If $ f $ is a scalar field and $ f(a) \neq 0 $, then there is an $ n $-ball in which $ f $ has the same sign as $ f(a) $.

Question

I’ve tried to prove the following theorem:

Let $ f $ be a scalar field that is continuous at an interior point $ a $ of $ S \subseteq \mathbb{R}^{n} $. If $ f(a) \neq 0 $, then there is an $ n $-ball $ \mathbb{B}(a) $ in which $ f $ has the same sign as $ f(a) $.

However, I’m not sure if my proof is correct, and honestly, I didn’t like the way that I proceeded, so I’m look for alternative proofs/proof methods for such theorems. Moreover, I’m looking for advice on the way that I wrote my proof.

Proof.

We will prove that $$ (\exists \delta > 0)[(0 < |x - a| < \delta) \to (|f(a) - f(x)| < |f(a)|)]. $$ As $$ (\forall \epsilon > 0)(\exists \delta_{1} > 0) [(0 < |x - a|< \delta_{1}) \to (|f(a) - f(x)| < \epsilon)], $$ if we pick $ \epsilon $ so that $ 0 < \epsilon < |f(a)| $, and $ 0 < \delta \leq \delta_{1} $, then the inequality becomes $$ (\exists \delta > 0)[(0 < |x - a| < \delta) \to (|f(a) - f(x)| < \epsilon)]. $$ The theorem is therefore proved.

Answer 1

Suppose $f(a)>0$ Then choose $\epsilon=\frac{f(a)}{2}$

Then $\frac{f(a)}{2}<f(x)<\frac{3f(a)}{2}$ for all $x\in B(a,\delta)$