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

- October 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.

- October 4th 2010, 10:03 AMemakarov
If is denumerable, there is a bijection between and . Take any . Show that and are equinumerous. Without loss of generality, one may assume that .

- October 4th 2010, 03:53 PMtonio