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Prove: The set $$A=\Big\{f \in C[0,1]: f(x)\neq 0,\forall x \in [0,1]\Big\}$$ is open in $C[0,1]$ with 'sup' metric via open set definition

The author gives the advice " how to choose $r$ with $B(f,r) \subset E$", namely $r=\vert f(0) \vert$

From geometrical point of view I know $B(f,r) \subset E$ but how to prove mathematically?

Ok! Start with , let $g \in B(f,r)$. Then $\text{sup}\;\vert f(x)-g(x) \vert<r$.

How to prove $g(x) \neq 0$ for all $x \in [0,1]$ ?

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2 Answers 2

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Take $f\in A$, you want to find $\epsilon$ such that $B(f,\epsilon)\subset A$. Since $|f|$ is continuous on the compact $[0,1]$ it has a nonzero minimum $m$, because this minimum will be attained and $f$ is never zero. Then if you take $\epsilon=\frac{m}{2}$ and $g\in B(f,m)$, $|g(x)|=|f(x)-(f(x)-g(x))|\geq |f(x)|-|f(x)-g(x)|\geq m-\frac{m}{2}=\frac{m}{2}> 0$, $\forall x\in [0,1]$. Then $g$ is in $A$ too.

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I disagree with the suggestion of using $r = |f(0)|$.

Instead, first realise that $f(x) > 0$ for all $x$ or $f(x) < 0$ for all $x$, by the intermediate value theorem. Without loss of generality, consider $f(x) > 0$ for all $x$ (by replacing $f$ with $-f$ if necessary).

By the extreme value theorem, $f$ must achieve its absolute minimum somewhere on the interval. In particular, there must be some $x_0 \in [0, 1]$ such that $0 < f(x_0) \le f(x)$ for all $x \in [0, 1]$. I suggest letting $r = f(x_0)$.

Then, if $g \in B(f; r)$, then $$f(x_0) > \sup_{x \in [0, 1]} |f(x) - g(x)|,$$ and hence using the definitions of $x_0$ and the supremum, for all $x \in [0, 1]$, $$f(x) > |f(x) - g(x)| \ge f(x) - g(x) \implies g(x) > 0.$$

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  • $\begingroup$ I understand your answer. But why you disagree with $r=\vert f(0) \vert$ ? Is this $r$ does't work ? $\endgroup$
    – user444830
    Oct 2, 2018 at 3:32
  • $\begingroup$ Ok! I'm waiting!. $\endgroup$
    – user444830
    Oct 2, 2018 at 3:49
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    $\begingroup$ Got it! That $r$ does not work because of the following: take $f(x)=-x+1.1$. Then $r=f(0)=1.1$. The function $g(x)=-x+0.11$ is in $B(f,r)$, however it is not in $A$. $\endgroup$
    – Laz
    Oct 2, 2018 at 3:55
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    $\begingroup$ If $f\colon x\mapsto(2-x)$, then $|f(0)|=2$ but $B(f,|f(0)|)$ is not contained in $A$ since $g\colon x\mapsto(1-x)$ belongs to $B(f,|f(0)|)$. $\endgroup$ Oct 2, 2018 at 3:57
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    $\begingroup$ You are welcome @TheoBendit. $\endgroup$
    – Laz
    Oct 2, 2018 at 3:57

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