Suppose $A$ and $B$ are sets and $A$ is finite. Prove that $A \sim B$ iff $B$ is also finite and $|A| = |B|$.
Notes on notation:
$A \sim B$ indicates that $A$ is equinumerous with $B$.
Suppose $A$ and $B$ are sets and $A$ is finite. Prove that $A \sim B$ iff $B$ is also finite and $|A| = |B|$.
Notes on notation:
$A \sim B$ indicates that $A$ is equinumerous with $B$.
What do you mean by 'finite' here? Something like $A$ is finite iff $A$ is empty or there is an $m$ such that there is a bijection between $A$ and $\{1, 2, \ldots m\}$??
Assuming so, suppose $A$ and $B$ are both non-empty but finite, i.e. there is an $m$ and $n$ such that $A$ and $B$ are equinumerous with $\{1, 2, \ldots m\}$ and $\{1, 2, \ldots n\}$ respectively. Now what happens when $m = n$, when $m \neq n$???
$A\sim B\Longleftrightarrow \,\exists\, f:A\to B\,\,,\,f\,\,\text{is a bijection}\Longleftrightarrow |A|=|B| $
so
$A\,\,\text{is finite together with the above}\,\Longleftrightarrow |B|=|A|<\infty$