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Proof

(related to Lemma: Functions Continuous at a Point and Non-Zero at this Point are Non-Zero in a Neighborhood of This Point)

• 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 $|x-a| < \delta.$
    ∎

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References

Bibliography

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