# Math Help - [SOLVED] is space of bounded continuous functions separable?

1. ## [SOLVED] is space of bounded continuous functions separable?

hi,

just to double check:

the space of bounded continuous functions **on the real line** is separable, right?

i want to take all rational points and all rational values within the range, but there's a slight problem with that most of those functions won't be continuous. however i feel it shouldn't be a bad obstruction, and since all i need is a "yes/no" i'm asking you people...

2. Originally Posted by choovuck
the space of bounded continuous functions **on the real line** is separable, right?
Do you mean separable for the topology of uniform convergence? If so, the answer is No.

Given a bounded sequence $(x_n)$, define a bounded continuous function f by $f(n) = x_n$ if n is an integer, and interpolating linearly between integer points. That shows that the map $f\mapsto(f(n))$ is surjective, and continuous, from the space CB(R) (continuous bounded functions on the line, with the uniform norm) to the space l^∞ of bounded sequences. But it is well known that l^∞ is not separable. Hence neither is CB(R).

The space of bounded continuous functions on a bounded closed interval is separable, because you can use Weierstrass's theorem to approximate functions uniformly by polynomials with rational coefficients.

3. Originally Posted by choovuck
hi,

just to double check:

the space of bounded continuous functions **on the real line** is separable, right?

i want to take all rational points and all rational values within the range, but there's a slight problem with that most of those functions won't be continuous. however i feel it shouldn't be a bad obstruction, and since all i need is a "yes/no" i'm asking you people...
You're considering the "sup" norm, right?
Then I would say yes since for instance, a countable dense subset seems to be given by the continuous piecewise linear functions with rational slopes and subdivisions at rational points. Where is the flaw? (Can anyone give me a bounded continuous function that is not well approximated by these functions?) I've got it: this set is not countable anyway... And neither was the set of the functions mapping the rationals to the rationals by the way.

4. Originally Posted by Opalg
Do you mean separable for the topology of uniform convergence?
yes, the sup norm

Originally Posted by Opalg
Given a bounded sequence $(x_n)$, define a bounded continuous function f by $f(n) = x_n$ if n is an integer, and interpolating linearly between integer points. That shows that the map $f\mapsto(f(n))$ is surjective, and continuous, from the space CB(R) (continuous bounded functions on the line, with the uniform norm) to the space l^∞ of bounded sequences.
nope, this map is clearly not a continuous one: the preimage of unit disk in l^∞ will contain lots of ugly functions which are nice at integers but have something wild in between

yes, Laurent's way works

5. Originally Posted by choovuck
yes, the sup norm

nope, this map is clearly not a continuous one: the preimage of unit disk in l^∞ will contain lots of ugly functions which are nice at integers but have something wild in between

yes, Laurent's way works, thanks.
No, it doesnt. See my edit... The fact is that the set of bounded continuous functions is just too big: the bounded continuous functions that equal 0 or 1 at the integers already provide an uncountable subset, and you can't approximate two of these functions at the same time at a distance less than a half... So you need uncountably many functions in a dense subset. This is just a re-writing of the proof of non-separability of $\ell^\infty$, as Opalg pointed out.

6. okay, the opalg's map was not continuous, but the same proof showing that l^\infty is not separable, shows that so is BC(R): assuming f_n are dense, we can define f to differ from f_n by at least 1 at the point x=n. then take it linear in between and we are screwed.

okay, not separable, thanks.

7. Originally Posted by choovuck
okay, the opalg's map was not continuous
It is: if we denote it by [tex]\Phi[/Math], then $\|\Phi(f)-\Phi(g)\|_{\ell^\infty}=\sup_{n\in \mathbb{N}} |f(n)-g(n)|\leq \sup_{x\in\mathbb{R}} |f(x)-g(x)|=\|f-g\|_{\infty}$.

8. ok, you're right, I was confused

9. Originally Posted by Opalg
The space of bounded continuous functions on a bounded closed interval is separable, because you can use Weierstrass's theorem to approximate functions uniformly by polynomials with rational coefficients.
Hi Opalg,

Are you sure? What about Laurent's example about functions that equal 0 or 1 but this time at rationals instead of integers. Let those functions be defined on a closed interval. Can't I map each function to a sequence of 0 and 1's and use Cantor's diagonalization to show that the set of those functions is uncountable?

Just because we have a HW problem where we have to show that B[a,b] is not separable...

10. we were talking about continuous functions there