# Proof verification for $\lim \inf x_n + \lim \sup y_n \le \lim \sup(x_n + y_n)$ for bounded $x_n, y_n$

Let $$\{x_n\}$$ and $$\{y_n\}$$ denote two bounded sequences. Prove that: $$\lim \inf x_n + \lim \sup y_n \le \lim \sup(x_n + y_n) \\$$

We know that both $$x_n$$ and $$y_n$$ are bounded hence is their sum: $$m \le x_n + y_n < M$$

Using that fact we may choose a subsequence in order to satisfy the following: $$\lim(x_{n_k} + y_{n_k}) = \lim\sup(x_n + y_n) \tag1$$

Since $$x_{n_k}$$ is bounded (as far as $$x_n$$ is) lets choose a convergent subsequence with indices $$n^\prime_k \ge n_k$$ such that: $$\exists \lim x_{n^\prime_k}$$

Now consider a sequence $$y_{n^\prime_k}$$ (note the index is $$n^\prime_k$$), since it is bounded we may choose a convergent subsequence from $$y_{n^\prime_k}$$ with indices $$n^{\prime\prime}_k \ge n^\prime_k$$ such that: $$\exists\lim y_{n^{\prime\prime}_k}$$

Since $$\{x_{n^{\prime\prime}_k}\}$$ is a subsequence of $$\{x_{n^\prime_k}\}$$ it is convergent to the same limit. Also $$\{y_{n^{\prime\prime}_k}\}$$ is a subsequence of $$\{y_{n^\prime_k}\}$$ and we've chosen $$\{y_{n^{\prime\prime}_k}\}$$ to be convergent. Based on that and on $$(1)$$ we may write: $$\lim(x_{n^{\prime\prime}_k} + y_{n^{\prime\prime}_k}) = \lim(x_{n_k} + y_{n_k}) = \lim\sup (x_n + y_n) \tag 2$$

By definition of limsup and liminf: $$\lim x_{n^{\prime\prime}_k} \ge \lim\inf x_n \\ \lim y_{n^{\prime\prime}_k} \le \lim\sup y_n$$

Or (multiply second inequality by $$-1$$): $$\lim x_{n^{\prime\prime}_k} \ge \lim\inf x_n \\ -\lim y_{n^{\prime\prime}_k} \ge -\lim\sup y_n$$

Subtract the inequalities: $$\lim x_{n^{\prime\prime}_k} + \lim y_{n^{\prime\prime}_k} \ge \lim\inf x_n + \lim\sup y_n \tag3$$

Limit of sum is just a sum of limits so: $$\lim(x_{n^{\prime\prime}_k} + y_{n^{\prime\prime}_k}) =\lim x_{n^{\prime\prime}_k} + \lim y_{n^{\prime\prime}_k}$$

So now using $$(2)$$ and $$(3)$$ we conclude that: $$\lim \sup(x_n + y_n) \ge \lim\inf x_n + \lim\sup y_n$$

Is this argument enough to consider the proof complete?

• You did not arrive at Equation 3 correctly. Subtracting inequalities is not possible to get that because the direction of inequality for one of them changes. So you just get the sum of the two inequalities before you multiplied the second by -1. Dec 14, 2018 at 16:12

However, there is a quicker way to answer the problem. By definition of infimum, we have that $$\displaystyle\inf_{k\geq n} x_k \leq x_j$$ for all $$j \geq n$$. Then we can add $$y_j$$ to both sides to obtain $$\displaystyle\inf_{k\geq n} x_k + y_j \leq x_j + y_j.$$
Taking the supremum of both sides leads us to $$\displaystyle\inf_{k\geq n} x_k + \sup_{j \geq n} y_j \leq \sup_{j \geq n} (x_j + y_j).$$ Lastly, taking the limit as $$n \to \infty$$ yields the result.
Notice that, in the solution, we never used the fact that the sequences are bounded. Thus, this result is actually stronger and can be proved for any real sequences (provided that none of the sums become $$\infty - \infty$$).
• The boundedness assumption is somewhat necessary. We might get $-\infty + \infty$ on the LHS. So long as both sides are well-defined, the theorem can be generalized. Dec 14, 2018 at 16:34