# Math Help - Is the Interior of the space of continuous functions empty?

1. ## Is the Interior of the space of continuous functions empty?

Let C be the space of continuous functions from [a,b] to R (real numbers) with the metric:

d(f,g) = sup lf-gl

Is the interior of C a non-empty set?

I say it is an empty set, but I'm not sure about some of my steps. Let me know what you all think:

Outline of Proof

Suppose the interior is not empty and contains f.

Then, for some e>0, the open ball around f of radius e is contained in C.

But, because f is continuous, then if we break [a,b] up into small enough intervals such that on each interval [Xi, X(i+1)], there is a number Ci whereby:

If x is in [Xi, X(i+1)) -->lf(x) - Cil < e

Define a function g as:

g(x) = Ci if x is in [Xi, X(i+1))

then d(f,g) < e

so g is in the ball of radius e around f

so g is in C, but g is not continuous...contradiction.

...thoughts?

2. Originally Posted by southprkfan1
Let C be the space of continuous functions from [a,b] to R (real numbers) with the metric:

d(f,g) = sup lf-gl

Is the interior of C a non-empty set?

I say it is an empty set, but I'm not sure about some of my steps. Let me know what you all think:

Outline of Proof

Suppose the interior is not empty and contains f.

Then, for some e>0, the open ball around f of radius e is contained in C.

But, because f is continuous, then if we break [a,b] up into small enough intervals such that on each interval [Xi, X(i+1)], there is a number Ci whereby:

If x is in [Xi, X(i+1)) -->lf(x) - Cil < e

Define a function g as:

g(x) = Ci if x is in [Xi, X(i+1))

then d(f,g) < e

so g is in the ball of radius e around f

so g is in C, but g is not continuous...contradiction.

...thoughts?
Clearly since $\left(\mathcal{C}\left[a,b\right]\right)^{\circ}\subseteq\mathcal{C}\left[a,b\right]$ we must merely show that no point of $\mathcal{C}[a,b]$ is an interior point.

So, let $f\in\mathcal{C}[a,b]$ and let $\varepsilon>0$ be given. Define, $g(x)=\begin{cases} f(x) & \mbox{if} \quad x\ne\frac{a+b}{2} \\ f(x)+\frac{\varepsilon}{2} & \mbox{if} \quad x=\frac{a+b}{2}\end{cases}$. Since $\frac{a+b}{2}\in\left([a,b]\right)^{\circ}$ we know that it is continuous if and only if $\lim_{x\to\frac{a+b}{2}}g(x)=g\left(\frac{a+b}{2}\ right)$, but $\lim_{x\to\frac{a+b}{2}}g(x)=\lim_{x\to\frac{a+b}{ 2}}f(x)=f\left(\frac{a+b}{2}\right)\ne g\left(\frac{a+b}{2}\right)$ and so $g\notin\mathcal{C}\left[a,b\right]$. But, $g-f=\begin{cases}0 & \mbox{if} \quad x\ne \frac{a+b}{2} \\ \frac{\varepsilon}{2} & \mbox{if} \quad x=\frac{a+b}{2}\end{cases}$ and so $\left\|g-f\right\|_{\infty}=\frac{\varepsilon}{2}$ and so $g\in B_{\varepsilon}(f)$. The conclusion follows.

Oh, I guess I tacitly assumed that $a\ne b$. Some authors, like the one of the book I'm reading now, allow $a=b$ in which case $[a,b]=\{a\}$ but in that case this is obvious.

3. Thanks,

So basically you agree with my method, its just that the way I got the non-continuous function g was way too complicated.

4. Originally Posted by southprkfan1
Let C be the space of continuous functions from [a,b] to R (real numbers) with the metric:

d(f,g) = sup lf-gl

Is the interior of C a non-empty set?

I say it is an empty set, but I'm not sure about some of my steps. Let me know what you all think:

Outline of Proof

Suppose the interior is not empty and contains f.

Then, for some e>0, the open ball around f of radius e is contained in C.

But, because f is continuous, then if we break [a,b] up into small enough intervals such that on each interval [Xi, X(i+1)], there is a number Ci whereby:

If x is in [Xi, X(i+1)) -->lf(x) - Cil < e

Define a function g as:

g(x) = Ci if x is in [Xi, X(i+1))

then d(f,g) < e

so g is in the ball of radius e around f

so g is in C, but g is not continuous...contradiction.

...thoughts?
As stated this makes no sense: A topological space $X$ is always open and so it is its own interior. Did you mean as a subspace of some larger space (ie. bounded functions on $[a,b]$ with sup norm)?

Also notice that both solutions are meaningless, since the functions constructed are not in the space you gave (thus the need for a larger one to fit in those).

5. Originally Posted by Jose27
As stated this makes no sense: A topological space $X$ is always open and so it is its own interior. Did you mean as a subspace of some larger space (ie. bounded functions on $[a,b]$ with sup norm)?

Also notice that both solutions are meaningless, since the functions constructed are not in the space you gave (thus the need for a larger one to fit in those).
I interpreted as being a subspace of bounded functions on $[a,b]$

6. Originally Posted by Drexel28
I interpreted as being a subspace of bounded functions on $[a,b]$
That's exactly right, I should have included that in the question. The actual question assumes we know this fact.

7. I have posted a question very similar to this one that I would appreciate any help on:

http://www.mathhelpforum.com/math-he...c-0-empty.html