Math Problem Statement

Prove the comparison principle. That is, for sequences (an)N,(bn)N with an ≤ bn ∀n, we have

  1. sup(an)N ≤ sup(bn)N

Solution

To prove the comparison principle for sequences (an)n=1N(a_n)_{n=1}^N and (bn)n=1N(b_n)_{n=1}^N such that anbna_n \leq b_n for all nn, we want to show that:

sup(an)n=1Nsup(bn)n=1N.\sup(a_n)_{n=1}^N \leq \sup(b_n)_{n=1}^N.

Definitions:

  • The supremum sup(an)n=1N\sup(a_n)_{n=1}^N is the least upper bound of the set {a1,a2,,aN}\{a_1, a_2, \ldots, a_N\}, i.e., it is the smallest real number α\alpha such that anαa_n \leq \alpha for all nn.
  • Similarly, sup(bn)n=1N\sup(b_n)_{n=1}^N is the least upper bound of the set {b1,b2,,bN}\{b_1, b_2, \ldots, b_N\}, i.e., the smallest real number β\beta such that bnβb_n \leq \beta for all nn.

Step-by-Step Proof:

  1. Given: anbna_n \leq b_n for all n{1,2,,N}n \in \{1, 2, \ldots, N\}.

  2. Let sup(an)=α\sup(a_n) = \alpha and sup(bn)=β\sup(b_n) = \beta. By the definition of supremum, we know that:

    • anαa_n \leq \alpha for all nn, and α\alpha is the smallest number with this property.
    • bnβb_n \leq \beta for all nn, and β\beta is the smallest number with this property.
  3. Upper bound property: Since anbna_n \leq b_n for all nn, and β\beta is an upper bound for the sequence (bn)n=1N(b_n)_{n=1}^N, it follows that β\beta is also an upper bound for the sequence (an)n=1N(a_n)_{n=1}^N, because for each nn, anbnβ.a_n \leq b_n \leq \beta. Therefore, β\beta is an upper bound for the set {a1,a2,,aN}\{a_1, a_2, \ldots, a_N\}.

  4. Smallest upper bound: Since α=sup(an)\alpha = \sup(a_n) is the least upper bound of {a1,a2,,aN}\{a_1, a_2, \ldots, a_N\}, and β\beta is an upper bound for this set, we must have αβ\alpha \leq \beta.

    In other words, sup(an)n=1Nsup(bn)n=1N.\sup(a_n)_{n=1}^N \leq \sup(b_n)_{n=1}^N.

Conclusion:

Thus, the comparison principle holds, and we have proven that if anbna_n \leq b_n for all nn, then sup(an)n=1Nsup(bn)n=1N.\sup(a_n)_{n=1}^N \leq \sup(b_n)_{n=1}^N.

Would you like more details or further clarification?

Here are some related questions to expand on this topic:

  1. How does the infimum behave under similar conditions anbna_n \leq b_n for all nn?
  2. What changes in the comparison principle when the sequences are infinite?
  3. Can you prove the comparison principle for integrals in a similar fashion?
  4. How does the comparison principle apply in the context of limits of sequences?
  5. Can you give an example of two sequences (an)(a_n) and (bn)(b_n) 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