# Prob. 5, Chap. 3 in Baby Rudin: $\lim\sup_{n\to\infty}\left(a_n+b_n\right)\leq\lim\sup_{n\to\infty}a_n+\lim\sup_{n\to\infty}b_n$.

Here's Prob. 5, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

For any two real sequences $$\left\{ a_n \right\}$$, $$\left\{ b_n \right\}$$, prove that $$\lim\sup_{n\to\infty}\left( a_n + b_n \right) \leq \lim\sup_{n\to\infty} a_n + \lim\sup_{n\to\infty} b_n,$$ provided the sum on the right is not of the form $$\infty-\infty$$.

My effort:

Let us put $$c_n \colon= a_n + b_n \tag{Definition A}$$ for all $$n \in \mathbb{N}$$, and let's put \begin{align} a^* & \colon= \lim\sup_{n\to\infty} a_n, \\ b^* & \colon= \lim\sup_{n\to\infty} b_n, \\ c^* & \colon= \lim\sup_{n\to\infty} c_n. \end{align} \tag{Definitions B}

We need to show that $$c^* \leq a^* + b^*. \tag{0}$$

So let's suppose that $$c^* \not\leq a^* + b^*.$$ Then we have $$c^* > a^* + b^*,$$ and so $$c^*- b^* > a^*,$$ and let's take a real number $$x$$ such that $$c^*- b^* > x > a^*. \tag{0}$$

Then as $$x > a^*$$, so by Theorem 3.17 (b) in Baby Rudin, we can find a natural number $$N_1$$ such that $$x \geq a_n \tag{1}$$ for all $$n > N_1$$.

Now from (0) above, as $$c^*-x > b^*,$$ so we can find a real number $$y$$ such that $$c^*-x > y > b^*. \tag{2}$$

Then as $$y > b^*$$, so again by Theorem 3.17 (b) in Baby Rudin, we can find a natural number $$N_2$$ such that $$y \geq b_n \tag{3}$$ for all $$n>N_2$$.

Now from (2) above we can conclude that $$c^* > x+y, \tag{4}$$ and hence from (1) and (3) we also have $$x+y \geq a_n + b_n = c_n,$$ for all $$n > \max \left\{\ N_1, N_2 \ \right\}$$, which in turn implies that $$x+y \geq \lim\sup_{n \to \infty} c_n,$$ that is [Please Refer to (Definition A) and (Definitions B) above.], $$x+y \geq c^*,$$ which, in view of (4) above, gives rise to a contradiction to our choice of $$c^*$$ as the limit superior of the sequence $$\left\{c_n\right\}$$.

Is this proof correct? If not, then where is it deficient?

An easier solution to this is to directly use the definition of $\limsup$.
Note, $\limsup_n(a_n+b_n)=\inf_n\sup_{k\geq n}(a_k+b_k)\le \inf_n (\sup_{k\geq n}a_k+\sup_{k\geq n}b_k)$
Now let $x_n=\sup_{k\geq n}a_k$ and $y_n=\sup_{k\geq n}b_k$. Then note that both $x_n$ and $y_n$ are decreasing, so $(x_n+y_n)$ is also decreasing. Hence $\inf_n (x_n+y_n)=\lim_n (x_n+y_n)=\lim_n x_n+\lim_n y_n=\inf_n x_n+\inf _n y_n$.
Therefore, $\inf_n (\sup_{k\geq n}a_k+\sup_{k\geq n}b_k)=\inf_n\sup_{k\geq n}a_k+\inf_n\sup_{k\geq n}b_k=\limsup_n a_n+\limsup_n b_n$.