Choosing proper sequence given a limit

Question

Suppose $X$ is a Banach space and $X_j$ are subsets with $X_1 \subseteq X_2 \subseteq X_3 \subseteq \ldots$ so that $X= \overline{\bigcup _{j=1}^{\infty} X_j }$. If $u \in X$ then $u= \lim_{n \rightarrow \infty} y_n$ with $y_n \in \bigcup _{j=1}^{\infty} X_j$. I want to replace the sequence $y_n$ with a sequence $u_n$ such that $u_n \in X_n$. I'm pretty sure this is possible, but not sure to what would be the most elegant way of doing this. Any help would be appreciated.

Answer 1

I assume you want that for any $u_{n}$ there exists a $y_{m}$ such that $u_{n}=y_{m}$. In that case this will be not be possible for a finite number of $u_{n}$, as for a general sequence $(y_{n})$ there can exists an $N\in\mathbb{N}$ such that $y_{n}\in X\setminus X_{N}$ for all $n$.

We can make a sequence which is close to your requirements. For $n\leq N$ we define $u_{n}=y_{n}$. For $n>N$ we define $Y_{n}=\{i\geq n:y_{i}\in X_{n}\}$, note that $Y_{n}$ has infinite cardinality. We can define $$u_{n}=y_{\min Y_{N}}.$$

Answer 2

Let $k(m) = \min \{ k \in \mathbb N \mid y_m \in X_k \}$ and $l(n) = \sum_{m=1}^n k(m)$. The sequence of integers $l(n)$ is strictly increasing such that $y_n \in X_{k(n)} \subset X_{l(n)}$.

Pick any $\xi \in X_1$. Then take $x_r = \xi$ for $r = 1,\dots,l(2)-1$ (which implies $x_r \in X_1 \subset X_r$) and for each $n \ge 2$ take $x_r = y_n$ for $r = l(n),\dots ,l(n+1)-1$ (which implies $x_r \in X_{l(n)} \subset X_r$).

Obviuously $x_n \to u$.