The binary relation $S= \phi \text{(empty set)}$ on a set $A = \left \{ 1,2,3 \right \}$ is

- Neither reflexive nor symmetric
- Symmetric and reflexive
- Transitive and reflexive
- Transitive and symmetric

### 4 Comments

If any one of them missing then the relation will not be reflexive, In this case, none of them are present inside the relation as the relation is empty. So the relation S = {} is not reflexive.

Now for symmetric property we have to check for each pair(a,b), (b, a) is present inside the relation or not. But here as the relation is empty so there is no (a,b) pair, so no need to check for (b, a) pair. Symmetric property is not violated so S is a symmetric relation.

Similarly, it is a transitive relation.

So it is Transitive and Symmetric but not Reflexive.

Answer : D

## 5 Answers

### 14 Comments

learncp A relation 'R' on set A said to be reflexive if ( x R x ) **∀x ∈ A** . It means all diagonal element should be present here .But here no (x,x) type relation is present in given empty set. thats why it is not reflexive.

@Digvijay Pandey @Arjun sir

If set A=ϕ(empty set) and also S=ϕ(empty set) then can we say S is reflexive on set A and hence S is an equivalence relation on set A ?

@ Sourav Basu here is a proof that "an empty relation on an empty set is equivalence."

https://proofwiki.org/wiki/Relation_on_Empty_Set_is_Equivalence

Relation S is defined as a set in which no element of A is related to any element.

This means that R is an empty relation.

**Empty relation with empty set** holds Reflexivity, Transitivity, Symmetric, Anti-Symmetric

This empty relation would match all conditions vacuously because there are no conditions to check(no elements)

**Empty relation with non-empty set** holds Transitivity, Symmetric, Anti-Symmetric but **not reflexivity **because set is non-empty and there are conditions to check for reflexive property.

In this particular question set A is non empty hence answer is D) **Transitive and symmetric**

Hence R= empty set implying it is symmetric,antisymmetric and transitive trivially but not reflexive since by definition any reflexive relation should contain all elements of the form (a,a) for all a in A

The answer is **D**.

When it comes to working with ϕ, keep it in mind that there are two possibilities:

1. The Relation R (or S in the question) is a relation on set A which itself is empty, i.e **R on A = { }**.

2. The Relation R (or S in the question) is a relation on set A which is not empty, i.e. **A have some element**s.

**Case 1: **

Here the very set on which our relation is based, A, is empty, so * there is nothing to check*, and if there is nothing to check, in mathematics and logic we take it to have all properties to be true for that. As it is, without any logical approval or explanation. And that is why we call it Vacuous Truth(wiki link), ( Vacuous means lack of intelligence! ).

So, here if any** R = { } is defined on A = { }, all the properties i.e Reflexivity, Transitivity, Symmetric, Anti-Symmetric, hold true.**

**Case 2: **

But as is asked in the question, the set** A is not empty**. So, * there are elements which need to be checked*.

So, lets take one by one...

a. **Reflexivity**: The only way Reflexivity is deemed true, is if the all the elements of must be related to each other and that is not the case here, so it is **NOT reflexive**.

b. **Transitivity and Symmetry**: These two properties are automatically implied True until and unless there is some reason for rejection. Here we have no elements at all, so there is no chance that someone can conjure up a reason to reject these. Hence, it is Symmetrical and Transitive.

Go to this link to have a detailed understanding.

There is another way of looking at this problem which uses first-order logic concepts. Recall the truth table of A implies B, or A => B.

The definition of **reflexivity** is that if element "a" belongs to A, then (a, a) must belong to S. This is an implication whose left side is true and right side is false, for all elements in A. Therefore the result is false.

For **symmetry**, for a pair of elements a,b in A, if (a, b) belongs to S, then (b, a) must also belong to S. The left side of this implication is false, so it does not matter if the right side is true or false, the result will always be true. Thus S is symmetric.

Similarly, for **transitivity**, given elements a,b,c in A, if (a, b) belongs to S and (b, c) belongs to S, then (a, c) must belong to S. The left side is once again false, so the result of the implication is true. S is transitive.

The same reasoning can be extended to anti-symmetry although it is not required here. Therefore the answer is (D).