Is the Set of Polynomials Dense in Continuous Function Space?

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In summary, the problem is to show that A⊂[C], which implies to proove that the set of all polynomials has polynomial functions ( i.e. P[X] = f(x) ) which are continuous
  • #1
tudor
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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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  • #2
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?
 
  • #3
the set of polynomials with rational coeficients has no "per say" order ... basically you take all plynomials and put them in a set ...
 
  • #4
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.
 
  • #5
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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  • #6
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.
 

FAQ: Is the Set of Polynomials Dense in Continuous Function Space?

What is a dense set?

A dense set is a subset of a larger set where every point in the larger set is either a member of the subset or is arbitrarily close to a member of the subset. In other words, every point in the larger set can be approximated by a member of the subset.

What are some examples of dense sets?

Examples of dense sets include the set of rational numbers within the set of real numbers, the set of integers within the set of real numbers, and the set of prime numbers within the set of natural numbers.

What is the importance of dense sets in mathematics?

Dense sets are important in mathematics because they allow us to approximate and approximate the behavior of a larger set. They also have applications in analysis, topology, and other areas of mathematics.

How are dense sets related to continuity?

Dense sets are closely related to continuity, as continuous functions map dense sets to dense sets. In other words, if every point in a set can be approximated by a point in another set, then the function mapping between the two sets is continuous.

How do you prove that a set is dense?

To prove that a set is dense, one must show that every point in the larger set can be approximated by a point in the subset. This can be done through various methods such as using the definition of density, using limit points, or using a proof by contradiction.

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