# Prove the set $\{x \in \mathbb{R} | f(x) > 0\}$ is open

Let $f: \mathbb{R} \to \mathbb{R}$ be continuous. I need to use the definition to prove that the set where $f(x) > 0$ is open. I think it's true that a set defined by $f(x)$ is closed if it contains its limit points. I'm not sure what do with the proof however.

• not sure how to do that by using the definition. With the sequential criterion, maybe, but not the definition – p3ngu1n Dec 9 '15 at 1:05
• Which definition? There are several definitions of continuity - which one are you using? – Carl Mummert Dec 9 '15 at 1:10
• Sorry! : We say that f is continuous at c if for all $\epsilon$ > 0, there is $\delta (\epsilon)$ > 0 such that if x $\in$ A and |x-c| < $\delta (\epsilon)$, then |f(x) - f(c)| < $\epsilon$. Two types of continuity: 1. c is isolated (not a cluster point) or 2. c is a cluster point – p3ngu1n Dec 9 '15 at 1:15
• Thanks - that will make it possible for people to write answers. – Carl Mummert Dec 9 '15 at 1:19

Approach 1: pick a point $x_0$ in $\{x \mid f(x)>0\}$. By the definition of continuity, some small neighborhood around $x_0$, say, $(x_0-\delta,x_0+\delta)$ is mapped by $f$ to a $(f(x)/2, 3f(x)/2) \subset (0,\infty)$. Thus, $(x_0-\delta,x_0+\delta) \subset \{x \mid f(x)>0\}$.
Approach 2: equivalently we can show the complement is closed. Let $x^* \in \mathbb{R}$ and suppose $(x_n)$ is a sequence in $\{x \mid f(x) \le 0\}$ such that $x_n \to x$. By continuity, $f(x_n) \to f(x^*)$, so $f(x^*) \le 0$ and thus $x^* \in \{x \mid f(x) \le 0\}$, so the set is closed.
Approach 2, without sequential characterization: Let $x^*$ be a limit point of $\{x \mid f(x) \le 0\}$. Suppose for sake of contradiction that $f(x^*) > 0$. Then some neighborhood of $x^*$ maps to $(f(x^*)/2,3f(x^*)/2) \subset (0,\infty)$, which is a contradiction because any neighborhood of $x^*$ contains a point of $\{x \mid f(x) \le 0\}$ which could not possibly map to $(f(x^*)/2,3f(x^*)/2)$.