# Show that every denumerable set is equivalent to a proper subset of itself

• Oct 4th 2010, 09:53 AM
dynas7y
Show 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 AM
emakarov
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 PM
tonio
Quote:

Originally Posted by dynas7y
I'm not sure how to do this. Help would be appreciated, thanks.

If a set A is denumerable (meaning: INFINITE denumerable), then we can write $\displaystyle A=\{a_1,a_2,a_3,\ldots\}$, so define

$\displaystyle f:A\rightarrow A-\{a_1\}\,,\,\,f(a_i):=a_{i+1}$ . Now just prove $\displaystyle f$ is a bijection...

Tonio