Let $E$ be a bounded subset of $\textbf{R}$, and let $s = \sup(E)$. Show $s$ is an adherent point of $E$, and is also an adherent point of $E^{c}$.

Question

Let $E$ be a bounded subset of $\textbf{R}$, and let $s = \sup(E)$ be the least upper bound of $E$. Show that $s$ is an adherent point of $E$, and is also an adherent point of $\textbf{R}\cap E^{c}$.

My solution

Since $E$ is bounded, we conclude that $E\subseteq[-M,M]$.

Due to the least upper bound property, $E$ admits a supremum $s = \sup(E)$.

We have to prove that for every $\varepsilon > 0$, ther exists a $y\in E$ such that $|s - y|\leq \varepsilon$.

Indeed, according to the properties of supremum, for every $\varepsilon > 0$, there is a real number $y\in E$ such that $s - \varepsilon < y \leq s < s + \varepsilon$.

That is to say, for every $\varepsilon > 0$ there is a real number $y\in E$ such that $|s-y|\leq\varepsilon$, and we are done with the first part.

As to the second part, I do not know how to proceed.

Could someone help me with this?

Answer 1

For every $n >0$, $s + 1/n$ is not in $E$ because $s$ is the supremum of all elements of $E$. So $s + 1/n \in E^c$.

Answer 2

You want to show that for every $\epsilon > 0 $ there exists a real number $x \in E^c$ such that $|x-s| < \epsilon$.

Assume by contradiction that this is not the case so there exists an $\epsilon > 0$ such that $(s-\epsilon, s + \epsilon) \cap E^c = \emptyset$ or equivalently $(s-\epsilon, s + \epsilon) \subset E$. Then $s + \frac{\epsilon}{2} \in E$, which contradicts that $s$ is the supremum.