# Stone-Weierstrass Theorem

In the proof of Stone-Weierstrass theorem provided in Rudin's Principles of Mathematical Analysis, why do we only need to show that there exists a sequence of polynomials $P_n$ that converges uniformly to the continuous complex function on $[0,1]$?

• Could you please elaborate a little? Maybe give the precise statement of the conclusion and say why you don't see how it follows? – Jonas Meyer Nov 16 '10 at 23:51
• Because you can stretch it and so on? – Jonas Teuwen Nov 16 '10 at 23:57

I think you are just asking why, "We may assume, without loss of generality, that $[a, b] = [0,1]$." The reason is that if $f$ is continuous on $[a,b]$, then we can uniformly approximate $f(a+(b-a)x)$ on $[0,1]$ with a sequence of polynomials $(p_n)_{n=1}^\infty$, and then $(p_n((x-a)/(b-a)))_{n=1}^\infty$ will converge uniformly to $f$.