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

[complexnumber]

Homework Statement

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

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

Homework Equations

The formulae for $$d_{\infty}$$ is
$$\begin{align*}<br /> d_{\infty}(f,g) = \max_{0 \leq t \leq 1} |f(t) - g(t)|<br /> \end{align*}$$

The Attempt at a Solution

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

 

[Office_Shredder]

Think about it like this: picking a function g(x) close to f(x) is like wiggling each individual value of f(x) a little bit (with some continuity constraints). It doesn't really matter how you wiggle any point except for at 0 (because that's the only place Y is defined). How can you change f(x) at 0 to make it leave Y? By making g(0)=0 of course. So to prevent that, what restriction should you put on your ability to wiggle f(x) at 0?

 

[JSuarez]

Ok, it's not entirely wrong, but it's messy:

f is an interior point of Y, then Y is open in (...)

This is not true: there are many sets with interior points that are not open sets, which means that all its points are interior. It's not clear if you're trying to say that you want to prove that all points are interior, or if you' are assuming that the set is open (you want to prove this).

After this, your argument needs a little cleaning, but it's basically correct.

 

[complexnumber]

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 $$\varepsilon > |f(0)|$$ ?

 

[Office_Shredder]

You can't. Your value for epsilon is too large. Think about my suggestion some more:

If you change the value of f(0) by f(0), you get g(0)=0 which doesn't lie in Y. So what should your restriction be on how much f(0) can change by?

 

[complexnumber]

You can't. Your value for epsilon is too large. Think about my suggestion some more:

If you change the value of f(0) by f(0), you get g(0)=0 which doesn't lie in Y. So what should your restriction be on how much f(0) can change by?

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

 

[Office_Shredder]

If f(0) can't be wiggled more than |f(0)|, then wouldn't it make sense to require that all of f(x) can't be wiggled by more than |f(0)|? Then what value of epsilon would that correspond to?

 

[complexnumber]

If f(0) can't be wiggled more than |f(0)|, then wouldn't it make sense to require that all of f(x) can't be wiggled by more than |f(0)|? Then what value of epsilon would that correspond to?

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