The question I am working on is:

Let $(a_n)$ be a fixed (and unspecified) strictly increasing sequence of real numbers. Find (with proof) a sequence $(b_n)$ whose set of cluster points is precisely $(a_n)$.

However, I don't see how this could be true. For example, let $a_n := \sum_{i = 0}^n 10^{-i}$. Then $(a_n)$ is strictly increasing, and it's limit is $\frac{10}{9} = L$. Suppose $(b_n)$ were such a sequence where it's cluster points were exactly the points in $(a_n)$.

Then because $a_n \to L$, for any $\varepsilon > 0$, we have some $N$ so that for all $n \geq N$, $d(a_n, L) < \varepsilon / 2$, and $d(b_i, a_n) < \varepsilon / 2$ for some $b_i$ (because $a_n$ is a cluster point of $(b_n)$), so $d(b_i, L) \leq d(b_i, a_n) + d(a_n, L) < \epsilon$. Picking such a $b_i$ for each $a_n$ gives a subsequence of $(b_n)$, $(b_{n_i})$, so that $b_{n_i} \to L$, so then $L$ is a limit point of $(b_n)$. Then $L$ must be a point in $a_i$. But there is no $n$ so that $\sum_{i = 0}^n 10^{-i} = \frac{10}{9}$.

So $(a_n)$ is a strictly increasing sequence of real numbers for which there can be no sequence whose cluster points are exactly $(a_n)$.

  • $\begingroup$ HINT: What got you in trouble was having the sequence $(a_n)$ converge in the first place! $\endgroup$ – Ted Shifrin Oct 9 '19 at 21:25
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    $\begingroup$ @TedShifrin But the question doesn’t say the sequence can’t converge? $\endgroup$ – Reed Oei Oct 9 '19 at 21:26
  • $\begingroup$ You're right. $(a_n)$ should "converge" to $+\infty$. $\endgroup$ – amsmath Oct 9 '19 at 21:27
  • $\begingroup$ You argued that if $a_n\to L$, then $L$ will have to be a cluster point of $(b_n)$. $\endgroup$ – Ted Shifrin Oct 9 '19 at 21:28
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    $\begingroup$ Your counterexample looks valid to me. There must be a missing hypothesis. $\endgroup$ – Bungo Oct 9 '19 at 21:43

You just need to design a sequence $(b_n)$ that gets arbitrarily close to every point in $(a_n)$. You have total freedom to choose the $b_n$'s. For example, the sequence $(b_n)$ could start out within $1$ of $a_1$; its next two terms could be within $\frac12$ of $a_1$ and $a_2$, the next three terms could be within $\frac13$ of $a_1$, $a_2$, $a_3$, etc. Your job is to show that such a $(b_n)$ satisfies the requirements of your problem.

EDIT: As discussed in the comments, any construction of $b_n$ will also pick up $\lim a_n$ as a cluster point of $(b_n)$, unless the sequence $(a_n)$ is unbounded. When $(a_n)$ is bounded, the OP's argument shows how $\lim a_n$ will be a cluster point of $(b_n)$.

  • $\begingroup$ But wouldn’t this have the same problem I mentioned above, where such a sequence may have additional cluster points that are not elements of the sequence $(a_n)$? $\endgroup$ – Reed Oei Oct 9 '19 at 21:43
  • $\begingroup$ This construction will ensure that the set of cluster points of $(b_n)$ includes every $a_n$, but this set may be a strict superset of $(a_n)$, e.g. if $(a_n)$ has cluster points not contained in $(a_n)$, as in the OP's example. $\endgroup$ – Bungo Oct 9 '19 at 21:44
  • $\begingroup$ @Bungo Quite right. It is hard to see how to avoid $L$ also being a limit point of any sequence $b_n$, unless the sequence $a_n$ is unbounded. $\endgroup$ – grand_chat Oct 9 '19 at 21:46
  • $\begingroup$ Even unboundedness wouldn't suffice in general. For example, $a_n$ could be something like $1, -1, 1/2, 2, -2, 1/3, 3, -3, 1/4, 4, -4, \ldots$ which is unbounded above and below but has zero as a cluster point that is not contained in the sequence, so $b_n$ will also cluster at zero. There's clearly a missing hypothesis, but I'm not sure what it should have been. $\endgroup$ – Bungo Oct 9 '19 at 21:48
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    $\begingroup$ @Bungo We can still assume $(a_n)$ is strictly increasing. $\endgroup$ – grand_chat Oct 9 '19 at 21:48

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