I want to build a convergent subsequence from two subsequences. Let $% \{n_{1_{l}}\}_{l=1}^{\infty }$ and $\{n_{2_{l}}\}_{l=1}^{\infty }$\ be increasing sequences of integers, not necessarily disjoint. Let $\{x_{n}\}$ have subsequences $\{x_{n_{1_{l}}}\}_{l=1}^{\infty }$ and $% \{x_{n_{2_{l}}}\}_{l=1}^{\infty }$ with $\lim_{l\rightarrow \infty }x_{n_{1_{l}}}$ $=$ $a$ and $\lim_{l\rightarrow \infty }x_{n_{2_{l}}}$ $=$ $a $. Now build a new index set $\tilde{n}_{l}$ by intertwining unique values from $\{n_{1_{l}},n_{2_{l}}\}$. For example, $\{n_{1_{l}}\}$ $=$ $% \{1,5,6,100,...\}$ and $\{n_{2_{l}}\}$ $=$ $\{1,5,56,90,...\}$ hence $\{% \tilde{n}_{l}\}$ $=$ $\{n_{1_{l}}\}\cup \{n_{2_{l}}\}$ $=$ $% \{1,5,6,56,90,100,...\}$. Then is $\lim_{l\rightarrow \infty }x_{\tilde{n}_{l}}$ $=$ $a$?
I cannot locate a matching thread here, and my apologies if other threads implicitly cover this. My argument is as follows. By convergence of the subsequences, $\forall \varepsilon $ $>$ $0$ $\exists (\mathcal{L}% _{1,\varepsilon },\mathcal{L}_{2,\varepsilon })$ $\in $ $\mathbb{N}$ such that \begin{equation*} \left\vert x_{n_{1_{l}}}-a\right\vert \leq \varepsilon \text{ }\forall l\geq \mathcal{L}_{1,\varepsilon }\text{ and }\left\vert x_{n_{2_{l}}}-a\right\vert \leq \varepsilon \text{ }\forall l\geq \mathcal{L}_{2,\varepsilon }. \end{equation*}
Now pick $\mathcal{L}_{\varepsilon }$ $=$ $\max \{\mathcal{L}_{1,\varepsilon },\mathcal{L}_{2,\varepsilon }\}$ to yield for the intertwined sequence $|x_{% \tilde{n}_{l}}$ $-$ $a|$ $\leq $ $\varepsilon $ $\forall l$ $\geq $ $% \mathcal{L}_{\varepsilon }$. Therefore $x_{\tilde{n}_{l}}$ $\rightarrow $ $a$.
Comments are appreciated.
