Beautiful Examples on Set , Relation and Functions

1. If A ={1,2,3,45,}, then number of Equivalence relation is ______

2. If a set S has {\color{Red} n} elements then the minimum and maximum cardinality of an equivalence relation is ________

3. If a set S has {\color{Red} n} elements then the total number of relations that are both reflexive and Symmetric is ________

4. An element of the set can be a subset of that set. True or False?  ______

5. if a set S has {\color{Red} n} elements, The total number of relations that are Antisymmetric and  Asymmetric is _____





Solution 1

Number of equivalence relations can be calculated using Bell Number: Please refer to the following link


Solution 2:

Set S has n element So: for equivalence relation, the reflexive property must be satisfied (we need to satisfied reflexive, symmetric and transitive property, I am talking in reference with the question )and we have already discussed if a set has n elements then there are n reflexive pair. 

So for a minimum cardinality at least n elements must be there 

     for maximum cardinality, all elements should be present: 2


Solution 3:-

If a set  S has 'n' elements then the total number of elements in S* S  (Available to participate in relation )is  :  n^{2} 

out of n^{2}  , there are  n diagonal elements which must be present if a relation has to reflexive. So the diagonal element has only one choice. 

Now remaining( n^{2}-n) element will form (n^{2}-n)/2  pair and each pair have two choices: Either they can participate in the realation or not. 

So the total number of relations that are Both Symmetric and Reflexive is   {\color{Red} 2^{(n^{2}-n)/2}}

Please refer to these links:


Solutions 4:-

False, An element of a set can't be the subset of that element : 

A subset is set itself but an element is not set. Example:


The question is asking whether 1 is a subset of A. No because 1 is just element but {1} is a set and hence subset of A.

Please refer to this example for more clarification.


Solution 5:

Asymmetric relations are stricter than Antisymmetric. Think Why??

Because in Antisymmetric we are allowing diagonal element (elements of type:  (1,1) , (2,2,).....(n,n)) but in Asymmetric, even diagonal elements are not allowed.

So diagonal element has only one choice they must not present.  

Remaining {\color{Red} n^{2}-n} form   ({\color{Red} n^{2}-n} ){\color{Red} /2} pair and each pair have three choices:

choice 1:  pair is not present in the relation.

choice 2:  if (a,b) is present then (b,a) is absent.

choice 3:  if (b,a) is present then (a,b) is absent 

So the total  number of relations that are both Asymmetric and  Antisymmetric is : 3^{({\color{Red} n^{2}-n} ){\color{Red} /2}} 


If a relation is Asymmetric it is definitely Antisymmnetic but the reverse is not true. You tell me the reason in Comment Box.