1. C_c(Â) = {f Î C₀(Â);   supp(f) =  the  closure  of  {x Î Â; f(x) ¹ 0}   is  compact  } is a dense ideal of C₀(Â). Note that the function f defined by

f(x) = ì ï ï ï ï í ï ï ï ï î 1 1+x x ³ 0 1 1-x x < 0

belongs to C₀(Â)- C_c(Â).

2. A = {f Î C([0,1]);f(0) = 0} is a closed subalgebra of C([0,1]) not containing the constant function 1. So A is a non-unital Banach algebra. Let f_°(t) = t  , t Î [0,1]. I = {f_° g ; g Î C[0,1]} is a proper ideal of A (since if

h(t) = ì ï ï í ï ï î tsin 1 t t Î (0,1] 0 t = 0

and for some g Î C[0,1], tg(t) = h(t) whenever t Î [0,1] then lim_{t® 0}sin[ 1/ t] = g(0), a contradiction). By the Stone-Weierstrass theorem, each f Î A is the uniform limit of a sequence (p_n) of polynomials with p_n(0) = 0. Moreover t\longmapsto [( p_n(t))/ t] belongs to C[0,1] and t[( p_n(t))/ t]® f(t) uniformly on [0,1]. So f belongs to the closure of I. Hence I is dense in A.