i have a few question, that i hope they will answered.(adsbygoogle = window.adsbygoogle || []).push({});

1) let w={0,1...,n,..}={0}UN, and let f:wxw->w such that the next requirements apply:

a) f(0,n)=n+1

b) f(m+1,0)=f(m,1)

c) f(m+1,n+1)=f(m,f(m+1,n).

i need to prove that for every n,m in w, the next statement applies:

f(m,n)<f(m,n+1).

(i tried with induction, but this was too much complicated for me to do).

2)let there be 2 sets, X and Y, prove with the help of AC that there exist a function from X onto Y iff |X|>=|Y|.

well for the second question, here what i did:

suppose, there exists f:X->Y which is onto Y, if we utilize AC then lets define P(Y)\{empty set} as a subset of X, then by AC f|P(Y)\{empty set} P{Y}\{EMPTY SET}->Y such that f|P(X)\{ES}(A) is in A for every A in P(Y)\{ES} which means Y is onto P(Y)\{ES} and therefore there is bijection between Y and a subset of X, therefore |X|>=|Y|, the second conditional here im not sure hoe to prove, i will appreciate your help.

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Axiom of choice and natural numbers.

Loading...

Similar Threads for Axiom choice natural | Date |
---|---|

A Axiom of Choice not self evident? | Sep 10, 2017 |

Axiom of Choice | Aug 23, 2015 |

Confusion over Axiom of Choice. | Jan 8, 2014 |

Question about the Axiom of Dependent Choice | Oct 13, 2013 |

The axiom of choice one a finite family of sets. | Aug 7, 2013 |

**Physics Forums - The Fusion of Science and Community**