Transitivity for set of 2 elements

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In a set of two elements, such as S={0, 1}, the transitivity axiom does not apply in the traditional sense since it requires at least three elements to evaluate the condition "if a < b and b < c, then a < c." The discussion concludes that the transitivity for this set is considered "vacuously true" because the premise can never be satisfied; there are no three distinct elements to test the condition. The term "vacuously true" means that a statement is considered true because its premise is false, regardless of the truth of the conclusion. Thus, in this case, since "a < b and b < c" can never hold, the transitive property is deemed vacuously true for the set. This understanding clarifies the nature of transitivity in sets with fewer than three elements.
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If we have a set of two elements, say S={0. 1} and we defined 0 to be less than 1, would this obey the transitivity axiom? (If a<b and b<c then a<c? )

To me, it seems looks like you need at least 3 elements but I'm not entirely sure.
 
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I would say that it is vacuously true.
 
Thanks...could you please elaborate a bit on what 'vacuous' means in this context?

EDIT: Nevermind, I wiki'd it and think I understand it. Of course, any additional thoughts would be appreciated.
 
For others who might be wondering, the implication "if P then Q" is true whenever P is false, whether Q is true or not. That is what is meant by "vacuously true". In this particular problem, because there are only two elements, "a< b and b< c" is never true, therefore the conclusion, that "<" for this set is transitive, is "vacuously true".
 
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