# When does subsequence convergence imply convergence?

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.

## 2 Answers

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}.

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

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 $$1$$.