Are you looking for something like this?
$$\Bbb Z_0 = \{0, \pm m, \pm2m, \pm3m, \ldots\}\\
\Bbb Z_1 = \{1, \pm m +1, \pm2m + 1, \ldots\}\\
\Bbb Z_2 = \{2, \pm m +2, \pm2m + 2, \ldots\}
$$
I'll do a concrete example, so you can see how it works. Say $m = 3$. Then
$$
\Bbb Z_0 = \{x \mid x = 3k\text{ for some }k\in \Bbb Z\}
$$
It is spelled out "$\Bbb Z_0$ is the set of all $x$ such that $x = 3k$ for some $k \in \Bbb Z$". This means that $15 \in \Bbb Z_0$, since $15 = 3\cdot 5$ (in that case, $k = 5$). However, $43 \notin \Bbb Z_0$, because there is no $k\in \Bbb Z$ so that $43 = 3k$. We get
$$
\Bbb Z_0 = \{0, \pm 3, \pm 6, \pm 9, \ldots\} = \{0, 3, -3, 6, -6, 9, -9,\ldots\}
$$
On the other hand, we have
$$
\Bbb Z_1 = \{x \mid x = 3k+1\text{ for some }k \in \Bbb Z\}
$$
which means that a number $x$ is an element of $\Bbb Z_1$ iff there is a $k\in \Bbb Z$ such that $x = 3k+1$. For instance, $43 \in \Bbb Z_1$, since there is a $k\in \Bbb Z$ that makes $43 = 3k + 1$ (it's $14$). However, $15 \notin \Bbb Z_1$, since there is no integer $k$ such that $15 = 3k + 1$. This gives
$$
\Bbb Z_1 = \{1, \pm3 + 1, \pm6 + 1, \ldots\} = \{1, 4, -2, 7, -5, \ldots\}
$$
Lastly, in exactly the same way, we get
$$
\Bbb Z_2 = \{2, \pm3 + 2, \pm 6 + 2, \ldots\} = \{2, 5, -1, 8, -4,\ldots\}
$$
Why polynomial functions $f(x)+g(x)$ is the same notation as $\mathsf (f+g\mathsf)(x)$?
Why not? It is convenient and brief. It lets us talk about the polynomial $f+g$ .
I've seen the sum of polynomials as $f(x)+g(x)$ before, but never saw a notation as with a operator in a parenthesis as $\mathsf(f+g\mathsf)(x)$. And author puts $\mathsf(f+g\mathsf)(x)$ at the first.
Well, the author is defining the notation; right there. It just introduced you to it. Shake hands and get to know it. Now you will know what the author means when it is used in future.
What do you suppose $\mathsf(f+g+h\mathsf)(x)$ means?
For that matter $\mathsf (f\cdot g\mathsf)(x)$ should be introduced soon too. Can you anticipate what that shall mean?
Best Answer
I’m not sure what it is that you don’t understand. We need to show that there is an injection $g:X\setminus\{x_0\}\to X\setminus\{x_0\}$ such that $g[X\setminus\{x_0\}]\subsetneqq X\setminus\{x_0\}$. The only thing that we know is that $X$ is infinite, i.e., that there is an injection $f:X\to X$ such that $f[X]\subsetneqq X$, so somehow we will have to use $f$ to construct the desired $g$. It turns out that how we do this depends on whether or not $x_0\in f[X]$.
If $x_0\notin f[X]$ we can just let $g$ be the restriction $f\upharpoonright X\setminus\{x_0\}$; that’s Case $\mathit 2$ of the proof. If $x_0\in f[X]$, however, this might not work. To see why, take $X=\Bbb N$ (which for me includes $0$), let $x_0=1$, and let
$$f:\Bbb N\to\Bbb N:x\mapsto x+1\,.$$
If $g=f\upharpoonright(\Bbb N\setminus\{1\})$, then
$$\begin{align*} g[\Bbb N\setminus\{1\}]&=f[\Bbb N\setminus\{1\}]\\ &=\{f(0)\}\cup\{f(n):2\le n\in\Bbb N\}\\ &=\{1\}\cup\{n\in\Bbb N:n\ge 3\}\\ &\not\subseteq\Bbb N\setminus\{1\}\,; \end{align*}$$
that is, the restriction of $f$ to $\Bbb N\setminus\{1\}$ is not a function from $\Bbb N\setminus\{1\}$ to $\Bbb N\setminus\{1\}$.
Case $\mathit 1$ of the proof shows how to construct $g$ in order to avoid this problem. We still take $g$ to be mostly the restriction of $f$ to $X\setminus\{x_0\}$, but if $x_1\ne x_0$, as is the case in my example, we have to define $g(x_1)$ in some other way to ensure that $g(x_1)\in X\setminus\{x_0\}$.
In my example $x_1$ is $0$, since $f(0)=1=x_0$, and we redefine $g(0)$ to be some $x_2$ that is not $1$ (since $1$ isn’t in $\Bbb N\setminus\{1\}$, the desired codomain of $g$) and is not in $f[\Bbb N]$ (so that it won’t conflict with $f(n)$ for any other $n\in\Bbb N\setminus\{1\}$). In this example the only possibility is $x_2=0$, since $f[\Bbb N]=\Bbb N\setminus\{0\}$. Thus, we let $g(0)=0$ and let $g(n)=f(n)=n+1$ if $n\ge 2$, so that
$$\begin{align*} g[\Bbb N\setminus\{1\}]&=\{g(0)\}\cup\{g(n):2\le n\in\Bbb N\}\\ &=\{0\}\cup\{f(n):n\ge 2\}\\ &=\{0\}\cup\{n\in\Bbb N:n\ge 3\}\\ &=\Bbb N\setminus\{1,2\}\\ &\subsetneqq\Bbb N\setminus\{1\}\,. \end{align*}$$