# Continuous Functions

• Feb 26th 2010, 08:11 PM
CrazyCat87
Continuous Functions
Let $\displaystyle f:\mathbb{R} \rightarrow \mathbb{R}$ be continuous at $\displaystyle c$ and let $\displaystyle f(c)>0$. How would you show that there is a neighborhood $\displaystyle V_{\delta}(c)$ of $\displaystyle c$ such that if $\displaystyle x \in V_{\delta}(c)$, then $\displaystyle f(x) > 0$
• Feb 26th 2010, 08:26 PM
southprkfan1
Forgive my lack of syntax here

if f is continuous at c, then for e> 0 there exists a d>0 st

0< l x - c l < d --> l f(x) - f(c) l < e

but l f(x) - f(c) l < e is the same thing as:

f(c) - e < f(x) < f(c) + e

so as long as 0 < e < f(c) we have f(x) > 0 for all 0 < l x - c l < d

and the conclusion follows
• Feb 26th 2010, 08:52 PM
Drexel28
Quote:

Originally Posted by CrazyCat87
Let $\displaystyle f:\mathbb{R} \rightarrow \mathbb{R}$ be continuous at $\displaystyle c$ and let $\displaystyle f(c)>0$. How would you show that there is a neighborhood $\displaystyle V_{\delta}(c)$ of $\displaystyle c$ such that if $\displaystyle x \in V_{\delta}(c)$, then $\displaystyle f(x) > 0$

The exact same concept as southprkfan1 but using open ball notation. Since $\displaystyle \{x_:x>0\}=(0,\infty)$ is an open set we know there exists some $\displaystyle B_{\varepsilon}(f(x))\susbseteq(0,\infty)$ but by continuity there exists some $\displaystyle B_{\delta}(x)$ such that $\displaystyle f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(x)\subseteq(0,\infty)$. The conclusion follows.