DM - Functions & Relations -Q19

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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. R\subseteq R', and

3. for any(every) relation R'' if R\subseteq R'' and R'' satisfies the property R, then 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). R' = \{ (0,0), (0,1),(1,1,),(2,2) \}

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

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

(D). None of these

asked Jun 24 in Discrete Maths by gbeditor (44,490 points)
edited 3 days ago by deepak-gatebook

1 Answer

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Best answer

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). R' = \{ (0,0), (0,1),(1,1,),(2,2) \}

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

(B). 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.

So, answer D.

answered 3 days ago by deepak-gatebook (112,390 points)
selected 3 days ago by deepak-gatebook
Does that imply that closure of a set should be unique?
Yes. Closure of a set should be unique if it exists.

//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 ///

Assume that Closure is Not unique. Then two closure exist, say R1 and R2. Then by 3rd property that is defined in the definition of closure, one of them has to be subset of other. If that so, say, R1 is subset of R2, then it will make that relation R1 a closure and R2 will no longer be a closure.
i did not get why option B is not correct, property 1 and 2 are holding for it. can you explain what is meant by property 3
Property 3 says that if two relations R1,R2 satisfy property 1 and 2 then one of them must be subset of other. Hence, basically property 3 says that clousre is unique if it exists.
For example, relation M = { (0,0),(0,1),(1,1),(2,2),(3,3) } also satisfies property 1 and 2. So, we have more than one relations who are candidate for being closure But since we don't have a unique relation which satisfies both property 1,2 and also is subset of ALL relations that satisfy property 1,2; so, No closure exists.
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