Question

For any set A, prove that,
(i) $A\cup A=A$
(ii) $A\cap A=A$

Answer 1

Hint:First of all take an element x belonging to $A\cup A$. Now from this, we get, $x\in A$, that means A belongs to $A\cup A$. Now, take the converse of it and prove that $A\cup A$ belongs to A. From these two results, prove that $A\cup A=A$. Similarly, take an element $x\in A\cap A$ and follow similar steps to prove the next result.

Complete step-by-step answer:
For any set A, we have to prove that,
(i) $A\cup A=A$
(ii) $A\cap A=A$
Let us prove that $A\cup A=A$. Let x be an element in the set $A\cup A$. So, we can write $x\in A\cup A$. We know that the set $A\cup B$ constitutes elements that are in set A or set B or both. So, if $x\in A\cup A$, then we get, $x\in A$ or $x\in A$. Hence we get, $x\in A$.
From this, we can say that A also belongs to $A\cup A$ or $A\subseteq A\cup A....\left( i \right)$
Conversely, if we take an element $x\in A$, we can also write $x\in A$ or $x\in A$. So, we get, $x\in A\cup A$. From this, we can say that $A\cup A$ also belongs to A or $A\cup A\subseteq A....\left( ii \right)$
We know that if any set x belongs to set y and set y belongs to set x, then set x and set y are equal. So from equation (i) and (ii), we get,
$A\cup A=A$
Hence proved.
Let us prove that $A\cap A=A$. Let x be an element in the set $A\cap A$. So, we can write $x\in A\cap A$. We know that the set $A\cap B$ constitutes elements that are in set A as well as in set B. So, if $x\in A\cap A$, then we get, $x\in A$ and $x\in A$. Hence we get, $x\in A$.
For this, we can say that A also belongs to $A\cap A$ or $A\subseteq A\cap A....\left( iii \right)$
Conversely, if we take an element $x\in A$, we can also write $x\in A$ and $x\in A$. So, we get, $x\in A\cap A$. From this, we can say that $A\cap A$ also belongs to A or $A\cap A\subseteq A....\left( iv \right)$
We know that if any set x belongs to set y and set y belongs to set x, then set x and set y are equal. So from equation (iii) and (iv), we get,
$A\cap A=A$
Hence proved.

Note: In this question, we can also verify the results by taking an example of set A as follows. Let set A be {1, 3, 5, 8, 9}. Now, if we find $A\cup A$ that is a set that contains the elements of both A or A. So, we get, A\cup A=\left\\{ 1,3,5,8,9 \right\\} that is equal to A.
Similarly, if we need to find $A\cap A$, that is a set that contains elements that are in A and A both. So, we get, A\cap A=\left\\{ 1,3,5,8,9 \right\\} that is equal to A.