# Prove that $\underline{\lim} \{a_n\}+ \underline{\lim} \{b_n\} \leq \underline{\lim} (a_n + b_n)$

In my real analysis class, my instructor proved that $\underline{\lim} \{a_n\}+ \underline{\lim} \{b_n\} \leq \underline{\lim} (a_n + b_n)$. (Note that $\underline{\lim}$ is the limit inferior of a sequence.) but a lot of details were left out and I want to make sure that I have a correct proof and that my logic is correct.

Note that $\underline{lim} \{a_n\} = lim_{n \rightarrow \infty} \inf\{a_k|k \geq n\}$ and in my class we define $t_n =\inf\{a_k|k \geq n\}$.

My proof:

Suppose that $\underline{\lim} \{a_n\} = a$ and $\underline{\lim} \{b_n\} = b$.

First consider $\underline{\lim} \{a_n\} = a$. This means that $\displaystyle \lim_{n \rightarrow \infty} t_n = a$. So for all $\epsilon > 0$ there exists $N_1 \in \mathbb{N}$ such that when $n \geq N_1$ we have that $| t_n - a | < \frac{ \epsilon } { 2}$. It follows that $a - \frac{\epsilon}{2} < t_n < a_n$.

By a similar argument, there exists $N_2$ such that for all $n \geq N_2$, we have $b - \frac{\epsilon}{2} < b_n$.

So if we choose $N = \max \{N_1, N_2 \}$, for all $n \geq N$ we have

$a - \frac{\epsilon}{2} < a_n$ and

$b - \frac{\epsilon}{2} < b_n$.

Combining these inequalities gives

$a_n + b_n > a+b-\epsilon$

Then since $a+b - \epsilon$ is a lower bound for the set $\{ a_n + b_n | n \geq N \}$, we must have

$\inf\{a_n + b_n | n \geq N \} \geq a+b- \epsilon$ and

$\displaystyle \lim_{N \rightarrow \infty} \inf\{a_n + b_n | n \geq N \} \geq a+b- \epsilon$ (because the ineq. was true for all $n \geq N$).

Then since $\epsilon$ was arbitrary,

$\underline{ lim} (a_n + b_n) \geq a+b$

• When using "N" to refer to the natural numbers, consider typesetting it as "\Bbb{N}", $\Bbb N$, or "\mathbb{N}", $\mathbb{N}$, so that we can distinguish it from $N$ :) Thanks! – RGS Jan 21 '17 at 16:48

It looks correct, assuming both inferior limits are finite and not $-\infty$ (otherwise there's nothing to prove).
Here's another approach. 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 $$t_n+u_n\le v_n$$