Is the Set of Polynomials Dense in Continuous Function Space?

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The discussion centers on the density of polynomials with rational coefficients in the space of continuous functions defined on the interval [a,b]. It asserts that this set of polynomials is dense in the continuous function space, meaning for any two values in that space, there exists a polynomial that fits between them. The conversation references the Stone-Weierstrass theorem, which supports the idea that the closure of rational polynomials includes real coefficient polynomials. Participants express a desire to prove these concepts independently, emphasizing the importance of understanding the underlying principles rather than relying on established theorems. The overall goal is to demonstrate that for any continuous function, a corresponding polynomial can be found that approximates it closely.
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The set of all polynomials with rational coefficients in dense in both spaces, the space of all continuous functions defined in [a,b] C_([a,b]) with the metric
ρ(f,g)=max┬(1≤t≤n)⁡〖|f(t)- g(t)|〗

( i hope you understand what i wrote ... prbl i will find a way to use mathml to write nicer ... :D )
Basically, if A is the set of all rational etc. , and C the countinous function space, the whole problem comes down to prooving A⊂[C], which implies to proove that the set of all polynomials has polynomial functions ( i.e. P[X] = f(x) ) which are continuous ( from now on i use the metric from the space C, and that's it )

am i write ?

p.s.
i don't want a demonstration, becouse i want to learn how to do it myself

Thanks !
 
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tudor said:
The set of all polynomials with rational coefficients in dense in both spaces, the space of all continuous functions defined in [a,b] C_([a,b]) with the metric
ρ(f,g)=max┬(1≤t≤n)⁡〖|f(t)- g(t)|〗

That's one space, what the other? And as i understand it, a set A is dense in a set B if for any x<y in B, there is an a in A with x<a<y. What is the order relation on your set of polynomials with rational coeff?
 
the set of polynomials with rational coeficients has no "per say" order ... basically you take all plynomials and put them in a set ...
 
It's a metric space, quasar - the question is to show that the closure of these polys in the metric topology is all of the space.

Are you aware of the Stone Weierstrass theorem? The closure of the set you wrote clearly contains the real coefficient polys.
 
i didn't want to look for some theorem or lemma or something else, becouse i like solving the problems myself.
but, thank you for your input, i will prbl look into that theorem ...

and now comes to proving that there is an open sphere containing the real coef poly and the rational coef poly among them. i think you would proove the contrary can not happen and then q.e.d. (... i think this is another approach to the problem but very very interesting !... )
 
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i think you should try to show that given a continuous function, you can find a polynomial that equals it at a lot of points, and that it does not change too much between these points.
 

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