If X is infinite and $x_0$ ∈ X , then X - {$x_0$} is still infinite proof

I am a student and I was reading this proof for the question
(book : zgkp you feng lin, shwu yeng_t lin set theory with applications) :

Note that definition 1 is Dedekind definition for infinite sets.

I cannot understand the basis of the proof. Why we consider two cases for this?

I need a deep explanation ;)

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*}

• Now I can imagine that , Complete and neat , Thanks! – program_craft Apr 12 at 2:38
• @program_craft: You’re welcome! – Brian M. Scott Apr 12 at 2:48