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Weierstrass theorem.

Lef $f$ be a defined and continuous function in $[a,b]$. Given $\epsilon>0,$ there exists a polynomial $P$ such that $\vert f(x)-P(x)\vert<\epsilon,$ for all $x\in[a,b].$

Stone-Weierstrass theorem.

Let $X$ be a topological compact space. If $F$ is an algebra of $C(X)$ that separate points and contains the constant functions then $F$ is dense with respect to the uniform convergence in $C(X).$

How is that an algebra $F$ of $C(X)$ that separate points and contains the constant functions is the generalization of $P(x)$ ?

Can someone shed some light on this?

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The set of polynomials defined on $[a,b]$ is an algebra of $C([a,b])$ that separate points (for all $x,y\in[a,b]$ such that $x\neq y$, there exists a polynomial $P$ such that $P(x)\neq P(y)$) and contains the constant functions.

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  • $\begingroup$ Thank you for your answer. In which part will this with respect to the uniform convergence in 𝐶(𝑋) be included? $\endgroup$ – Isa Mar 17 at 0:19
  • $\begingroup$ The Weierstraß theorem states that there exists a sequence of polynomials which converges uniformly on $[a,b]$ to $f$. Indeed, you can rewrite your statement in this way: $\exists$ polynomial $P$ such that $\sup_{x\in[a,b]}\vert f(x)-P(x)\vert\le\varepsilon$. $\endgroup$ – Will Mar 17 at 8:54
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Hint:

For $x_0, y_0 \in [a,b]$ such that $x_0 \ne y_0$ we consider the polynomial $p(t) = t-x_0$. Then $p(x_0) = 0$ but $p(y_0) = y_0 - x_0 \ne 0$.

Hence polynomials on $[a,b]$ separate the points of $[a,b]$.

Also, constant functions are polynomials so...

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