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

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

\[
\sup(a_n)_{n=1}^N \leq \sup(b_n)_{n=1}^N.
\]

### Definitions:
- The **supremum** \( \sup(a_n)_{n=1}^N \) is the least upper bound of the set \( \{a_1, a_2, \ldots, a_N\} \), i.e., it is the smallest real number \( \alpha \) such that \( a_n \leq \alpha \) for all \( n \).
- Similarly, \( \sup(b_n)_{n=1}^N \) is the least upper bound of the set \( \{b_1, b_2, \ldots, b_N\} \), i.e., the smallest real number \( \beta \) such that \( b_n \leq \beta \) for all \( n \).

### Step-by-Step Proof:

1. **Given**: \( a_n \leq b_n \) for all \( n \in \{1, 2, \ldots, N\} \).

2. Let \( \sup(a_n) = \alpha \) and \( \sup(b_n) = \beta \). By the definition of supremum, we know that:
   - \( a_n \leq \alpha \) for all \( n \), and \( \alpha \) is the smallest number with this property.
   - \( b_n \leq \beta \) for all \( n \), and \( \beta \) is the smallest number with this property.

3. **Upper bound property**: Since \( a_n \leq b_n \) for all \( n \), and \( \beta \) is an upper bound for the sequence \( (b_n)_{n=1}^N \), it follows that \( \beta \) is also an upper bound for the sequence \( (a_n)_{n=1}^N \), because for each \( n \),
   \[
   a_n \leq b_n \leq \beta.
   \]
   Therefore, \( \beta \) is an upper bound for the set \( \{a_1, a_2, \ldots, a_N\} \).

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

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

### Conclusion:
Thus, the comparison principle holds, and we have proven that if \( a_n \leq b_n \) for all \( n \), then
\[
\sup(a_n)_{n=1}^N \leq \sup(b_n)_{n=1}^N.
\]

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 \( a_n \leq b_n \) for all \( n \)?
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 \( (a_n) \) and \( (b_n) \) to illustrate this principle?

**Tip**: When dealing with supremum or infimum, always consider whether the sequence is bounded and whether the bounds are attained.

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

Explore the proof of the comparison principle for sequences where a_n ≤ b_n for all n. This problem covers key real analysis concepts, including the supremum of sequences and inequality relations. Suitable for undergraduate students studying mathematical analysis.

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