As just said we will prove the result assuming first

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As just said, we will prove the result assuming first that Z = { p 0 } is a singleton set. In this case consider the set ˜ A = { c + f : c R , f ∈ A} . It is immediate that ˜ A is an algebra and, since (for example) the constant function 1 is in it, it satisfies the non-vanishing property. It also separates points. By the Stone Weierstrass Theorem it is dense in C ( X ); in other words, given ² > 0, f C ( X ), there exists c R , g ∈ A such that | f ( p ) - ( c + g ( p )) | < ²/ 2 for all p X . If f ( p 0 ) = 0, then letting p = p 0 in | f ( p ) - ( c + g ( p )) | < ²/ 2 we get | c | < ²/ 2. Thus | f ( p ) - g ( p ) | ≤ | f ( p ) - ( c + g ( p )) | + | c | < ² 2 + ² 2 = ² for all p X . The case Z = { p 0 } is done. For the general case, we have to be a bit more devious. We can start with a result which can either be seen as a consequence of what we just did, or proved using the same idea we used here from Weierstrass’ Approximation Theorem. Theorem 1 (Weierstrass2) Let f C ([ a, b ]) , where -∞ < a < 0 < b < and assume f (0) = 0 . For every ² > 0 there exists a polynomial p with 0 constant term such that | f ( x ) - p ( x ) | < ² for all x [ a, b ] . Proof. Proof #1. Let ² > 0 be given. By the Theorem of Weierstrass there exists a polynomial q , q ( x ) = a 0 + a 1 x + · · · + a n x n , such that | f ( x ) - q ( x ) | < ²/ 2 for all x [ a, b ]. Evaluating at 0 we get | a 0 | < ²/ 2. Letting p ( x ) = a 1 x + · · · + a n x n , we have that | f ( x ) - p ( x ) | = | f ( x ) - p ( x ) - a 0 + a 0 | = | f ( x ) - q ( x ) + a 0 | ≤ | f ( x ) - q ( x ) | + | a 0 | < ² 2 + ² 2 = ². Proof #2. The set P = { p : p is a polynomial and p (0) = 0 } satisfies properties (a)-(d) above with Z = { 0 } . The result is now immediate from the Z = singleton set case. Assume now A is a subalgebra of C ( X ) (Properties (a), (b) above hold), the set Z = ∩{ f - 1 ( { 0 } ) : f ∈ A} 6 = , and p X \ Z , q X , p 6 = q implies there exists f ∈ A such that f ( p ) 6 = f ( q ). (Properties (c), (d)). Incidentally, if p / Z , there is f sA such that f ( p ) 6 = 0; this f automatically separates p from all points of Z . Property (d) could as well have been stated as: If p, q X \ Z , p 6 = q , then there exists f ∈ A such that f ( p ) 6 = f ( q ). We will imitate the proof of Stone’s Theorem. We denote by ¯ A the closure of A in the norm metric. Step 1. If p, q X \ Z , p 6 = q , then for every a, b R there exists f ∈ A such that f ( p ) = a , f ( q ) = b . If p X \ Z , for every a R there exists f ∈ A such that f ( p ) = a . The second statement is immediate from the definition of Z . If p / Z , there exists g ∈ A such that g ( p ) 6 = 0, then f = ( a/g ( p )) g ∈ A and f ( p ) = a . The first statement can be proved exactly as in the proof of Stone Weierstrass. First of all,there is g ∈ A such that g ( p ) 6 = 0 6 = g ( q ). In fact, by the definition of Z , there exist g 1 , g 2 ∈ A such that g 1 ( p ) 6 = 0 6 = g 2 ( q ); then g = g 2 1 + g 2 2 sA and foots the bill. Because A separates points not in Z , there exists h ∈ A such that h ( p ) 6 = h ( q ). Then det g ( p ) g ( p ) h ( p ) g ( q ) g ( q ) h ( q ) = g ( p ) g ( q )( h ( q ) - h ( p )) 6 = 0 , 3
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thus there exist c, d R such that cg ( p ) + dg ( p ) h ( p ) = a cg ( q ) + dg ( q ) h ( q ) = b.
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