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?
A subsequential limit is the limit of a subsequence. So he wants to know: if A is a closed subset of , is there a sequence such that for every , there is a subsequence that converges to a.
Let S be the set of all intervals of the form where and are rational. This is a countable set. If you choose to be a point from 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 are intervals of rational numbers then if A is a singleton set of one irrational number your algorithm will fail. If you mean that 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.
A subsequential limit of a sequence is a limit of some subsequence of . 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 with range equal to this set. Since the rationals are dense in R, every point in [0,1] is a subsequential limit of .
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 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.
I meant to choose an element of A in . 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 , 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 . Choose any point (the same construction works for ). There are infinitely many intervals with , so choose one, and let 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 that is the limit of a subsequence of my sequence.
- Hollywood