Prove that liming(a_n + b_n) ≤ liminf a_n + limsup b_n

[looserlama]

Homework Statement

Consider two bounded sequences {a_n} and {b_n} and put r = liminf(a_n)$_{n→\infty}$, s = limsup(b_n)$_{n→\infty}$ and t = liminf(a_n + b_n)$_{n→\infty}$

Show that t ≤ r + s, i.e., that
liminf(a_n + b_n)$_{n→\infty}$ ≤ liminf(a_n)$_{n→\infty}$ + limsup(b_n)$_{n→\infty}$ .

Homework Equations

In the part of the problem before we needed to show that for every ε>0 there are an infinite number of integers n such that a_n ≤ r + ε and b_n ≤ s + ε

The Attempt at a Solution

I tried to do a proof by contradiction, so:

Assume that t > r + s

Therefore $\exists$x s.t. t > x > r + s$\forall$ε>0 a_n + b_n ≤ t - ε for finitely many terms

Therefore, let ε = t - x

Therefore a_n + b_n ≤ x for finitely many terms.However, from previous problem we know that

$\forall$ε>0
a_n ≤ r + ε/2 for infinitely many terms
b_n ≤ s + ε/2 for infinitely many terms

Therefore a_n + b_n ≤ r + s + ε

Therefore, let ε = x - r - s

Therefore a_n + b_n ≤ x for infinitely many terms

Therefore there is a contradiction.

So t ≤ r + s.I think that makes sense, but does that cover all possible values for r and s? What if they are ±$\infty$ ?

 

[Adrmay]

Also, I am not sure if my proof by contradiction is correct. Any help/suggestions would be appreciated!