Give an example of a bounded sequence in R that has no supremum, and prove that it has no supremum.

To provide an example of a bounded sequence in $\mathbb{R}$ that has no supremum, we must first understand the definitions involved. A sequence is bounded if there exists a real number $M$ such that all elements of the sequence are less than or equal to $M$. The supremum (or least upper bound) of a sequence is the smallest real number that is greater than or equal to every element of the sequence.

Example of a Bounded Sequence with No Supremum

Consider the sequence $a_n = (-1)^n + 1$ for $n = 1, 2, 3, \ldots$. This sequence alternates between 0 and 2:

• $a_1 = 0$
• $a_2 = 2$
• $a_3 = 0$
• $a_4 = 2$
• ...

Proof that the Sequence Has No Supremum

1. Boundedness:

   1. The sequence is bounded because all terms are either 0 or 2. Thus, we can choose $M = 2$ as an upper bound.

2. Non-existence of Supremum:

   1. Assume, for contradiction, that the sequence has a supremum $S$.
   2. Since the sequence contains the number 2, $S$ must be at least 2.
   3. If $S = 2$, then 2 should be an upper bound for the sequence. However, 2 is also a term in the sequence, and for a supremum, there should be no greater number in the sequence. But since 2 is part of the sequence, it cannot be strictly greater than all other terms.
   4. If $S > 2$, then $S$ is not the least upper bound because 2 is already an upper bound, contradicting the definition of supremum.

Conclusion

The sequence $a_n = (-1)^n + 1$ is bounded but does not have a supremum because any candidate for the supremum either equals an element of the sequence or is not the least upper bound. Thus, this sequence serves as an example of a bounded sequence in $\mathbb{R}$ without a supremum.