I know that for a sequence $\{x_n\}$, if $\{x_{2n}\}$ and $\{x_{2n+1}\}$ converge to the same limit, then $\{x_n\}$ converges and to the same limit.

My question is if I know that from any $x_i$ I can construct a subsequence of $\{x_n\}$, and these all converge to $X$, then does this imply that $\{x_n\}\to X$?

Some additional pieces: $x_n \in [0, \alpha]$, and I show the $x_i$ subsequences converge by showing that $\exists m > i : x_m < x_i,\ \forall i$. In words: that going further along the sequence, you'd eventually find a $x_m$ that is definitely smaller.


A fairly general condition is that if you have finitely many subsequences converging to the same limit $x$ such that there is some $N$ such that all $n > N$ is one of the indexes appearing in one of your subsequences.

To see that this works: let $n_i^j$ be the index in $\{x_n\}$ of the $i$-th term of the $j$-th subsequence.

For any $\varepsilon > 0$, for each $j$-th subsequence, there is a $N_j \in \mathbb{N}$ such that for all $n_i^j > N_j$, we have $|x_{n_i^j} - x| < \varepsilon$. Thus, for all $n > \max\limits_j\{N,N_j\}$ (note that that maximum exists and is finite since there are finitely many $N_j$), $n = n_i^j$ for some $i,j$, and $n_ > N_j$ so $|x_n - x| < \varepsilon$.


It is not true if the partition of $\mathbb{N}$ into infinite sets has infinitely many elements.

Let $\{A_{n}\mid n\in\mathbb{N}\}$ be a partition of $\mathbb{N}$ into infinitely many countable sets. This is possible as shown in

Partitioning an infinite set

For each $n$ enumerate $A_{n}$, say $A_{n}=\{x_{n,k}\}$. Without loss of generality we will assume that this enumeration is such that $x_{n,k}<x_{n,k+1}$.

For each $n$ we will construct a sequence $(y_{x_{n,k}})$ in $[0,1]$ that converges to $1$, but that the sequence $(N_{n})$ in $\mathbb{N}$ is strictly increasing where $N_{n}$ is the natural number such that, for a given $\epsilon>0$, for $M\geq N_{n}$ we have $|y_{x_{n,M}}-1|<\epsilon$.

Defining these sequences is easy. Let $n\in\mathbb{N}$ be given and define

$$y_{x_{n,k}}=\left\{\begin{array}{ll} 0 & k<n\\ 1 & k\geq n \end{array} \right.$$

Then each sequence $(y_{x_{n,k}})$ is well define and certainly converges to $1$. Moreover, for any $\epsilon>0$ such that $\epsilon<1$ we have that $N_{n}=n$.

Now define the sequence $(z_{m})$ in $[0,1]$ by defining $z_{m}=(y_{x_{n,k}})$ if $m=x_{n,k}$. This defines $(z_{m})$ entirely. (Hopefully) it is clear that $(z_{m})$ does not converge to $1$, although it of course has several subsequences that converge to $1$.


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