Question

Let R be the ring of all real valued continuous functions on the closed interval [0,1] .Let M = ={f∈R∶f(1/3)=0} Show that M is non- empty.

Answer 1

Answer:

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Step-by-step explanation:

Let I be a proper ideal of the ring C([0,1]) of continuous functions on [0,1]. We prove that

Z(I):=⋂f∈If−1({0})

is non-empty. This proves the existence of a point x satisfying f(x)=0 for all f∈I.

Assume otherwise that Z(I)=∅. Then {f−1(R∖{0}):f∈I} is an open cover of [0,1], and by the compactness, there exists a finite set J⊂I such that {f−1(R∖{0}):f∈J} is also an open cover of [0,1]. Now define g:[0,1]→R by

g(x)=∑f∈Jf(x)2.

Then g∈I and g is non-zero everywhere. Since I is ideal and 1/g∈C([0,1]), this implies that 1=g⋅(1/g)∈I as well, which in turn implies that I=C([0,1]), a contradiction.

Finally, if M is a maximal ideal, then for each z∈Z(M) we have M⊂Mz, where Mz is as in OP's notation. Thus by the maximality, M=Mz (and, in particular, Z(M) is a singleton)