# Exercise $1.9.3$ of Howie's “Fundamentals of Semigroup Theory” follow up

In the following link there is an answer to a question I am working on but I'm nut sure I understand it fully.

Exercise 1.9.3 of Howie's “Fundamentals of Semigroup Theory”.

The second question:

Let $$X$$ be a countably infinite set and let S be the set of one-to-one maps $$\alpha:X\rightarrow X$$ with the property that $$X\setminus X\alpha$$ is infinite.
(b) Show that for all $$\alpha\in S$$ there exists a bijection between $$X\setminus X\alpha$$ and $$X\alpha \setminus X\alpha^2$$.

(b) In general, $$X\alpha\setminus X\alpha^2\subseteq (X\setminus X\alpha)\alpha.$$
The reverse inclusion holds because α is injective. α restricts to a bijection $$X \setminus X\alpha\rightarrow X\alpha \setminus X\alpha^2$$.

I'm unsure as to why the first line in general is true?

• I have tried to type your question properly. Is this edit correct? – MRT Oct 30 '18 at 18:28
• @MRT Sorry to say that your edit made the question incomprehensible. You confused the symbols $\setminus$ and $/$. – J.-E. Pin Nov 9 '18 at 15:18

The first line follows from general properties of maps between sets. Let $$\alpha:E \to F$$ be a map (not necessarily injective) and let $$A, B$$ be subsets of $$E$$. Then $$A = (A \setminus B) \cup B$$, whence $$A\alpha = (A \setminus B)\alpha \cup B\alpha$$ and finally $$A\alpha \setminus B\alpha \subseteq (A \setminus B)\alpha$$.
The inclusion $$X\alpha\setminus X\alpha^2\subseteq (X\setminus X\alpha)\alpha$$ now follows by taking $$A = X$$ and $$B = X\alpha$$.