The statement
$\overline{\lim} (a_n+b_n) = \overline{\lim} a_n + \overline{\lim} b_n $, for bounded sequences $(a_n)$ and $(b_n)$
is false for arbitrary bounded sequenecs. What is true generally however is subadditivity:
$\overline{\lim} (a_n+b_n) \leq \overline{\lim} a_n + \overline{\lim} b_n $, for bounded sequences $(a_n)$ and $(b_n)$.
Knowing that, I am not able to spot the error in the arguments I've written up below that seemingly aver the statement of equality. Can someone see where exactly the error is present?
The definition of limit superior that I'm working with here is:
$$ \overline{\lim} (a_n + b_n) = \sup A = \sup \{ x \in \mathbb{R}: x < a_n + b_n \ \text{ infinitely many times} \} $$
Putative proof:
We know there exist subsequences $(a_{n_k})$ and $(b_{n_k})$ that converge to $\overline{\lim} a_n$ and $\overline{\lim} b_n$ respectively.
That means:
For each real $\varepsilon > 0$ there is a positive integer $N_1$ such that $k \geq N_1$ implies
$ \overline{\lim} a_n – \varepsilon < a_{n_k} < \overline{\lim} a_n + \varepsilon $
And similarly for the other sequence, there is a positive integer $N_2$ such that $k \geq N_2$ implies
$ \overline{\lim} b_n – \varepsilon < b_{n_k} < \overline{\lim} b_n + \varepsilon $
Consequently, for each integer $k \geq \max(N_1,N_2)$ we have
$ \overline{\lim} a_n + \overline{\lim} b_n – 2\varepsilon < a_{n_k} + b_{n_k} < \overline{\lim} a_n + \overline{\lim} b_n + 2\varepsilon $
This means that the number $\overline{\lim} a_n + \overline{\lim} b_n – 2 \varepsilon $ has above it infinitely many terms from the sequence $(a_n + b_n)_{n\geq 1}$.
Therefore it is an element of the set $A$ and bounded above by the supremum $\overline{\lim} (a_n + b_n)$, that is
$ \overline{\lim} a_n + \overline{\lim} b_n – 2\varepsilon \leq \overline{\lim} (a_n+b_n) $
If $x \in A$, then $x < a_n + b_n$ infinitely many times. What does that mean, "infinitely many times"? Well it must mean that given any positive integer $m$, there exists some other positive integer $p > m$ such that $x < a_p + b_p$ (otherwise there would be finitely many terms above $x$, which would be false).
So there must exist some integer $m > \max(N_1, N_2)$ such that $x < a_m + b_m$. But then since $m > \max (N_1, N_2)$, this means $x < a_m + b_m < \overline{\lim} a_n + \overline{\lim} b_n + 2 \varepsilon$.
Consequently $A$ is bounded above by the number $\overline{\lim} a_n + \overline{\lim} b_n + 2 \varepsilon$, so this number must greater or equal to $\sup A$. That is,
$ \overline{\lim} ( a_n + b_n ) \leq \overline{\lim} a_n + \overline{\lim} b_n + 2 \varepsilon $
All together we have shown for each real $\varepsilon > 0$ that
$ \overline{\lim} a_n + \overline{\lim} b_n – 2 \varepsilon \leq \overline{\lim} ( a_n + b_n ) \leq \overline{\lim} a_n + \overline{\lim} b_n + 2 \varepsilon $
This implies $\overline {\lim} (a_n + b_n) = \overline{\lim} a_n + \overline {\lim} b_n$. $\blacksquare$
But of course, that is false! Yet I can't manage to find the error in my arguments above.
Best Answer
The problem is that the sequences $a_{n_k}$ and $b_{n_k}$ don't necessarily follow the same subsequences. Instead, replace $b_{n_k}$ with, say, $b_{m_k}$.
Eventually, this part:
$\overline{\lim} a_n + \overline{\lim} b_n - 2\varepsilon < a_{n_k} + b_{n_k} < \overline{\lim} a_n + \overline{\lim} b_n + 2\varepsilon$
should be
$\overline{\lim} a_n + \overline{\lim} b_n - 2\varepsilon < a_{n_k} + b_{m_k} < \overline{\lim} a_n + \overline{\lim} b_n + 2\varepsilon$
and this fact does not guarantee that
"This means that the number $\overline{\lim} a_n + \overline{\lim} b_n - 2 \varepsilon$ has above it infinitely many terms from the sequence $(a_n + b_n)_{n\geq 1}$."
Even as just a trivial example, suppose $n_k$ are odd integers and $m_k$ are even integers. Then actually the subsequences form a partition.