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complexnumber
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Homework Statement
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].
Homework 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]
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?