Functional analysis: Shoe set is not dense in C([a,b])

[Mixer]

Homework Statement

Let $[a,b] \subset \mathbb{R}$ be a compact interval and t₀ $\in [a,b]$ fixed. Show that the set $S = {f \in C[a,b] | f(t_0) = 0}$ is not dense in the space $C[a,b]$ (with the sup-norm).

Homework Equations

Dense set: http://en.wikipedia.org/wiki/Dense_set

sup - norm: http://mathworld.wolfram.com/SupremumNorm.html

The Attempt at a Solution

I tried to take function f from S and function g from C[a,b] and calculate the sup-norm of the difference of the functions and make it bigger than some number. However I am not able to do so.. I'm not even sure if my approach is correct here. What should be my strategy?

 

[HallsofIvy]

For a given number Y, take g(x)= Y+ 1, a constant function. What is d(f, g)?

 

[Mixer]

Thank you for reply!

So are you saying that I should take g(t) = t₀ + 1 for all t. Then

$\left\|f - g\right\| = sup_{t \in [a,b]} |f(t) - g(t)| \geq |f(t_0) - g(t)| = |0 - t_0 -1| = |t_0 + 1|$

Therefore set S is not dense in C[a,b] ?

 

[LCKurtz]

Thank you for reply!

So are you saying that I should take g(t) = t₀ + 1 for all t. Then

$\left\|f - g\right\| = sup_{t \in [a,b]} |f(t) - g(t)| \geq |f(t_0) - g(t)| = |0 - t_0 -1| = |t_0 + 1|$

Therefore set S is not dense in C[a,b] ?

I don't think that is quite what he is saying. And you could accidentally have $t_0=-1$ which would wreck your argument. Why don't you just use a similar argument to show that if $f(x) \equiv 1$ that no $g$ in your subset gets close to it in sup norm?