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

Question

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?

Answer 1

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, $\inf_n (\sup_{k\geq n}a_k+\sup_{k\geq n}b_k)=\inf_n\sup_{k\geq n}a_k+\inf_n\sup_{k\geq n}b_k=\limsup_n a_n+\limsup_n b_n$.