Schur Theorem implies Finite Schur Theorem

Question

$\text{If } \mathbb{N}=C_{1} \cup C_{2} \cup \ldots \cup C_{r} \text{ is a finite partition, then for some } j \in\{1,2, \ldots, r\}, \text{ there exist } a, b, c \in C_{j} \text{ with } a+b=c.$

$\Rightarrow \text { Given } r \in \mathbb{N}, \text { there exists a positive integer } N= N(r) \text { such that if }\{1,2, \ldots, N\}=C_{1} \cup C_{2} \cup \ldots \cup C_{r}, \text { is a finite partition, then for some } j \in\{1,2, \ldots, r\}, C_{j} \text { contains } a, b, c \text { with } a+b=c $

I am trying to understand the following proof to the given statement.

Suppose not. Then for each $N \in \mathbb{N},$ there is a coloring of $\{1,2, \ldots, N\}$ with $r$ colors that does not contain any monochromatic Schur triple. Use any of the $r$ colors to color the integers $\{N+1, N+$ $2, \ldots\},$ thus extending each coloring of $\{1,2, \ldots, N\}$ to a coloring of $\mathbb{N}$ with $r$ colors. Thus for each $N \in \mathbb{N}$, we have a corresponding coloring of $\mathbb{N}$. (*) Since only finitely many colors were used, by the Pigeonhole Principle there are infinitely many $N \in \mathbb{N}$ such that the corresponding coloring uses the same color for the integer $1 .$ Consider only this subsequence of colorings and note that we are still left with an infinite set of colorings. We can now repeat this process in this subsequence. By passing to a subsequence, if necessary, we can assume that all the colorings use the same color for the integer 2 . Inductively, we can pass to a subsequence of the original list of colorings that fixes each $n \in \mathbb{N}$ with a single color. Taking the limit along this subsequence, we obtain a coloring of $\mathbb{N}$ that does not contain a monochromatic Schur triple, a contradiction.

I am stuck at (*). How does the author use the Pigeonhole Principle? What exactly is the process of taking sub-sequences?

There is a very similar question here which was helpful to me in proving what I want but I want to understand this proof too.

Answer 1

For each $N\in\Bbb N$ we have a coloring $\mathscr{C}_N$ of $\Bbb N$ that uses $r$ colors and does not have any monochromatic Schur triple in the set $\{1,2,\ldots,N\}$. Let $c_N$ be the color given to the number $1$ by the coloring $\mathscr{C}_N$. Since each of the colorings $\mathscr{C}_N$ uses the same $r$ colors, there are only $r$ possibilities for the colors $c_N$. But there’s a $c_N$ for every $N\in\Bbb N$, so at least one of those $r$ colors must be used for infinitely many of the $c_N$. Thus, there are an increasing sequence $\langle N_k:k\in\Bbb N\rangle$ of positive integers and a color $c(1)$ such that $c_{N_k}=c(1)$ for every $k\in\Bbb N$. That is, we have an infinite subsequence $\langle\mathscr{C}_{N_k}:k\in\Bbb N\rangle$ of the original sequence $\langle\mathscr{C}_N:N\in\Bbb N\rangle$ of colorings with the property that every coloring $\mathscr{C}_{N_k}$ in this subsequence colors $1$ with the color $c(1)$.

Now start with this subsequence $\langle\mathscr{C}_{N_k}:k\in\Bbb N\rangle$ and do the same thing. Every coloring $\mathscr{C}_{N_k}$ gives $2$ one of the same $r$ colors, so there must be an infinite subsequence of it, $\langle\mathscr{C}_{N_{k_i}}:i\in\Bbb N\rangle$, consisting of colorings that all give $2$ the same color, say $c(2)$. We continue repeating this process. At stage $n+1$ we start with an infinite sequence of colorings that all color $1$ with color $c(1)$, $2$ with color $c(2)$, and so on $n$ with color $c(n)$. We repeat the argument of the first paragraph to conclude that an infinite subsequence of this sequence must all color $n+1$ with the same color, which we’ll call $c(n+1)$. This recursive construction can be carried out to produce a color $c(n)$ for each $n\in\Bbb N$.

Let $\mathscr{C}$ be the coloring defined by the colors $c(n)$ for $n\in\Bbb N$: we color $1$ with $c(1)$, $2$ with $c(2)$, and so on for all $n\in\Bbb N$. Suppose that $\mathscr{C}$ had a monochromatic Schur triple $\langle u,v,w\rangle$, with $u+v=w$. At stage $w$ in the construction we had an infinite subsequence of the original sequence of colorings such that every coloring in the subsequence gave every positive integer $n<w$ the color $c(n)$, and we found an infinite subsequence $\sigma$ of that consisting of colorings that gave $w$ the color $c(w)$. The triple $\langle u,v,w\rangle$ is monochromatic, so $c(u)=c(v)=c(w)$. Since $\sigma$ is an infinite subsequence of $\langle\mathscr{C}_N:N\in\Bbb N\rangle$, there is an $N\in\Bbb N$ such that $N>w$ and $\mathscr{C}_N$ is one of the sequences in the subsequence $\sigma$. And now we have our contradiction: $\mathscr{C}_N$ has no monochromatic triple in the set $\{1,2,\ldots,N\}$, but $\langle u,v,w\rangle$ is a monochromatic triple in that set.

This contradiction shows that $\mathscr{C}$ cannot have a monochromatic Schur triple, and that contradicts the Schur theorem. This final contradiction then shows that no such sequence as $\langle\mathscr{C}_N:N\in\Bbb N\rangle$ can exist, and therefore there must be some $N\in\Bbb N$ such that any $r$-coloring of $\{1,2,\ldots,N\}$ has a monochromatic Schur triple.