Answered

Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

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:

We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.