I'm not sure how to do this. Help would be appreciated, thanks.

- Oct 4th 2010, 09:53 AMdynas7yShow that every denumerable set is equivalent to a proper subset of itself
I'm not sure how to do this. Help would be appreciated, thanks.

- Oct 4th 2010, 10:03 AMemakarov
If $\displaystyle A$ is denumerable, there is a bijection $\displaystyle f$ between $\displaystyle A$ and $\displaystyle \mathbb{N}$. Take any $\displaystyle a\in A$. Show that $\displaystyle \displaystyle\mathbb{N}$ and $\displaystyle \mathbb{N}\setminus\{f(a)\}$ are equinumerous. Without loss of generality, one may assume that $\displaystyle f(a)=0$.

- Oct 4th 2010, 03:53 PMtonio