# DM - Functions & Relations -Q17

+1 vote

Let "Closure of a Relation R with respect to a Property P" be defined as follows:

let R be a relation on a set A. Then a relation R' is the closure of the relation R with respect to property P if and only if

1. R' satisfy the property P, and

2. $\dpi{100} R\subseteq R'$, and

3. for any(every) relation R'' if $\dpi{100} R\subseteq R''$ and $\dpi{100} R''$ satisfies the property R, then $\dpi{100} R'\subseteq R''$

i.e if there is a relation R' with property P containing R such that R' is a subset of every relation R'' with property P containing R, then R' is called the closure of R with repect to P

Which of the following is False ?

(A). For any non-empty relation R on a non-empty set A, reflexive closure i.e. closure of R with respect to property “reflexivity”, can always exist.

(B). For any non-empty relation R on a non-empty set A, symmetric closure i.e. closure of R with respect to property “symmetricity”, can always exist.

(C). For any non-empty relation R on a non-empty set A, irreflexive closure i.e. closure of R with respect to property “irreflexivity”, can always exist.

(D). For any non-empty relation R on a non-empty set A, transitive closure i.e. closure of R with respect to property “transitivity”, can always exist.

edited 4 days ago

+1 vote

Statement 1 :

For any non-empty relation R on a non-empty set A, reflexive closure i.e. closure of R with respect to property “reflexivity”, can always exist.

It is true. Reflexive closure of R can be found like this

$R'$ = $R \cup \{ (a,a) | a \in A \}$

Statement 2 :

For any non-empty relation R on a non-empty set A, symmetric closure i.e. closure of R with respect to property “symmetricity”, can always exist.

It is true. Symmetric closure of R can be found like this

$R'$ = $R \cup R^{-1}$

Statement 3 :

For any non-empty relation R on a non-empty set A, irreflexive closure i.e. closure of R with respect to property “irreflexivity”, can always exist.

It is false. Consider this relation : $R = \{ (1,1) \}$

Now, we can not make it irreflexive by ADDING something to it.

Statement 4 :

For any non-empty relation R on a non-empty set A, transitive closure i.e. closure of R with respect to property “transitivity”, can always exist.

This is true.

https://en.wikipedia.org/wiki/Transitive_closure#:~:text=Informally%2C%20the%20transitive%20closure%20gives,337).

answered 4 days ago by (112,390 points)