# axiom of choice

• Sep 13th 2009, 01:49 PM
dori1123
axiom of choice
Define an injective map $\displaystyle f: \mathbb{Z}^+ \rightarrow \{0,1\}^\omega$ without using the axiom of choice.

I would define an injective function by $\displaystyle f(n)=x$, where $\displaystyle x_i = 1$ if $\displaystyle i=n$ and $\displaystyle x_i=0$ if $\displaystyle i \not= n$. But how would I define the map without using the axiom of choice?
• Sep 13th 2009, 02:01 PM
ThePerfectHacker
Quote:

Originally Posted by dori1123
Define an injective map $\displaystyle f: \mathbb{Z}^+ \rightarrow \{0,1\}^\omega$ without using the axiom of choice.

I would define an injective function by $\displaystyle f(n)=x$, where $\displaystyle x_i = 1$ if $\displaystyle i=n$ and $\displaystyle x_i=0$ if $\displaystyle i \not= n$. But how would I define the map without using the axiom of choice?

If $\displaystyle n>0$ let $\displaystyle n\mapsto \left< 1,\underbrace{1,1,...,1}_{n\text{ times }},0,0,0,... \right>$

If $\displaystyle n<0$ let $\displaystyle n\mapsto \left< 0, \underbrace{1,1,...,1}_{|n|\text{ times }},0,0,0,... \right>$

If $\displaystyle n=0$ let $\displaystyle n\mapsto \left<0,0,0,0,0,0,...\right>$