Let $\left(a_{n}\right)_{n=1}^{\infty},\left(b_{n}\right)_{n=1}^{\infty}$ be bounded sequences.
Prove that $\underset{n\rightarrow\infty}{\lim\inf}\left(a_{n}\right)+\underset{n\rightarrow\infty}{\lim\inf}\left(b_{n}\right)\leq\underset{n\rightarrow\infty}{\lim\inf}\left(a_{n}+b_{n}\right)$.
I know there are diffrent ways to prove it but I was asked to prove it using the proposition:
For the given sequences there exist increasing sequence of index in $\mathbb{N}\quad\left(n_{k}\right)_{k=1}^{\infty} $ s.t. $
\left(a_{n_{k}}\right)_{k=1}^{\infty},\left(b_{n_{k}}\right)_{k=1}^{\infty}$ converge.
Thoughts if both $\left(a_{n_{k}}\right)_{k=1}^{\infty},\left(b_{n_{k}}\right)_{k=1}^{\infty}$ converge, I can choose them as the infimum sequences and then it easy, but how can I choose the index sequence that way?
I found this answer
Set $t_n=\inf\{a_k\mid k\ge n\}$, $u_n=\inf\{b_k\mid k\ge n\}$ and $v_n=\inf\{a_k+b_k\mid k\ge n\}$.
If $k\ge n$, then $t_n\le a_k$ and $u_n\le b_k$, so $t_n+u_n\le a_k+b_k$. Since $k$ is arbitrary, we have
$$
t_n+u_n\le v_n
$$
My problem is that $t_n,u_n$ are not necessarily part of $a_n,b_n$.
Best Answer
Let $$c=\liminf (a_n+b_n),\quad a=\liminf a_n,\quad b=\liminf b_n$$ There exists a subsequence $a_{n_k}+b_{n_k}$ convergent to $c.$ Since $a_{n_k}$ is bounded, we can choose a subsequence $a_{n_{k_l}},$ which is convergent, say to $\tilde{a}.$ Hence $\tilde{a}\ge a.$ Next as $b_{n_{k_l}}$ is bounded, we can choose a subsequence $b_{n_{k_{l_m}}},$ which is convergent, say to $\tilde{b}.$ Then $\tilde{b}\ge b.$ Moreover $a_{n_{k_{l_m}}}\to \tilde{a}. $ Therefore $$c=\lim_{m\to\infty}\left [a_{n_{k_{l_m}}}+b_{n_{k_{l_m}}}\right ]=\tilde{a}+\tilde{b}\ge a+b$$