Can anyone give me a hint on how to prove that subadditivty of the limit supremum of two sequences? Everything I can find says "it's trivial." I'm working through some analysis textbooks on my own, so I don't want the full answer. I'm only looking for a hint on this problem.
I'm trying to prove the subadditivity of the limit supremum, i.e.
$$
\limsup_{n \to \infty} (x_n + y_n) \le \limsup_{n \to \infty} (x_n) + \limsup_{n \to \infty} (y_n)
$$
My book and the notes that accompany it say I should be able to do this without resorting to properties that relate the infimum to the supremum, so I'm assuming my approach shouldn't follow this question (which doesn't actually address this particular statement directly).
Here's my attempt at a proof that uses contradiction:


*

*Assume that $\limsup_{n \to \infty} (x_n + y_n) > \limsup_{n \to \infty} (x_n) + \limsup_{n \to \infty} (y_n)$

*Consider the sequences $(x_n) = (-1)^n$ and $(y_n) = (-1)^{n+1}$. I know that $\forall n, \sup(x_n) = \sup(y_n) = 1$, but $\sup(x_n + y_n) = 0$.

*This is a contradiction because I can't have $0 > 1$, so the proof is complete. 
Is this a valid proof for this? 
If this isn't a valid proof, is there a simple hint that may put me on the right track? I'm a bit stuck on this one. 

EDIT:
Here's what I can think of so far, but I'm really stuck on this. 


*

*I know that the supremums of subsequences form a monotonically decreasing sequence, and because that sequence converges, it's bounded. 

*I know that $\limsup_{n \to \infty} (x_n)$ and $\limsup_{n \to \infty} (y_n)$ aren't necessarily upper bounds on the entire sequence; they're just upper bounds on some subsequence in the limit. 
I'm really at a loss where to go from here. Can anyone give me some nudge in the right direction? After staring at this proof for days and seeing a lot of resources that say "it's trivial" which is pretty disheartening.
 A: First note that the claim is not quite true as stated: you have to exclude the case that 
 $\limsup_{n \to \infty} x_n = +\infty$ and $\limsup_{n \to \infty} y_n = -\infty$ or vice versa, in which case the right side is undefined.
Suggestion:


*

*Let $a_m = \sup_{n \ge m} (x_n + y_n)$, let $b_m = \sup_{n \ge m} x_n$, $c_m = \sup_{n \ge m} y_n$.  We want to show $\lim_{m \to \infty} a_m \le \lim_{m \to \infty} b_m + \lim_{m \to \infty} c_m$.

*Note that for any $n \ge m$ we have $x_n \le b_m$ and $y_n \le c_m$.  Conclude that for any $n \ge m$ we have $x_n + y_n \le b_m + c_m$.  

*Show that $a_m \le b_m + c_m$.

*By standard properties of limits, $\lim_{m \to \infty} a_m \le \lim_{m \to \infty} (b_m + c_m)$.

*By standard properties of limits, $\lim_{m \to \infty} (b_m + c_m) = \lim_{m \to \infty} b_m + \lim_{m \to \infty} c_m$.  (Note that we have to assume, as I mentioned at the top, that the limit on the right side is not of the form $\infty - \infty$.)
A: You have only given one example of sequences where the opposite statement (sub-additivity fails) is false, by contradiction. A full proof by contradiction would have to consider arbitrary sequences where sub-additivity fails and then somehow derive a contradiction.
