# Proof of $\varliminf a_n + \varlimsup b_n \leq \varlimsup (a_n + b_n)$ using the subsequential definition

$$a_n$$ and $$b_n$$ are assumed to bounded sequences. The definition I'm strictly adhering to is

$$\varliminf a_n = \inf\{\text{subsequential limits of } a_ n \}$$

$$\varlimsup a_n = \sup\{\text{subsequential limits of } a_ n \}$$

Below is an attempt I made; would highly appreciate any feedback on it. I have both seen and done other proofs of this statement, using alternative definitions of limit superior/inferior, so I wanted to write up a new one on my own.

Let $$L_2$$ be any subsequential limit of the sequence $$b_n$$. Then some $$b_{n_k} \to L_2$$.

Study the sequence $$a_{n_k}$$; this is bounded and thus has a convergent subsequence $$a_{l_k}$$. Denote the limit $$L_1$$. We know $$\varliminf a_n \leq L_1$$.

Looking at exactly the $$l_k$$-indices of $$b_{n_k}$$, we know $$b_{l_k}$$ is a subsequence of $$b_{n_k}$$ and hence also converges to $$L_2$$.

But then $$a_{l_k} + b_{l_k}$$ must converge to $$L_1 + L_2$$. But then of course $$L_1 + L_2 \leq \varlimsup (a_n + b_n)$$.

The statement we have is then: $$\varliminf a_n + L_2 \leq L_1 + L_2 \leq \varlimsup (a_n +b_n)$$, that is

$$L_2 \leq \varlimsup (a_n +b_n) - \varliminf a_n$$

So we have found an upper bound to the set of subsequential limits of $$b_n$$, which finally leaves us with:

$$\varlimsup b_n \leq \varlimsup (a_n + b_n) - \varliminf a_n$$. We are done $$\blacksquare$$.

Your proof is fine. I only would use a different notation for the “iterated subsequence” (that is what caused my initial confusion): If $$(b_{n_k})$$ is a convergent subsequence of $$(b_n)$$ with limit $$L_1$$ then we can choose a convergent subsequence $$(a_{n_{k_j}})$$ of $$(a_{n_k})$$. Then $$L_2 = \lim_{j \to \infty} a_{n_{k_j}} \ge \varliminf a_n$$ and $$L_1 + L_2 = \lim_{j \to \infty} (a_{n_{k_j}} + b_{n_{k_j}}) \le \varlimsup (a_n + b_n)$$ so that $$L_1 = (L_1 + L_2) - L_2 \le \varlimsup (a_n + b_n) - \varliminf a_n$$ This holds for all convergent subsequences of $$(b_n)$$, which implies $$\varlimsup b_n \le \varlimsup (a_n + b_n) - \varliminf a_n \, .$$