Summing well ordered sets $A$ and $B$ are well-ordered sets. $A, B \subseteq \mathbb{R}$
$$
  C := \{ n+m : n \in A , m \in B \}
  $$
How do i now prove that $C$ is well ordered?
It seem logical to me, but I have to prove that ever $S \subset C$ has a minimum.
 A: Hint: Let's assume that $C$ is not well-ordered, then there is in an infinite strictly decreasing sequence $c_1 > c_2 > \ldots$. Each of those is of the form $a_i + b_j$, so at least one of the $\{a_i\}_i$ or $\{b_j\}_j$ has to contain infinite strictly decreasing sequence, contradiction.
Edit: According to Andres Caicedo (as pointed out in the comments), the follow-up is a standard technique in Ramsey theory.
A: For those who, like me, found the completion of the proof not immediately obvious from the accepted answer and its comments:
We have a strictly decreasing sequence $(c_n)_{n\in\mathbb{N}}$ and two sequences $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$ such that $a_n+b_n=c_n$ for all $n\in\mathbb{N}$. Define $n_0=0$ and, for $k\geq 1$, choose $n_k>n_{k-1}$ such that
$$
a_{n_k}=\min\{a_n:n>n_{k-1}\}.
$$
The subsequence $(a_{n_k})_{k\in\mathbf{N}}$ thus defined is increasing. Since $(c_{n_k})_{k\in\mathbf{N}}$ is strictly decreasing, it follows that $(b_{n_k})_{k\in\mathbf{N}}$ must be strictly decreasing.
