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

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SUMMARY

The discussion centers on the concept of set partitions, specifically whether the sets {S_a, S_b, S_c, S_null} form a partition of the power set P(S) where S = {a, b, c}. The conclusion is definitive: {S_a, S_b, S_c, S_null} does not constitute a partition of P(S) because the sets are not mutually disjoint. Additionally, while it is true that S is not equal to the union of these sets, this fact does not affect the partition criteria. The key takeaway is that mutual disjointness is essential for a valid partition.

PREREQUISITES
  • Understanding of set theory concepts, particularly power sets.
  • Familiarity with the definitions of subsets and partitions.
  • Knowledge of mutual disjoint sets.
  • Basic mathematical notation and symbols.
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  • Study the properties of power sets in set theory.
  • Learn about the criteria for set partitions and mutual disjointness.
  • Explore examples of valid and invalid partitions in set theory.
  • Investigate advanced topics in combinatorics related to subsets and partitions.
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Students of mathematics, educators teaching set theory, and anyone interested in combinatorial mathematics will benefit from this discussion.

mr_coffee
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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) because 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.
 
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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].
 
Thanks for the help AKG! :biggrin:
 

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