Proving that $\mathcal{L}(x_n+y_n) \subset {\cal L}(x_n)+{\cal L}(y_n)$.

Question

Let $(x_n)$ and $(y_n)$ be two bounded real sequences. Let: $${\cal L}(x_n) =\{ L \in [-\infty,+\infty] \mid L \text{ is the limit of some subsequence }(x_{n_k}) \},$$ and the same for ${\cal L}(y_n)$ and $\mathcal{L}(x_n+y_n)$. I want to prove that $$\mathcal{L}(x_n+y_n) \subset {\cal L}(x_n)+{\cal L}(y_n).$$

If I take $L \in \mathcal{L}(x_n+y_n)$, then I have $\lim_k (x_{n_k}+y_{n_k}) = L$ for some convenient subsequence. I wanted to write $L = \lim_k x_{n_k}+ \lim_k y_{n_k}$ to conclude that $L \in {\cal L}(x_n)+{\cal L}(y_n)$, however I do not know how to guarantee (or even if it is really true) that these last two limits do exist. Can someone give me a hand, please? Thanks.

Answer 1

Use the Bolzano-Weierstrass theorem, take a converging sub-sequence of $x_{n_k}$, say $x_{n_{k_j}}\to X$. Then it must be the case that $y_{n_{k_j}}\to L-X$.
So $L=X+(L-X)\in {\cal L}(x_n)+{\cal L}(y_n)$.

Answer 2

If by 'limited' you mean bounded then from your original sequence $x_n$ s.t. $x_n + y_n \rightarrow L$ you can extract a convergent subsequence $x_{n_k}$. take another sub-subsequence from $y_{n_k}$ and you're done. Is this what you meant?