Limit inferior of bounded sequence

Question

I found an interesting problem that I can't tackle as I am studying real analysis on my own. Let there be bounded sequences $(a_n)$ and $(b_n)$. Proove that $\varliminf_{n\to\infty} (\min \{ a_n , b_n \} ) = \min \{ \varliminf_{n\to\infty} a_n , \varliminf_{n\to\infty} b_n \}$. So far I have figured out that I should use lower limit monotonicity and then use definition of subsequential sequence. Am I on the right track? I have been stuck on this problem for quite some time.

Answer 1

• Since $$\min\{a_n,b_n\}\leq a_n\quad \text{and}\quad \min\{a_n,b_n\}\leq b_n,$$ for all $n$, the inequality $$\liminf_{n\to \infty }\min\{a_n,b_n\}\leq \min\{\liminf_{n\to \infty }a_n,\liminf_{n\to \infty }b_n\},$$ is straightforward.

• For the converse inequality, notice first that sequences $\left(\inf_{k\geq n}a_k\right)_{n\geq 1}$ and $\left(\inf_{k\geq n}b_k\right)_{n\geq 1}$ are increasing. Let $m\in\mathbb N$ and $N_m\in \mathbb N$ s.t. $$\liminf_{n\to \infty }a_n\leq \inf_{k\geq n}a_k+\frac{1}{m}\quad \text{and}\quad \liminf_{n\to \infty }b_n\leq \inf_{k\geq n} b_k+\frac{1}{m},$$ for all $n\geq N_m$. Since $$\inf_{k\geq n}a_k\leq a_n\quad \text{and}\inf_{k\geq n}b_k\leq b_n,$$ for all $n$, we get $$\liminf_{n\to \infty }a_n\leq a_n+\frac{1}{m}\quad \text{and}\quad \liminf_{n\to \infty }b_n\leq b_n+\frac{1}{m},$$ for all $n\geq N_m.$ Therefore, $$\min\{\liminf_{n\to \infty }a_n,\liminf_{n\to \infty }b_n\}\leq \min\{a_n,b_n\}+\frac{1}{m},$$ for all $n\geq N_m$, and thus $$\min\{\liminf_{n\to \infty }a_n,\liminf_{n\to \infty }b_n\}\leq \liminf_{n\to \infty }\min\{a_n,b_n\}+\frac{1}{m}.$$ Taking $m\to \infty $ gives the wished result.