set of subsequential limits

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November 13th 2012, 03:30 PM
mobydick

set of subsequential limits

Let A be a closed subset of R. Prove that there exists a sequence of real numbers s.th. if B is the set of all subsequential limits, then B=A.

I know how to do it for a closed interval, but not for an arbitrary closed set. Any ideas?
November 13th 2012, 05:19 PM
Plato

Re: set of subsequential limits

Quote:

Originally Posted by mobydick

Let A be a closed subset of R. Prove that there exists a sequence of real numbers s.th. if B is the set of all subsequential limits, then B=A.

That statement is nonsense.
Please restate the question in a clearly readable format.
November 13th 2012, 05:27 PM
mobydick

Re: set of subsequential limits

Quote:

Originally Posted by Plato

That statement is nonsense.
Please restate the question in a clearly readable format.

What exactly don't you understand. A is an arbitrary closed subset of R. I need to prove that there exists some sequence {Xn} of real numbers, such that the set of subsequential limits of {Xn} is equal to A.
November 13th 2012, 09:39 PM
hollywood

Re: set of subsequential limits

A subsequential limit is the limit of a subsequence. So he wants to know: if A is a closed subset of $\mathbb{R}$ , is there a sequence $(x_n)_{n=1}^\infty$ such that for every $a\in{A}$ , there is a subsequence $(x_{n_k})_{k=1}^\infty$ that converges to a.

Let S be the set of all intervals of the form $(r_1,r_2)$ where $r_1$ and $r_2$ are rational. This is a countable set. If you choose $x_n$ to be a point from $A\cap{(r_1,r_2)}$ if one exists, then you can prove that any neighborhood of a contains an element from the sequence. You would also have to prove that a limit point of the sequence is in A.

- Hollywood
November 14th 2012, 01:32 AM
mobydick

Re: set of subsequential limits

Quote:

Originally Posted by hollywood

A subsequential limit is the limit of a subsequence. So he wants to know: if A is a closed subset of $\mathbb{R}$ , is there a sequence $(x_n)_{n=1}^\infty$ such that for every $a\in{A}$ , there is a subsequence $(x_{n_k})_{k=1}^\infty$ that converges to a.

Let S be the set of all intervals of the form $(r_1,r_2)$ where $r_1$ and $r_2$ are rational. This is a countable set. If you choose $x_n$ to be a point from $A\cap{(r_1,r_2)}$ if one exists, then you can prove that any neighborhood of a contains an element from the sequence. You would also have to prove that a limit point of the sequence is in A.

- Hollywood

Thanks, man but I don't think that works. What are the elements of the sequence?
If you mean that $(r_1,r_2)$ are intervals of rational numbers then if A is a singleton set of one irrational number your algorithm will fail. If you mean that $(r_1,r_2)$ are the intervals of real numbers between these rational numbers, then if A=[0,1] your algorithm will not construct the required sequence, which is a sequence of all the rationals in the interval [0,1].
Maybe I am missing something.
November 14th 2012, 04:17 AM
Plato

Re: set of subsequential limits

Quote:

Originally Posted by mobydick

A is an arbitrary closed subset of R. I need to prove that there exists some sequence {Xn} of real numbers, such that the set of subsequential limits of {Xn} is equal to A.

Well that is certainly a clear discrimination.
You want the set $B$ to be countable, a sequence, the closure of which is $A$ .

You said you can do it for a closed interval. If you show us what you have in mind we may be able you to generalize that method.
November 14th 2012, 04:30 AM
mobydick

Re: set of subsequential limits

Quote:

Originally Posted by Plato

Well that is certainly a clear discrimination.
You want the set $B$ to be countable, a sequence, the closure of which is $A$ .

You said you can do it for a closed interval. If you show us what you have in mind we may be able you to generalize that method.

A subsequential limit of a sequence ${x_n}$ is a limit of some subsequence of ${x_n}$ . B is the set of all such subsequential limits and it is not necessarily countable.

For the interval [0,1]:
The set of all rational numbers in the interval [0,1] is countable, so you can construct a sequence ${x_n}$ with range equal to this set. Since the rationals are dense in R, every point in [0,1] is a subsequential limit of ${x_n}$ .
November 14th 2012, 05:45 AM
Plato

Re: set of subsequential limits

Quote:

Originally Posted by mobydick

A subsequential limit of a sequence ${x_n}$ is a limit of some subsequence of ${x_n}$ . B is the set of all such subsequential limits and it is not necessarily countable.

In the OP you asked for a sequence. A sequence is a countable set by definition. That is the confusing part of this question. Every point of $A$ is the limit of a sequence of rationals. And you are right that the collection of those sequences may not the countable. But then it is not a sequence.
November 14th 2012, 07:25 AM
hollywood

Re: set of subsequential limits

Quote:

Originally Posted by mobydick

Thanks, man but I don't think that works. What are the elements of the sequence?
If you mean that $(r_1,r_2)$ are intervals of rational numbers then if A is a singleton set of one irrational number your algorithm will fail. If you mean that $(r_1,r_2)$ are the intervals of real numbers between these rational numbers, then if A=[0,1] your algorithm will not construct the required sequence, which is a sequence of all the rationals in the interval [0,1].
Maybe I am missing something.

I meant to choose an element of A in $(r_1,r_2)$ . If no such element exists, skip that interval and move to the next one.

In the case of A={a}, where a is irrational, you would choose a (the only possible choice) if $r_1<a<r_2$ , and otherwise you would skip. There are infinitely many intervals containing a, so the sequence would be a, a, a, ....

In the case of A=[0,1], yes, the result is not an enumeration of the rationals in A, but that's not the only possible sequence. Consider a point $a\in{A}$ . Choose any point $x_{n_1}<{a}$ (the same construction works for $x_{n_1}>{a}$ ). There are infinitely many intervals $(r_1,r_2)$ with $x_{n_1}<r_1<r_2<a$ , so choose one, and let $x_{n_2}$ be the choice from that interval. In this way, a subsequence approaching a can be found.

In fact, I think those two examples generalize to isolated points and limit points, so that's all the points of A.

You would still need to prove that there is no $b\not\in{A}$ that is the limit of a subsequence of my sequence.

- Hollywood
November 14th 2012, 08:03 AM
mobydick

Re: set of subsequential limits

Quote:

Originally Posted by hollywood

I meant to choose an element of A in $(r_1,r_2)$ . If no such element exists, skip that interval and move to the next one.

In the case of A={a}, where a is irrational, you would choose a (the only possible choice) if $r_1<a<r_2$ , and otherwise you would skip. There are infinitely many intervals containing a, so the sequence would be a, a, a, ....

In the case of A=[0,1], yes, the result is not an enumeration of the rationals in A, but that's not the only possible sequence. Consider a point $a\in{A}$ . Choose any point $x_{n_1}<{a}$ (the same construction works for $x_{n_1}>{a}$ ). There are infinitely many intervals $(r_1,r_2)$ with $x_{n_1}<r_1<r_2<a$ , so choose one, and let $x_{n_2}$ be the choice from that interval. In this way, a subsequence approaching a can be found.

In fact, I think those two examples generalize to isolated points and limit points, so that's all the points of A.

You would still need to prove that there is no $b\not\in{A}$ that is the limit of a subsequence of my sequence.

- Hollywood

This might work but we need to make it more precise. It is not at all clear how you choose the elements from the intervals.
November 14th 2012, 08:25 PM
hollywood

Re: set of subsequential limits

Since we don't know what $A\cap{(r_1,r_2)}$ looks like, it's hard to be more specific. But I think the construction works regardless of which choice we make.

- Hollywood