# Proof

• Let $$D$$ be a subset of real numbers $$\mathbb R$$.
• Let $$a\in D$$.
• Let $$f:D\to\mathbb R$$ be continuous at $a$.
• Let $f(a)\neq 0.$
• We want to show that there exists a real number $\delta > 0$ such that $f(x)\neq 0$ for all $x\in D$ with $|x-a| < \delta.$
• Set $\epsilon=|f(a)| > 0,$ where $|f(a)|$ denotes the absolute value.
• According to the $\epsilon-\delta$-definition of continuity at $a$, there exists a $\delta > 0,$, such that $|f(x)-f(a)| < \epsilon$ for all $x\in D$ with $|x-a| < \delta.$
• From $|f(x)-f(a)| < \epsilon=|f(a)|,$ and the rules of calculation for inequalities it follows that $|f(a)|-|f(x)-f(a)| > 0$ for all $x\in D$ with $|x-a| < \delta.$
• Thus, $|f(a)|\ge |f(a)|-|f(x)-f(a)| > 0$ for all $x\in D$ with $|x-a| < \delta.$
• Thus, $|f(a)| > 0$ for all $x\in D$ with $|x-a| < \delta.$
• Thus, $f(a)\neq 0$ for all $x\in D$ with $|x-a| < \delta.$

Github: ### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983