If $\{x_n\}, \{z_n\}$ are $\mathbb{R}$ sequences such that $\{x_n\}$ converges to $x$ and $lim sup_{n\to\infty} z_n = z$, how can we show that $lim sup_{n\to\infty}(x_n+z_n) \leq x+z$?
The definitions I'm using here:
A sequence $\{x_k\}$ of real numbers converges to a real number $x$ if for every $\epsilon>0,\exists K>0$ s.t. $|x_k-x|<\epsilon, \forall k\geq K$.
The supremum of a set A of real numbers is defined as the smallest extended real number $x$ s.t. $x\geq y, \forall y\in A$. $lim sup_{k\to\infty} x_k$ is defined to be $lim_{m\to\infty} sup\{x_k|k\geq m\}$.
I'm not sure how to reason about the sum of the sequences, $x_n +z_n$ -- any rigorous explanation from these first principles would be appreciated!
Let $l:=lim sup_{n\to\infty}(x_n+z_n)$. Then there is a subsequence $(x_{n_k}+z_{n_k})$ such that
$x_{n_k}+z_{n_k} \to l$ for $k \to \infty$.
Since $(x_n)$ converges to $x$, we have: $x_{n_k} \to x$ for $k \to \infty$.
Hence
$z_{n_k} \to l-x$ for $k \to \infty$.
Since $z=lim sup_{n\to\infty} z_n$ , it follows that $l-x \le z$ and therefore
$$l \le x+z.$$