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Let $\{a_{n}\}$ be a sequence of real numbers, where $0<a_{n}<1$, such that $\lim_{n\to \infty} a_{n}=0$, (then every subsequence will converges to zero). Is there any way to find a subsequence of $a_{n}$ which is decreasing to 0?

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Yes: since the sequence converges to $0$, for every $\epsilon\gt 0$ there exists $N\in\mathbb{N}$ such that for all $n\geq N$, we have $0\lt a_n\lt\epsilon$.

So define the sequence recursively: take $a_1$. Then let $\epsilon = \frac{a_1}{2}$; we know there is an $n_1\gt 1$ such that $a_{n_1}\lt \frac{a_1}{2}$. Now take $\epsilon = \frac{a_{n_1}}{2}$; we know there exists $n_2\gt n_1$ such that $a_{n_2}\lt \frac{a_{n_1}}{2}$.

Lather, rinse, and repeat.

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According to your proof this result is true for any sequence $\{a_{n}\}\subset (0,\infty)$ which converges to 0, not just because $a_{n}\in(0,1)$, I'm right!? – Stanley Jun 11 '12 at 2:04
Tiny variable name fix (it is not letting me edit it because the edit is not long enough). Where you say "... we know there is an $n_2 > 1$ such ...", that supposed to be $n_1 > 1$ rather than $n_2 > 1.$ – Souparna P Jun 11 '12 at 2:27
@Stanley: Yes, you are right. – Arturo Magidin Jun 11 '12 at 7:02

Yes. Put $n_1 = 1$. Suppose $n_1<n_2 < \cdots n_k$ are chosen. Choose $n_{k+1}> n_k$ so that $a_{n_{k+1}} < \min\{a_{n_j}, 1\le j \le k\}$. Such a sequence will fulfill your specification.

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Can be simplified slightly, since $\min \{a_{n_j}, 1 \le j \le k\} = a_{n_k}$. – Robert Israel Jun 11 '12 at 1:59

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