Why is max{n,k}= Ө(n+k)
Assuming n and k to be non-negative,
$n\leq n+k$ and $k\leq n+k$. Hence,$\max(n, k) \in \mathcal{O}(n+k)$.
Next, $n+k≤2\max(n,k)$. Hence, $\max(n,k)\in\mathcal{\Omega}(n+k)$.
Hence, we get that $\max(n,k)\in \mathcal{\Theta}(n+k)$.