ISRO CSE 2007 | Question: 29

Question

The set of all Equivalence Classes of a set A of Cardinality C

• A. is of cardinality $2^c$
• B. have the same cardinality as A
• C. forms a partition of A
• D. is of cardinality $C^2$

Answer 1

The set of all Equivalence Classes of a set A of Cardinality  C  forms a partition of A.

     option (c) is correct.

Comments

answer is C

u may refer: https://www.quora.com/What-is-an-equivalence-class-of-an-equivalence-relation

Answer 2

Equivalence classes of A form a partition of A. Option C

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Let's get into details:-

For the concept of equivalence classes to even be mentioned, the relation must be equivalence first. An equivalence relation is a relation that is reflexive, symmetric and transitive — all the three.

In contrast, a relation that is reflexive, anti-symmetric and transitive is a partial order.

Let's take a set $A=\left \{ 1,2,3,4 \right \}$

Let's define a relation R on A. $R=\left \{ (1,1),(2,2),(3,3),(4,4),(1,2),(2,1) \right \}$

Clearly, R is equivalence.

To find equivalence classes of each element, we find what element is our element personally related to.

(x,y) means x is related to y NOT y is related to x.

1 is related to 1 and 2. So, $\left \{ 1,2 \right \}$

2 is related to 2 and 1. So, $\left \{ 1,2 \right \}$

3 is related to 3. So, $\left \{ 3 \right \}$

4 is related to 4. $\left \{ 4 \right \}$

5 is related to 5. $\left \{ 5 \right \}$

These are all the equivalence classes, and hence the partitions of A.

 

PS: Remember that union of all equivalence classes is the original set, and intersection of all is $\phi$

Comments

@ JashanArora

Here Cardinalty of A is 6 right ?

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No. Cardinality of A is 4.

Cardinality of R is 6.

Answer 3

None of these. The question is again wrong from ISRO.

An equivalence class is formed by an equivalence relation which groups all elements related to one another and as per definition of equivalence relation this grouping naturally partitions a given set. That is, the set of all equivalence classes of a set A forms a partition of A.

But when we say "of Cadinality C", this becomes wrong. What is C? there can even be no equivalence class of cardinality C where C is any number $\leq |A|$.

Comments

@Arjun. They have mentioned 'C' as cardinality of set A and not equivalence class(This can be guessed even from options). Also,

A={1,2,3} and the equivalence relation is R= { (11) (22) (33) (12) (21) } .

Then equivalence classes are (12) (3). Now, cardinality of (12) =2 < |A| .

Where is the issue?

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yes, I parsed the statement wrong..