Given a bounded sequence $\{x_n\}$ prove that it may be split into countably many sequences having the same limit.

The question is stated as above, without any other constraints on the sequence. I'm not sure how to follow with the proof, but here are some thoughts of mine.

First of all $\{x_n\}$ is bounded meaning: $$ \exists M\in\Bbb R: |x_n| \le M, \forall n\in\Bbb N $$

Recall that a bounded sequence does have subsequential limits. Let's denote the set of subsequential limits as $\Bbb L$: $$ \Bbb L = \{L_1, L_2, L_3, \dots \} $$

Now from each of the subsequences, we may go one level down and choose subsequences such that their limit is equal to $L_k$, namely: $$ \lim_{p\to\infty}(x_n)_p^{(k)} = L_k,\ \forall k \in\Bbb N $$

Now I was thinking about using an approach similar to proving the countability of rational numbers.

$$ \begin{array}{|c|c|} \hline L_1 & L_2 & \cdots & L_k \\\hline (x_n)_1^{(1)}& (x_n)_1^{(2)} &\cdots &(x_n)_1^{(k)} \\\hline (x_n)_2^{(1)} & (x_n)_2^{(2)} &\cdots &(x_n)_2^{(k)} \\\hline (x_n)_3^{(1)} & (x_n)_3^{(2)} &\cdots &(x_n)_3^{(k)} \\\hline \cdots & \cdots &\cdots &\cdots \\\hline (x_n)_m^{(1)} & (x_n)_m^{(2)} &\cdots &(x_n)_m^{(k)} \\\hline \end{array} $$

So the first column contains all possible subsequences converging to $L_1$, the second all possible subsequences converging to $L_2$ and etc. We can now assign an index to each element in the table, in the same manner, as it is done while proving countability of rationals.

The problem is I'm not sure whether this approach even works. Especially in the case when $\{x_n\}$ is a sequence such that the set of its limit point forms a dense interval.

In case all of the above is a non-sense could someone please suggest a proof of what's in problem statement? Thank you!

  • $\begingroup$ Well the number of such parts has to be countable as the number of terms in whole sequence is countable. $\endgroup$ – Paramanand Singh Mar 20 '19 at 13:34

A substantial hint, but not an answer:

You're sort of on the right track. First, pick a subsequence (any one!) that you like, and which converges to some limit $L$. Call the items in this subsequence "good" and the remaining items "bad".

Looking at the good subsequence, can you think of a convergent sub-sequence? Can you think of two disjoint convergent subsequences? If you could, let's call them $a_n$ and $b_n$ (I'm skipping the sequence notation here) and suppose that $P$ and $Q$ are two "bad" items. Then the sequences $$ P, a_1, a_2, \ldots, \\ Q, b_1, b_2, \ldots,$$

both converge. In fact, if you put "P" in the right place among the $a$s (i.e., in the place it occurs within the $x_n$ sequence), that'd still be find, and the same for $Q$.

Of course, this only deals with two of the "bad" elements, but you have lots of them...but perhaps you can generalize from here.


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