- #1

complexnumber

- 62

- 0

## 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?