# Real Analysis: Sequence whose set of cluster points is another (strictly increasing) sequence

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 $$(a_n)$$.

• HINT: What got you in trouble was having the sequence $(a_n)$ converge in the first place! – Ted Shifrin Oct 9 '19 at 21:25
• @TedShifrin But the question doesn’t say the sequence can’t converge? – Reed Oei Oct 9 '19 at 21:26
• You're right. $(a_n)$ should "converge" to $+\infty$. – amsmath Oct 9 '19 at 21:27
• You argued that if $a_n\to L$, then $L$ will have to be a cluster point of $(b_n)$. – Ted Shifrin Oct 9 '19 at 21:28
• Your counterexample looks valid to me. There must be a missing hypothesis. – 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 $$(b_n)$$.
• 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 $(a_n)$? – Reed Oei Oct 9 '19 at 21:43
• 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. – Bungo Oct 9 '19 at 21:44
• @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. – grand_chat Oct 9 '19 at 21:46
• 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. – Bungo Oct 9 '19 at 21:48
• @Bungo We can still assume $(a_n)$ is strictly increasing. – grand_chat Oct 9 '19 at 21:48