prove or disprove using inequality properties of Infimum and supremum that Max (A+B) is less or equals to Max(A) + Max(B)prove or disprove using inequality properties of Infimum and supremum that Max (A+B) is less or equals to Max(A) + Max(B)
The Correct Answer and Explanation is :
The statement we are trying to prove or disprove is:
[
\text{Max}(A + B) \leq \text{Max}(A) + \text{Max}(B)
]
where ( A ) and ( B ) are sets, and ( \text{Max}(A) ) denotes the maximum element of set ( A ), and similarly for ( B ). To analyze this, we will use properties of the supremum and infimum.
1. Definitions
- Supremum (sup): The least upper bound of a set.
- Infimum (inf): The greatest lower bound of a set.
- Max: The maximum value in a set, which is the supremum of the set if the set has a maximum element.
For any two sets ( A ) and ( B ), we can think of their elements as real numbers. Let’s assume that the sets ( A ) and ( B ) have a maximum element, meaning that the supremum of ( A ), denoted ( \sup(A) ), is equal to ( \max(A) ), and similarly for ( B ).
2. Proving the Statement
We need to prove the inequality:
[
\text{Max}(A + B) \leq \text{Max}(A) + \text{Max}(B)
]
Let’s break down the sets and analyze their sum. The sum ( A + B ) represents the set of all possible sums of elements ( a \in A ) and ( b \in B ):
[
A + B = {a + b \mid a \in A, b \in B}
]
Now, for any ( a \in A ) and ( b \in B ), we know that ( a \leq \text{Max}(A) ) and ( b \leq \text{Max}(B) ), because ( \text{Max}(A) ) and ( \text{Max}(B) ) are the largest elements in their respective sets.
Therefore:
[
a + b \leq \text{Max}(A) + \text{Max}(B)
]
Since this holds for every pair of ( a \in A ) and ( b \in B ), we conclude that:
[
\text{Max}(A + B) \leq \text{Max}(A) + \text{Max}(B)
]
3. Conclusion
The statement is true. By the properties of supremum and infimum, we have shown that the maximum value of the sum of two sets is less than or equal to the sum of the maximum values of the individual sets.