Question

Let $A$ be a nonempty subset of $\mathbf{R}$ that is bounded above. Show that there is a sequence $\left(x_n\right)$ of elements of $A$ that converges to $\sup A$.

   Let $A$ be a nonempty subset of $\mathbf{R}$ that is bounded above. Show that there is a sequence $\left(x_n\right)$ of elements of $A$ that converges to $\sup A$.

Real analysis

N. L. Carothers 1st Edition

Chapter 1, Problem 4

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- A set $A \subseteq \mathbf{R}$ is bounded above if there exists some real number $M$ such that $a \leq M$ for all $a \in A$. - The supremum (or least upper bound) of $A$, denoted $\sup A$, is the smallest real number that is an upper bound of $A$. - We need to Show more…

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Let $A$ be a nonempty subset of $\mathbf{R}$ that is bounded above. Show that there is a sequence $\left(x_n\right)$ of elements of $A$ that converges to $\sup A$.

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