symmetry + transitivity => reflexivity, or not?

Question

Professor has posted a question that if one relation has the property of symmetry, such that if aRb, then bRa, and also transitivity, aRb, bRc so aRc, so can we say that because aRb and bRa, we can get aRa which is reflexivity?

Answer 1

Basically, the observation is that reflexivity has to be true for all elements.

For example, consider, S={1,2,3} and R={(1,2), (2,1), (1,1), (2,2)}.

R is symmetric as well as transitive, but not reflexive, since (3,3) is not in R.

Therefore, a relation my be symmetric as well as transitive, and it still may not be reflexive.

Comments

I got it. So reflexivity has to be applied on all elements but symmetry and transitivity can be satisfied as long as all the relationship is symmetry and transitivity.

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Hi Prof, since (3,3) is not in the relation, then can we say that R is not applied to set = {1,2,3}, rather it is applied on set {1,2}? I am confused on the definition that R  is a relation on set. However R has no relation with 3. Is it a correct statement to say that R is a relation on set {1,2,3}?

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Yeah, I have the same question. R should be applied to every elements of S so we can say R is symmetric and transitive on set S, right?

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Oh!I was confused symmetric and transitive with reflexive. Now I get it.

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There is no requirement that a relation include every element of the set that the relation is defined on.

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Thank you Professor I am now clear about it

Answer 2

I agree with the answer above and just want to add something which may be helpful for understanding.

Given a set S = {1,2,3} and a binary relation R = {(1,2), (2,1) , (1,1), (2,2)};

"Reflexivity is not an internal property of a relation. Given a relation R we don't have a sufficient information to decide whether or not it is reflexive. The reason is that we say that R is reflexive on S rather than just reflexive."

To see if  relation R is reflexive on S, we need to see for every element a in S, if  (a,a) is an element of  R, as Prof said, (3,3) is not an element of R, thus symmetry + transitivity !=> reflexivity

Comments

Your explanation is clear enough. thank you!

Answer 3

After consideration and search online, i have find that the prove of wrong of this deduction is following:

Assume a set of elements {1,2,3}

Relation R can be apply on this set so that we have

R = {(1,2), (1,3), (2,1), (3,1), (2,3), (3,2)}

The relation (1,2) and (2,1) satisfy symmetry, (1,2), (2,3), (3,1) satisfy transit etc.

So the main point is that all the element in the set is satisfied with symmetry and transit. But (1,1), (2,2), (3,3) is not defined in the set so it is not existed ever in the set. So even if we say that relation R is symmetry transit, but it is not necessarily reflexive. 

But my confusion is that since R has symmetry and transit property, can we say that (1,1), (2,2), (3,3) is the hidden element in relation R? Its like before Faraday, we just know electricity and magnetivity and they are independent. But actually the hidden property is that electricity and magnetic can be generated by each other, it just has not been recognised among our physical system yet. Or take another example is that for primitive man, they only know egg is egg, and fire drop down by thunder. However actually egg + fire we can make sunrise eggs. So after induction can we say that reflexivity is actually existing in this system? 

Comments

I don't think this relation satisfies the requirement. For example, if (1,2) and (2,1) are in R, and since R is supposed to be transitive, therefore (1,1) should also be in R.