Math Problem Statement
Prove the comparison principle. That is, for sequences (an)N,(bn)N with an ≤ bn ∀n, we have
- sup(an)N ≤ sup(bn)N
Solution
To prove the comparison principle for sequences and such that for all , we want to show that:
Definitions:
- The supremum is the least upper bound of the set , i.e., it is the smallest real number such that for all .
- Similarly, is the least upper bound of the set , i.e., the smallest real number such that for all .
Step-by-Step Proof:
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Given: for all .
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Let and . By the definition of supremum, we know that:
- for all , and is the smallest number with this property.
- for all , and is the smallest number with this property.
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Upper bound property: Since for all , and is an upper bound for the sequence , it follows that is also an upper bound for the sequence , because for each , Therefore, is an upper bound for the set .
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Smallest upper bound: Since is the least upper bound of , and is an upper bound for this set, we must have .
In other words,
Conclusion:
Thus, the comparison principle holds, and we have proven that if for all , then
Would you like more details or further clarification?
Here are some related questions to expand on this topic:
- How does the infimum behave under similar conditions for all ?
- What changes in the comparison principle when the sequences are infinite?
- Can you prove the comparison principle for integrals in a similar fashion?
- How does the comparison principle apply in the context of limits of sequences?
- Can you give an example of two sequences and to illustrate this principle?
Tip: When dealing with supremum or infimum, always consider whether the sequence is bounded and whether the bounds are attained.
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Math Problem Analysis
Mathematical Concepts
Real Analysis
Sequences
Supremum
Inequalities
Formulas
sup(a_n) ≤ sup(b_n)
Theorems
Comparison Principle
Supremum and Infimum Definitions
Suitable Grade Level
Undergraduate Mathematics