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More set fun, can u see if i'm right?

  1. Oct 11, 2006 #1
    Hello everyone 'im not sure if this is right or not, its about 4:00am so bare with me on typos.
    Let S = {a,b,c} and let S_a be the set of all subsets of S that contain a, let S_b be the set of all subsets of S that contain b, let S_c be the set of all subsets of S that contain c, and let S_null be the set whoese only element is Null. Is {S_a, S_b, S_c, S_NULL} a partion of P(S).

    I said:

    No, {S_a, S_b, S_c, S_NULL} is not a partion of P(s) becuase S != S_a U S_b U S_C U S_NULL. All the sets must also be mutally disjoint to be a partition so it can't be a parition of P(S).

    P stands for power set.

    Thanks, if i'm wrong can you tell me what i'm mess up on or misunderstanding. There arn't any examples like this in the book to help me.
     
  2. jcsd
  3. Oct 11, 2006 #2

    AKG

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    It's true that [itex]S \neq S_a \cup S_b \cup S_c \cup S_{\emptyset }[/itex] but this is irrelevant. It so happens that [itex]\mathcal{P}(S) = S_a \cup S_b \cup S_c \cup S_{\emptyset }[/itex], but this is also irrelevant. Since the sets in [itex]\{S_a,\, S_b,\, S_c,\, S_{\emptyset }\}[/itex] are not mutually disjoint, they do not partition [itex]\mathcal{P}(S)[/itex].
     
  4. Oct 11, 2006 #3
    Thanks for the help AKG! :biggrin:
     
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