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The following argument claims to prove that the requirement that an equivalence relation be reflexive is redundant. In other words, it claims to show that if a relation is symmetric and transitive, then it is reflexive. Find the mistake in the argument.

Proof: Let R be a relation on a set A and suppose R is symmetric and transitive. For any two elements, x and y in A, if x R y, then y R x since R is symmetric. But then it follows by transitivity that x R x. Hence R is reflexive.

Sagot :

aalsuraimi6

Answer:

2

If R is a relation that is transitive and symmetric, then R is reflexive on dom(R)={a∣(∃b)aRb}: if a∈dom(R), then there is b such that aRb, thus bRa by symmetry, so aRa by transitivity.

Note that if R is symmetric, then dom(R)=range(R)={b∣(∃a)aRb}.

Hence, to get an example of a relation R on a set A that is transitive and symmetric but not reflexive (on A), there has to be some a∈A which is not R-related to any b∈A. There are many examples of this:

A={0,1} and R={(0,0)},

not reflexive on A because (1,1)∉R,

A={0,1,2} and R={(0,0),(0,1),(1,0),(1,1)},

not reflexive on A because (2,2)∉R.

Step-by-step explanation:

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