2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ not sure how to do that by using the definition. With the sequential criterion, maybe, but not the definition $\endgroup$ – p3ngu1n Dec 9 '15 at 1:05
  • 1
    $\begingroup$ Which definition? There are several definitions of continuity - which one are you using? $\endgroup$ – Carl Mummert Dec 9 '15 at 1:10
  • $\begingroup$ 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 $\endgroup$ – p3ngu1n Dec 9 '15 at 1:15
  • 1
    $\begingroup$ Thanks - that will make it possible for people to write answers. $\endgroup$ – Carl Mummert Dec 9 '15 at 1:19
2
$\begingroup$

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 $(f(x^*)/2,3f(x^*)/2)$.

$\endgroup$
  • $\begingroup$ How would I use approach 2 without using sequential characterization? $\endgroup$ – p3ngu1n Dec 9 '15 at 1:17
  • 1
    $\begingroup$ @p3ngu1n See my edit. It ends up being much like approach 1. $\endgroup$ – angryavian Dec 9 '15 at 1:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.