Map from $\phi_N: \{1, ..., N\} \rightarrow \{1, ..., r\}$, subsequence of $(\phi_N)_{N=1}^\infty$ that stabilises for each $k \in \mathbb{N}$. In these lecture notes on MIT Opencourseware, on top of page 14 (which is page 2 really), a rough proof is provided for the equivalence of theorems 1.1 and 1.2. These are Schur's theorems but you really don't need to know any of that to understand my problem. You don't even need to read the theorems. I must be missing something really basic here. Theorem 1.2 clearly implies theorem 1.1, but in showing the converse (that theorem 1.1 implies theorem 1.2), the lecturer uses proof by contradiction and negates theorem 1.2. In it, he argues, suppose that given $r \in \mathbb{N}$ (the number of colours), for any $N \in \mathbb{N}$ there exists a colouring $\phi_N: \{1, ..., N\} \rightarrow \{1, ..., r\}$ of these $N$ natural numbers such that there is no monochrome $x, y, z$ (that is, all of the same colour) such that $x + y = z$. Then he considers the infinite sequence $(\phi_N)_{N=1}^\infty$, and in particular an infinite subsequence of it, such that for every $k \in \mathbb{N}$, the value of $\phi_N(k)$ stabilises as $N$ increases. That's the sentence I have a problem with.
How do we know that for every $k \in \mathbb{N}$, we can find a subsequence of $(\phi_N)_{N=1}^\infty$ such that $\phi_N(k)$ stabilises? I mean, let's take $k=1$ for example, and suppose $\exists N \in \mathbb{N}$ such that $\phi_N(1) = s$ for some $s \in \{1, ..., r\}$. Then how do we know that there even exists $M > N$ such that $\phi_M(1) = s$? In plain words, how do we know that some $\phi_N$ will map $1$ onto $s$ ever again (after it's done so once for some $N$)? From what I can see, it's entirely possible that the $\phi_N$ may never map $1$ onto $s$ ever again, while maintaining the condition that the mappings (colourings) $\phi_N$ never permits a monochrome solution $x, y, z \in \mathbb{N}$ to $x + y = z$. What am I missing (other than brain cells)?
Please let me know if I haven't been clear.
 A: For some $s_1$, there are infinitely many $N$ such that $\phi_N(1)=s_1$.   Let $A_1$ be the set of such $N$, and let $N_1$ be its first element.  Now since $A_1$ is infinite, there is some $s_2$ such that there are infinitely many $N\in A_1$ such that $\phi_N(2)=s_2$.  Let $A_2$ be the set of such $N$, and let $N_2$ be the first element of $A_2$ that is greater than $N_1$.  Now since $A_2$ is infinite, there is some $s_3$ such that there are infinitely many $N\in A_2$ such that $\phi_N(3)=s_3$.  Let $A_3$ be the set of such $N$, and let $N_3$ be the first element of $A_3$ that is greater than $N_2$.
Continuing in this fashion, we get an increasing sequence $(N_k)$ and a sequence of elements $s_k\in\{1,\dots,r\}$ such that $\phi_{N_\ell}(k)=s_k$ for all $\ell\geq k$.  In particular, $\phi_{N_\ell}(k)$ stabilizes for each $k$.

Here's a more abstract perspective.  We may extend each $\phi_N$ in some arbitrary way to assume they defined on all of $\mathbb{N}$.  So, $(\phi_N)$ is then a sequence in the space $\{1,\dots,r\}^{\mathbb{N}}$, which is compact and metrizable in the product topology.  Thus, $(\phi_N)$ has a convergent subsequence.  But convergence in the product topology is just pointwise convergence, and convergence in $\{1,\dots,r\}$ is just stabilization, so this means there is a subsequence on which $\phi_N(k)$ stabilizes for each $k$.
