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?
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 $\displaystyle \mathbb{R}$, is there a sequence $\displaystyle (x_n)_{n=1}^\infty$ such that for every $\displaystyle a\in{A}$, there is a subsequence $\displaystyle (x_{n_k})_{k=1}^\infty$ that converges to a.
Let S be the set of all intervals of the form $\displaystyle (r_1,r_2)$ where $\displaystyle r_1$ and $\displaystyle r_2$ are rational. This is a countable set. If you choose $\displaystyle x_n$ to be a point from $\displaystyle A\cap{(r_1,r_2)}$ 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 $\displaystyle (r_1,r_2)$ are intervals of rational numbers then if A is a singleton set of one irrational number your algorithm will fail. If you mean that $\displaystyle (r_1,r_2)$ 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.
Well that is certainly a clear discrimination.
You want the set $\displaystyle B$ to be countable, a sequence, the closure of which is $\displaystyle A$.
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 $\displaystyle {x_n}$ is a limit of some subsequence of $\displaystyle {x_n}$. 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 $\displaystyle {x_n}$ with range equal to this set. Since the rationals are dense in R, every point in [0,1] is a subsequential limit of $\displaystyle {x_n}$.
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 $\displaystyle A$ 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.
I meant to choose an element of A in $\displaystyle (r_1,r_2)$. 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 $\displaystyle r_1<a<r_2$, 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 $\displaystyle a\in{A}$. Choose any point $\displaystyle x_{n_1}<{a}$ (the same construction works for $\displaystyle x_{n_1}>{a}$). There are infinitely many intervals $\displaystyle (r_1,r_2)$ with $\displaystyle x_{n_1}<r_1<r_2<a$, so choose one, and let $\displaystyle x_{n_2}$ 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 $\displaystyle b\not\in{A}$ that is the limit of a subsequence of my sequence.
- Hollywood