I'm stuck on this question, not really sure how to go about it, so it'd be good if you could do it but explain what you're doing, etc. Thanks. Any help would greatly be appreciated! 🙂
Question
Suppose $f:X\rightarrow\mathbb{R}$ is continuous at some $c\in X$. Suppose also that $f(c)>0$. Prove that there exists $\delta>0$ such that $f(x)>0$ for all $x\in(c-\delta, c+\delta)$.
Working
I haven't done much, as I said I'm not really 100% how to do it, but here is my current working (sorry it isn't a lot) 🙂
Fix $\epsilon>0$ then $\exists\delta>0 : |x-c|<\delta \implies |f(x)-f(c)|<\epsilon$
Thanks again!
Best Answer
With $\epsilon=\frac{f(c)}{2}$ in the $\epsilon$-$\delta$ definition of continuity, we get the following.
We can find a $\delta>0$ such that if $x\in X$ and $|x-c|<\delta$ then $$|f(x)-f(c)|<\frac{f(c)}{2}.$$ We have $$|x-c|<\delta\iff x\in(c-\delta,c+\delta)$$ and $$|f(x)-f(c)|<\frac{f(c)}{2}\iff \overbrace{-\frac{f(c)}{2}<f(x)-f(c)}^{*}<\frac{f(c)}{2}.$$ Thus, for all $x\in X$ such that $x\in(c-\delta,c+\delta)$, we get (by using $*$) $$f(x)>f(c)-\frac{f(c)}{2}=\frac{f(c)}{2}>0.$$