Prove open set in C[0,1] function space

1. The problem statement, all variables and given/known data

Let [tex]X = C[0,1][/tex] under the metric [tex]d_{\infty}[/tex]. Let
[tex]
\begin{align*}
Y = \{ f \in C[0,1] : f(0) \ne 0 \}
\end{align*}
[/tex]

Prove that [tex]Y[/tex] is open in [tex](C[0,1], d_{\infty})[/tex].

2. Relevant equations


The formulae for [tex]d_{\infty}[/tex] is
[tex]
\begin{align*}
d_{\infty}(f,g) = \max_{0 \leq t \leq 1} |f(t) - g(t)|
\end{align*}
[/tex]

3. The attempt at a solution

For any [tex]f \in Y[/tex], which means [tex]f \in C[0,1] \wedge f(0) \ne 0[/tex], if
[tex]f[/tex] is an interior point of [tex]Y[/tex], then [tex]Y[/tex] is open in [tex](C[0,1],
d_{\infty})[/tex]. [tex]f[/tex] is an interior point if there is an [tex]\varepsilon
> 0[/tex] such that the neighborhood [tex]O_{\varepsilon}(f) \subseteq Y[/tex], meaning all [tex]g \in C[0,1][/tex] satisfying [tex]\displaystyle d_{\infty}(f,g)
= \max_{0 \leq t \leq 1} f(t) - g(t)| < \varepsilon[/tex] must
belong to [tex]Y[/tex], which says [tex]g(0) \ne 0[/tex]. Suppose to the contrary there
exists a [tex]g \in C[0,1][/tex] such that [tex]\displaystyle d_{\infty}(f,g) =
\max_{0 \leq t \leq 1} |f(t) - g(t)| < \varepsilon[/tex] and [tex]g(0) =
0[/tex]. Then [tex]|f(0) - g(0)| = |f(0)|
> 0[/tex] since [tex]f(0) \ne 0[/tex] and [tex]\displaystyle d_{\infty}(f,g) = \max_{0
\leq t \leq 1} f(t) - g(t)| \geq |f(0)|[/tex]. For [tex]\varepsilon
\leq |f(0)|[/tex] this results in a contradiction with the assumption
that [tex]d_{\infty}(f,g) < \varepsilon[/tex]. What about [tex]\varepsilon > |f(0)|[/tex]? Is this on the right track at all?

 

Thank you to both of you. But I still don't know how to finish the proof.

If the argument is correct. How can I show a contradiction for the case [tex]\varepsilon > |f(0)|[/tex] ?

 

The restriction should be that [tex]f(0)[/tex] cannot be wiggled as much as
[tex]|f(0)|[/tex]. But this restriction is what the original set builder for [tex]Y[/tex] specified. How does this directly prove that every [tex]g[/tex] inside some [tex]\varepsilon > 0[/tex] neighborhood of [tex]y[/tex] is inside [tex]Y[/tex]? I must be missing some piece of the logic.

 

Aghh, so [tex]\varepsilon[/tex] has to be less than [tex]|f(0)|[/tex] and I don't need to show a contradiction for the case of [tex]\varepsilon > |f(0)|[/tex]. This is the missing logic in my solution. Thank you so much for your patience and time.