How to decompose a number sequence into convergent subsequences?

Question

Let ${a_n}$ be a real number sequence. Let $I$ be a subset of $mathbb N_+$, use the symbol ${a_n}_I$ to denote the subsequence of ${a_n}$ indexed by $I$. Does there exist a countable family of subsets $I_1,I_2,cdots subseteq mathbb N_+$, that $bigcuplimits_{n=1}^infty I_n=mathbb N_+$, and for every $ineq j$, $I_icap I_j=emptyset$, and $forall iin mathbb N_+$, ${a_n}_{I_i}$ is a convergent sequence? We consider divergent to positive or negative infinity as a special type of convergence.
If in general it doesn't exist, which condition we need for ${a_n}$ for its existence?

Saaqib Mahmood · Answer

For a real sequence, we can have only the following three situations:
(1) the sequence is convergent,
(2) the sequence diverges to $pm infty$.
(3) the sequence is bounded but not convergent (i.e. oscillates between a set of values)
Case 1. If the sequence is convergent, then we can consider
$$
I_p colon=  { p, p^2, p^3, cdots }
$$
for all primes $p$.
Case 2. If the sequence diverges to $pm infty$, then we can assume that some monotonic subsequence diverges to $pm infty$, and in that case we can of consider the sets
$I_p$ of Case 1 for such a monotonically divergent subsequence.
Case 3. If our sequence is bounded but not convergent, then we can consider it to be a sequence in some closed and bounded interval $[a, b]$ on the real line for some real numbers $a$ abd $b$ such that $a < b$. Since our sequence is in the compact metric space $[a, b]$, therefore this sequence necessarily has a convergent subsequence. Now we apply Case 1 to this convergent subsequence.
Thus what you're asking for is always possible.
Hope this helps.

How to decompose a number sequence into convergent subsequences?

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