# $A=\Big\{f \in C[0,1]: f(x)\neq 0,\forall x \in [0,1]\Big\}$ is open in $C[0,1]$

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.

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

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.

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 $$x \in [0, 1]$$, $$f(x) > |f(x) - g(x)| \ge f(x) - g(x) \implies g(x) > 0.$$

• I understand your answer. But why you disagree with $r=\vert f(0) \vert$ ? Is this $r$ does't work ?
– user444830
Oct 2, 2018 at 3:32
• Ok! I'm waiting!.
– user444830
Oct 2, 2018 at 3:49
• 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 $A$.
– Laz
Oct 2, 2018 at 3:55
• 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 $B(f,|f(0)|)$. Oct 2, 2018 at 3:57
• You are welcome @TheoBendit.
– Laz
Oct 2, 2018 at 3:57