Question

Show that a set can't be a neighborhood of its supremum.

Answer 1

Task 1. Show that a set can't be a neighborhood of its supremum.

Solution. Suppose that a set $X \subset \mathbb{R}$ is a neighborhood of $x \in \mathbb{R}$. By definition this means that there is an open set $G \subset X$, which contains $x$. Therefore, $x \in (x - \varepsilon, x + \varepsilon) \subset G \subset X$ for some $\varepsilon > 0$, where $(x - \varepsilon, x + \varepsilon)$ denotes the open interval with bounds $x - \varepsilon$ and $x + \varepsilon$. Hence, $X$ contains the elements, which are strictly greater, than $x$, for example, the elements from $(x, x + \varepsilon)$. Thus, $x$ cannot be a supremum of $X$.