it seems I have to. Here's what I have now.

\cap^{\infty}_{n=1}(0,1-\frac{1}{2^{n-1}}+\varepsilon]

Will that suffice, or is there a better example?

I don't understand what you are trying to do. A sigma algebra is a collection of sets (that is closed under unions and complements). You are saying that you want to give an example of a collection of sigma algebras whose intersection is not a sigma algebra but your example is a single collection of sets (NOT a sigma algebra) and you are taking the intersection of the sets in that single collection of sets.

 

[Take_it_Easy]

As far as I know you don't have to think that intersection as a limit just because the \infty symbol appears.
Whenever you have a NOT EMPTY collection of sets \cal A (for a collection of sets I mean a set \cal A whose elements are sets themselves... for istance, given a set A, the set {\cal P}(A) of the subsets of A is a collection of sets) you can define the \bigcap {\cal A} as the following set.

\bigcap {\cal A} = \left\{ x \in Y\, |\,\, \forall X \in {\cal A}, \ x \in X \right\}

where Y is any element of the collection of sets(that is non empty).

The definition do not depend on the choice of Y and you don't require any idea of convergence or limit.