Proving the Boundedness of Finite Sequences of Real Numbers in Real Analysis

Introduction to Boundedness and Order Relations

In the realm of real analysis, a fundamental concept is that of a bounded set. A set ( S ) of real numbers is said to be bounded above if there exists a real number ( M ) such that ( x leq M ) for every ( x in S ). Similarly, ( S ) is bounded below if there exists a real number ( m ) such that ( x geq m ) for every ( x in S ). When a set is both bounded above and below, it is called bounded. This concept is crucial in real analysis, particularly when dealing with sequences and series.

Boundedness of Finite Sequences

In this article, we will delve into proving a specific claim: any finite sequence of real numbers has a maximum element and a minimum element. We will achieve this by leveraging the comparability of the order relation on the set of real numbers, ( mathbb{R} ), and mathematical induction.

Claim: A Finite Sequence of Real Numbers has a Maximum and Minimum

The claim is as follows: A finite sequence of real numbers, say ( x_1, x_2, ldots, x_n ), has a maximum element and a minimum element. To prove this, it suffices to demonstrate that, for a sequence of distinct real numbers, the maximum and minimum elements can be found.

Let ( x_1, x_2, ldots, x_n ) be a finite sequence of distinct real numbers. We will prove this by induction on the length of the sequence.

Base Case: ( n 2 )

For a sequence of two elements ( x_1 ) and ( x_2 ), the comparability of the order relation on ( mathbb{R} ) implies that either ( x_1 leq x_2 ) or ( x_2 leq x_1 ). Without loss of generality, assume ( x_1 leq x_2 ). In this case, we have ( x_1 min{x_1, x_2} ) and ( x_2 max{x_1, x_2} ).

Inductive Step

Assume the claim holds for a sequence of ( k ) elements, i.e., ( x_1, x_2, ldots, x_k ) has a minimum ( x_1 ) and a maximum ( x_k ). We need to show that a sequence of ( k 1 ) elements, ( x_1, x_2, ldots, x_k, x_{k 1} ), also has a minimum and a maximum.

Consider the set ( {x_1, x_2, ldots, x_k, x_{k 1}} ). By the inductive hypothesis, the set ( {x_1, x_2, ldots, x_k} ) has a minimum ( x_1 ) and a maximum ( x_k ). Now, consider the pair ( x_k ) and ( x_{k 1} ). By the comparability of the order relation on ( mathbb{R} ), either ( x_k leq x_{k 1} ) or ( x_{k 1} leq x_k ).

Case 1: ( x_k leq x_{k 1} )

Then, from the inductive hypothesis, ( x_1 ) is the minimum of ( {x_1, x_2, ldots, x_k, x_{k 1}} ). Additionally, since ( x_k leq x_{k 1} ), ( x_k ) is the maximum of this set.

Case 2: ( x_{k 1} leq x_k )

Then, from the inductive hypothesis, ( x_1 ) is the minimum of ( {x_1, x_2, ldots, x_k, x_{k 1}} ). Additionally, since ( x_{k 1} leq x_k ), ( x_{k 1} ) is the minimum of this set (which is already ( x_1 )) and ( x_k ) remains the maximum of this set.

Hence, the inductive step is complete. By the principle of mathematical induction, the claim holds for all finite sequences of distinct real numbers.

Boundedness of the Union of Bounded Sets

Another important concept in real analysis is the boundedness of the union of bounded sets. We will first prove a lemma stating that the union of two bounded sets is bounded, and then extend this result to a finite number of sets using induction.

Lemma: Union of Two Bounded Sets is Bounded

Let ( A ) and ( B ) be two bounded sets of real numbers. There exist real numbers ( M_A ) and ( M_B ) such that ( x leq M_A ) for all ( x in A ) and ( y leq M_B ) for all ( y in B ). Let ( M max{M_A, M_B} ). Then, for all ( z in A cup B ), we have ( z leq M ). Hence, ( A cup B ) is bounded above by ( M ).

Boundedness of the Union of a Finite Number of Bounded Sets

By induction, we can extend this result to any finite number of bounded sets. Suppose ( A_1, A_2, ldots, A_n ) are bounded sets. We can prove this by induction on ( n ).

Base Case: ( n 1 )

If ( n 1 ), the set ( A_1 ) is bounded, and we are done.

Inductive Step

Assume that the union of bounded sets ( A_1, A_2, ldots, A_k ) is bounded. Let ( M max{max{A_1}, max{A_2}, ldots, max{A_k}} ) and ( m min{min{A_1}, min{A_2}, ldots, min{A_k}} ). Then, for all ( x in A_1 cup A_2 cup cdots cup A_k ), we have ( x leq M ) and ( x geq m ). Therefore, the union ( A_1 cup A_2 cup cdots cup A_k cup A_{k 1} ) is also bounded.

Hence, by the principle of mathematical induction, the union of a finite number of bounded sets is bounded.

Conclusion

We have proven that any finite sequence of real numbers has a maximum and a minimum element, and we have also demonstrated that the union of a finite number of bounded sets is itself bounded. These results are fundamental in real analysis, providing a solid foundation for further study in sequences, series, and set theory.