Exercise 3.3.1

Question

Show that if $K$ is compact and nonempty, then sup $K$ and $\inf K$ both exist and are elements of $K$.

Ulisse Mini · Accepted Answer

Let $s = \sup K$, since $s$ is the least upper bound for every $\epsilon > 0$ there exists an $x \in K$ with $s - \epsilon < x$. Picking $\epsilon_n = 1/n$ and $x_n$ such that $s - \epsilon_n < x_n$ we get that $(x_n) 	o s$ since $(\epsilon_n) 	o 0$, and thus $s \in K$.

A similar argument applies to $\inf K$.

Exercise 3.3.1

Answers

Comments

Exercise 3.3.1

Answers

Comments

Add answer