# DM - Functions & Relations -Q19

+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

Let  relation R on a set $A = \{ 0,1,2 \} \ \ and \ \ R = \{ (0,0), (0,1),(1,1,),(2,2) \}$

Let the property "P" be "has an odd number of elements".

The closure of R with respect to property P is :

(A). $\dpi{100} R' = \{ (0,0), (0,1),(1,1,),(2,2) \}$

(B). $\dpi{100} R' = \{ (0,0), (0,1),(1,1,),(2,2), (0,2) \}$

(C). $\dpi{100} R' = \{ (0,0), (0,1),(1,1,),(2,2),(0,2), (1,2),(1,0) , (2,0), (2,1) \}$

(D). None of these

edited 3 days ago

Let the property "P" be "has an odd number of elements".

Let  relation R on a set $A = \{ 0,1,2 \} \ \ and \ \ R = \{ (0,0), (0,1),(1,1,),(2,2) \}$

Since R doesn't have property P so we have to add something to it and make R' that willl satisfy property P.

The closure of R with respect to property P is :

(A). $\dpi{100} R' = \{ (0,0), (0,1),(1,1,),(2,2) \}$

This doesn't have property P. So, False.

(B). $\dpi{100} R' = \{ (0,0), (0,1),(1,1,),(2,2), (0,2) \}$

This has property P BUT  Other relation R'' with property P is also possible such that R'' is Not a superset of R' and R'' also covers R and satisfies P. For example, $\dpi{100} R'' = \{ (0,0), (0,1),(1,1,),(2,2), (1,0) \}$

So, this is also false and we can now see that No such closure exists w.r.t. this given property for this given relation.