TRWBW | noway-: is this the same problem as yesterday? |

noway- | Yes :-/ http://www.mathbin.net/8735 Should I use the fact that every element of A is also a subset of A. Then since P(A) is all the subsets of A, every element in P(A) will a set with elements from A. Then relate the two? will be a* |

TRWBW | noway-: s in P(A) implies s sub A. that's just the definition of P(A). x in S sub A => x in A => x sub A from the hypothesis. x sub A => x in P(A), again by definition of P(A). so x in S => x in P(A), which implies S sub P(A) noway-: i think the key thing here is to see: S in P(A) <=> S sub A. they are completely the same thing. |

noway- | ok is see it more clear now, thank you |

TRWBW | np |

noway- | the second part, x in S sub A => x in A => x sub A, was kinda catching me up a bit |

raz0 | number E-12 is number * 10^(-12), isn't it? |

LoganCapaldo | usu. |

raz0 | got it |

koro | kercyr, i'm here |

prettytoney | Hi. Did that guy prove his set theory thing? |

kercyr | There is a countable abelian group where every element has order 2. |

koro | yeah i saw the backlog |

prettytoney | He was on earlier, I forget the question exactly. S \in P(A) => S \subset P(A), where P is a power set. |

koro | so it's not true that every abelian group of order 2 is isomorphic to Z_2^n for some n? |

kercyr | right. |

koro | er, with every element of order 2 i mean |

TRWBW | PRETTYTONEY: he had been working on it for 2 days so i gave him the answer. i ran out of hints. |

prettytoney | haha, ok. |

kercyr | But it's true that the group is isomorphic to a direct sum of Z_2's. |

Knight_Lord | Can I numerically compute df/dx^2 without computing df/dx? |

koro | oh? |

Knight_Lord | From a 2D function |

koro | i thought Z_2^n was the same as Z_2 + ... + Z_2 (n times) whatever that means when n is infinite |

TRWBW | koro: think of operating over P(R) with A+B=C given by C=(AuB)-(AnB) |

kercyr | Z_2^n is the set of functions from {0,1,..., n-1} to Z_2. |

Knight_Lord | How to compute df/dxy numerically? |

koro | kercyr: oh so direct sum means with all but finitely many coordinates being nonzero? |

kercyr | right. |

prettytoney | How did it go, trwbw? Got a hint for me? |

kercyr | uh... |

TRWBW | kercyr: given any set S, that operation i gave over the powerset of S is isomporphic to Z_2^S |

koro | oh well i told you my algebra was shitty |