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I am asked to prove that there is no sequence such that it contains subsequences converging to every number in $(0,1)$ and no other number. A previous question in the book told me that such a sequence is possible if the interval is closed, i.e, its $[0,1]$ instead of $(0,1)$. The way I reason it out is:

Suppose we have a sequence $x_n$, such that it contains subsequences converging to every number in $(0,1)$ and no other number. Then, for any $a_1 \in (0,1)$, there is a subsequence converging to it, but there is also a subsequence converging to some $a_2$ such that $1>a_2>a_1$. Existence of such an $a_2$ is guaranteed by density of reals, and this will hold for every number in $(0,1)$. Now it seems clear that there has to be at least one subsequence, maybe the main sequence $x_n$ itself, that converges to or is bounded by $1$ proper. Similar reasoning seems to hold for lower bound of $0$.

I want to make this reasoning rigorous, if it right in the first place. The main idea is that no number strictly less than $1$ can be largest possible limit of the subsequences since there will always be some larger. Thus $1$ has to be the largest limit and thus the interval needs to be closed. This seems like "limit of limits" situation, although I don't know if that's technically even a thing, but I hope I have conveyed my general reasoning.

Aniruddh
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  • The set of all subsequential limits is a closed set and $(0, 1) $ is not closed. Hence $(0, 1) $ can't be a set of subsequential limit of any sequence. – Sourav Ghosh Jul 05 '22 at 05:46

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Take a number in the sequence pretty close to, say, $0.9$. Then take a later number in the sequence twice as close to, say, $0.95$. Then the next twice as close as that to $ 0.975$. Etc. A little triangle inequality magic shows that this subsequence converges to $1$.

More generally, you can show that the set of subsequential limits is closed.

Sourav Ghosh
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Ethan Dlugie
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    Does that argument need countable choice? – J.G. Jul 05 '22 at 05:58
  • Perhaps, I'm not an expert on choice axioms. I suppose that when we want to "pick an element close to $x$", there are countably infinitely many such elements. So maybe yes it does require this. But I think this kind of thing is a very standard argument in analysis. – Ethan Dlugie Jul 05 '22 at 06:34