Sequence

[bghalmeida]

Given an infinite collection [MATH]A_n, n= 1,2,\ldots[/MATH] of intervals of the real line, their intersection is defined to be [MATH]\bigcap_{n=1}^{\infty}A_n = \{x \, | \, (\forall n)(x \in A_n)\}[/MATH] Give an example of a family of intervals [MATH]A_n, n= 1,2,\ldots[/MATH] such that [MATH]A_{n+1} \subset A_n[/MATH] for all [MATH]n[/MATH] and [MATH]\bigcap_{n=1}^{\infty}A_n = \emptyset[/MATH]. Prove that your example has the stated property.

 

[Dr.Peterson]

To start, ignore the requirement that the intersection be empty. Can you just give an example of a family of intervals, each a subset of the one before?

 

[bghalmeida]

Yes. [MATH]\left(0, \frac{1}{2^{n+1}}\right)\subseteq\left(0, \frac{1}{2^{n}}\right)[/MATH]. The problem is that I am not sure about how can I prove the statement.

To start, ignore the requirement that the intersection be empty. Can you just give an example of a family of intervals, each a subset of the one before?

 

[JeffM]

Can you show that

[MATH]\left ( 0,\ \dfrac{1}{4} \right ) \subseteq \left ( 0,\ \dfrac{1}{2} \right ).[/MATH]
If so, what kind of proof does that suggest?

 

[bghalmeida]

Can you show that

[MATH]\left ( 0,\ \dfrac{1}{4} \right ) \subseteq \left ( 0,\ \dfrac{1}{2} \right ).[/MATH]
If so, what kind of proof does that suggest?

I think I will have to find an arbitrary number in the subset and prove that this number is also in the other set.

 

[pka]

Yes. [MATH]\left(0, \frac{1}{2^{n+1}}\right)\subseteq\left(0, \frac{1}{2^{n}}\right)[/MATH].
The problem is that I am not sure about how can I prove the statement.

Let's use your sequence: \(a_n=\left(0,\frac{1}{2^n}\right)\)
To prove that \(\bigcap\limits_{n = 1}^\infty {{a_n}} = \emptyset \) you need some knowledge sequence convergence.
The sequence \(\left(\frac{1}{2^n}\right)\) is decreasing and has limit zero. That means \((t > 0)(\left( {\exists n} \right)\left[ {\frac{1}{{{2^n}}} < t}\right]\).
That last bit means that nothing positive can be in every \(a_n\) but \((\forall n)[0\notin a_n)\) so the intersection is \(\emptyset\).

 

[bghalmeida]

Let's use your sequence: \(a_n=\left(0,\frac{1}{2^n}\right)\)
To prove that \(\bigcap\limits_{n = 1}^\infty {{a_n}} = \emptyset \) you need some knowledge sequence convergence.
The sequence \(\left(\frac{1}{2^n}\right)\) is decreasing and has limit zero. That means \((t > 0)(\left( {\exists n} \right)\left[ {\frac{1}{{{2^n}}} < t}\right]\).
That last bit means that nothing positive can be in every \(a_n\) but \((\forall n)[0\notin a_n)\) so the intersection is \(\emptyset\).

I'm having trouble understanding the logic, do you mind explaining it to me again?

 

[Dr.Peterson]

Yes. [MATH]\left(0, \frac{1}{2^{n+1}}\right)\subseteq\left(0, \frac{1}{2^{n}}\right)[/MATH]. The problem is that I am not sure about how can I prove the statement.

To prove this statement, show that if [MATH]0<x<\frac{1}{2^{n+1}}[/MATH], then [MATH]0<x<\frac{1}{2^{n}}[/MATH].

Or do you mean, you don't know how to prove that [MATH]\bigcap_{n=1}^{\infty}A_n = \emptyset[/MATH]?

For that, use the definition they gave for this intersection. Suppose that [MATH](\forall n) (x \in A_n)[/MATH], and obtain a contradiction, so that there is no such x.

 

[pka]

I'm having trouble understanding the logic, do you mind explaining it to me again?

First, have you had any course that studied sequence convergence? If not I do not understand your being asked to do this question.
O.K. the fact is \(\left(\frac{1}{2^n}\right) \downarrow 0\) or the sequence decreases to \(0\).
No positive number is less that every term of that sequence.
Thus the 'space' between \(0\) and the terms of the sequence shrinks to nothing( approaches the emptyset.)
That is a lot of hand-waving. But without the theories of convergence it is the best I can do.

 

[bghalmeida]

First, have you had any course that studied sequence convergence? If not I do not understand your being asked to do this question.
O.K. the fact is \(\left(\frac{1}{2^n}\right) \downarrow 0\) or the sequence decreases to \(0\).
No positive number is less that every term of that sequence.
Thus the 'space' between \(0\) and the terms of the sequence shrinks to nothing( approaches the emptyset.)
That is a lot of hand-waving. But without the theories of convergence it is the best I can do.

I haven't, but I am studying mathematical thinking so I have to prove this statemente, but I think I got it, thank's for your help.