Let S = {x ∈ R : x ^ 3 < 5}. Prove that S is bounded above and find its supremum.

To prove that the set $S = \{ x \in \mathbb{R} : x^3 < 5 \}$ is bounded above and to find its supremum, we will follow these steps:

1. Break Down the Problem

   We need to show that there exists an upper bound for the set $S$ and then determine the least upper bound, known as the supremum.

2. Relevant Concepts

   • Upper Bound: A number $M$ is an upper bound for a set $S$ if for all $x \in S$, $x \leq M$.
   • Supremum: The supremum of a set $S$ is the smallest upper bound of $S$.

3. Analysis and Detail

   1. Finding an Upper Bound:

      • Consider the inequality $x^3 < 5$.
      • We need to find a number $M$ such that for all $x$ satisfying $x^3 < 5$, $x \leq M$.
      • If $x^3 < 5$, then $x < \sqrt[3]{5}$.
      • Therefore, $\sqrt[3]{5}$ is an upper bound for $S$.

   2. Proving Supremum:

      • We need to show that $\sqrt[3]{5}$ is the least upper bound.
      • Assume there exists a smaller upper bound $M < \sqrt[3]{5}$.
      • Then there exists some $x$ such that $M < x < \sqrt[3]{5}$ and $x^3 < 5$.
      • This contradicts the assumption that $M$ is an upper bound, as $x$ is in $S$ but $x > M$.
      • Hence, no such $M$ can exist, and $\sqrt[3]{5}$ is the supremum.

4. Verify and Summarize

   • We verified that $\sqrt[3]{5}$ is an upper bound and no smaller upper bound exists.
   • Therefore, the supremum of $S$ is $\sqrt[3]{5}$.

Final Answer

The set $S = \{ x \in \mathbb{R} : x^3 < 5 \}$ is bounded above, and its supremum is $\sqrt[3]{5}$.