Transitivity for set of 2 elements

[autodidude]

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.

 

[CompuChip]

I would say that it is vacuously true.

 

[autodidude]

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.

 

[HallsofIvy]

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".

 

[blue_raver22]

I can confirm that the transitivity axiom does indeed require at least three elements. In the example given, we have two elements, 0 and 1, and the relationship defined between them is that 0 is less than 1. However, this does not provide enough information to determine the relationship between 0 and any other element, as there is no defined relationship between 1 and any other element. Therefore, the transitivity axiom cannot be applied in this case.

In order for the transitivity axiom to hold, there must be a clear and defined relationship between each element in the set. For example, if we have a set of three elements, say {0, 1, 2}, and we define the relationships as 0<1 and 1<2, then we can apply the transitivity axiom to conclude that 0<2. This is because the relationships between all three elements are defined, allowing us to make a logical conclusion based on the transitivity axiom.

In conclusion, the transitivity axiom requires a set of at least three elements with clearly defined relationships between each element in order to hold true. In the given example with only two elements, the axiom cannot be applied.