Can someone help me with this homework? Thank you.
Let $a \notin A$ and $b \notin P(A)$. If $A \cup \{a\} \sim A$, prove that $P(A) \cup \{b\} \sim P(A)$. (dont use the Axiom of choice).
Can someone help me with this homework? Thank you.
Let $a \notin A$ and $b \notin P(A)$. If $A \cup \{a\} \sim A$, prove that $P(A) \cup \{b\} \sim P(A)$. (dont use the Axiom of choice).
Sets which have the property $|A|+1=|A|$ are called Dedekind-infinite, the question in essence asks to show that if $A$ is Dedekind-infinite then $P(A)$ is Dedekind-infinite. The canonical example for a Dedekind-infinite set is $\mathbb N$.
In fact, an equivalent definition for Dedekind-infinite sets are $A$ such that $\mathbb N\lesssim A$. The axiom of choice is brought into play because assuming it we have that every infinite set is Dedekind-infinite. It is consistent, however, that without the axiom of choice there are sets which are infinite and have no countable subset.
Let $f\colon A\cup\{a\}\to A$ be a bijection. Define $g\colon P(A)\cup\{b\}\to P(A)$ as:
$g(X)=\begin{cases} X & |X|\neq 1, X\neq b,\\ \{f(a)\} & X=b\\ \{f(x)\} & X=\{x\}, x\in A\end{cases}$
Show that $g$ is a bijection, which follows from the fact that $f$ is a bijection.