# Property on $n$-ball

I am having trouble with the following exercise.

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

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Use the $\epsilon-\delta$ definition of continuity.

Choose some $\epsilon$ such that all $y$ satisfying $||f(a)- y|| \lt \epsilon$ are the same sign as $f(a)$.

Then there is some $\delta$ such that all $x$ satisfying $||x-a|| \lt \delta$ have $||f(a) - f(x)|| \lt \epsilon$ and so have the same sign as $f(a)$.

Also, since $a$ is an interior point of $S$, there is some $r$ such that $B(a;r) \subset S$.

Let $m = \min(\delta; r)$, then $B(a; m)$ is an $n$-ball $\in S$ in which $f$ has the same sign as $f(a)$.

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Nice approach Amzoti + –  B. S. Mar 16 '13 at 17:15
@BabakS.: Thank you - have a great day! –  Amzoti Mar 16 '13 at 17:26
+1 and another $\color{green}{\bf \large \checkmark}$ –  amWhy Apr 22 '13 at 0:23
Hint: If $f(a)$ is not zero, this means that $f(a)\in (-\infty,0)$ or $(0,\infty)$. By continuity, you can choose a small enough neighborhood of $a$ which maps into either of these open sets.