Proving Continuity at a Point with Non-Zero Value: A Case Study

[FallArk]

Problem: Let $f$ be defined on an open interval $I$, and $c \in I$. Prove that, if $f$ is continuous at $c$ and $f(c) ≠ 0$, then there is an open interval $J \subset I$ such that $c \in J$ and $f(x) ≠ 0$, for any $x \in J$.

I was thinking another function $g$ defined on the subset $J$ and $g = f$, then I can find a way to verify that $g$ is also continuous at $c$. But I am currently stuck on how to prove that.

 

[I like Serena]

Problem: Let f be defined on an open interval I, and c ∈ I. Prove that, if f is continuous at c and f(c) ≠ 0, then there is an open interval J ∈ (is a subset, I can't find the symbol) I such that c ∈ J and f(x) ≠ 0, for any x ∈ J.
I was thinking another function g defined on the subset J and g = f, then I can find a way to verify that g is also continuous at c. But I am currently stuck on how to prove that.

Hi FallArk! ;)

Let's take the definition of continuity into account:
$$\lim_{x\to c} f(x) = f(c) \ne 0$$
This means that for any $\epsilon>0$ there exists a $\delta >0$ such that for any $x$ with $|x-c|<\delta$ we have that $|f(x)-f(c)|<\epsilon$.
Let's pick $\epsilon = \frac 12 f(c)$, then $(c-\delta, c+\delta)$ satisfies the requirements for the interval $J$.

That is because any $x$ in that interval has an $f(x)$ that deviates at most $\frac 12 f(c)$ from $f(c)\ne 0$.
Therefore $f(x)\ne 0$ for any such $x$.

 

[Euge]

Let's pick $\epsilon = \frac 12 f(c)$, then $(c-\delta, c+\delta)$ satisfies the requirements for the interval $J$.

That is because any $x$ in that interval has an $f(x)$ that deviates at most $\frac 12 f(c)$ from $f(c)\ne 0$.
Therefore $f(x)\ne 0$ for any such $x$.

It's a great idea, but for this argument to work, $f(c)$ needs to be positive. It's possible to avoid cases by setting $\epsilon = \frac12\lvert f(c)\rvert$, and establish that $\frac12 \lvert f(c)\rvert < \lvert f(x)\rvert < \frac32\lvert f(c)\rvert$ in a subinterval of $I$. In particular, the inequalities imply $f(x)$ is nonzero for all $x$ in that subinterval.

 

[FallArk]

Thank you so much! I knew there was something that the book did not mention, and there it is!

- - - Updated - - -

It's a great idea, but for this argument to work, $f(c)$ needs to be positive. It's possible to avoid cases by setting $\epsilon = \frac12\lvert f(c)\rvert$, and establish that $\frac12 \lvert f(c)\rvert < \lvert f(x)\rvert < \frac32\lvert f(c)\rvert$ in a subinterval of $I$. In particular, the inequalities imply $f(x)$ is nonzero for all $x$ in that subinterval.

so two cases? f(c)>0 and f(c)<0?

 

[Euge]

so two cases? f(c)>0 and f(c)<0?

Yes.