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.