# Show that the set of points at which a function is bounded is open

Need help: A function f : R → R is said to be bounded at a point $x_0$ provided that there are positive numbers $\epsilon$ and M so that $\mid f(x)\mid$ < M for all x $\in (x_0 − \epsilon, x_0 + \epsilon)$. Show that the set of points at which a function is bounded is open... Here is what I tried.

Proof: Let E be the set of bounded points of the function f : R → R . Let $x_0\in E$. Then by the definition of bounded points, for some positive numbers, $\epsilon$ and M, $\mid f(x)\mid$ < M for all x $\in (x_0 − \epsilon, x_0 + \epsilon)$. Let $y_0$ be an arbitrary point in $(x_0 − \epsilon, x_0 + \epsilon)$. Then there exists c such that the interval $(y_0-c,y_0+c)\in (x_0 − \epsilon, x_0 + \epsilon)$. WLOG, let $y \in (y_0-c,y_0+c)$, then y is also in $(x_0 − \epsilon, x_0 + \epsilon)$. Thus $\mid f(y) \mid$ < M. Therefore, $y_0$ is a bounded point. Therefore $y_0 \in E$. Since our choice of $y_0$ was arbitrary, every point in the interval $(x_0 − \epsilon, x_0 + \epsilon)$ is in E. Thus $(x_0 − \epsilon, x_0 + \epsilon) \subset E.$ Therefore $x_0$ is an interior point so every point of E is an interior point of E. Thus by the definition of open sets, E is open.

QED...?

Almost correct. At the end, what you meant was that $E$ is open, not closed.