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.
Best Answer
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 $|x_n - x| < \varepsilon$.