Is this statement an aspect of the Hairy Ball or Fixed Point Theorem?

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SUMMARY

The discussion centers on the relationship between a statement regarding mutually exclusive classes and the Hairy Ball and Fixed Point Theorems. It establishes that for any class of non-null mutually exclusive classes, there exists at least one class with a unique term in common with each class. This concept parallels the axiom of choice, where selecting an element from disjoint sets leads to a union that shares exactly one element with each set. The analogy with hair whorls and crumpled bingo sheets illustrates the underlying mathematical principles effectively.

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  • Understanding of the Axiom of Choice in set theory
  • Familiarity with the Hairy Ball Theorem
  • Knowledge of Fixed Point Theorems in topology
  • Basic concepts of mutually exclusive classes in mathematics
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  • Investigate Fixed Point Theorems and their relevance in mathematical analysis
  • Examine examples of mutually exclusive classes and their properties
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Mathematicians, students of advanced mathematics, and anyone interested in set theory and topology will benefit from this discussion.

swampwiz
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“Given any class of mutually exclusive classes, of which none is null, there is at least one class which has exactly one term in common with each of the given classes…”

The reason this statement sounds like one of those theora is that I recall reading a Time-Life book on Mathematics, and there was a discussion about the fact that a hair whorl always has at least 1 point where the hair doesn't whorl, and as well that if a sheet of bingo paper is crumpled up and placed on an identical un-crumpled up sheet, there will be at least 1 bingo number for which that of the crumpled one will be on top of that of the un-crimpled one, and it seemed that this was the same idea as the hair whorl.
 
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This looks like an alternate statement of the axiom of choice: if the family of disjoint sets is ##\{E_{\alpha}\}_{\alpha\in A}## and for each ##\alpha\in A,## we select an element ##a_{\alpha}\in E_{\alpha}## [using axiom of choice here] then the set ##\bigcup_{\alpha\in A}\{a_{\alpha}\}## has exactly one element in common with each ##E_{\alpha}.##

It also implies the (usual statement of the) axiom of choice, because given the family ##\{E_{\alpha}\}_{\alpha\in A}## and such a set ##E##, then ##E\cap E_{\alpha}## selects one element from each set.
 
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