Question

2. IR aera l Cer) hat separate a Separa an alse bre hent vanishes atnPon

Answer 1

By Stone-Weierstrass theorem, ${\bf a}$ must vanish at some point $x_0\in X$. Thus,

${\bf a}\subseteq \{f\in C(X):f(x_0)=0\}$

Note that the subalgebra

$A:= \{f\in C(X):f(x_0)=0\}$

is closed in
C(X)
. Since $A\supseteq {{\bf a}}$ , we have $A\supseteq {\overline{\bf a}}$ ; thus, in order to show that $A= {\overline{\bf a}}$ , we just need to show
A C
.

Consider $B={\bf a}+\mathbb R1$ . If $b_1=a_1+c_1$ and
(l
are elements in $B={\bf a}+\mathbb R1$ , where
a1, a2 E a
and
C1, C2 ER
, then

$b_1b_2=(a_1a_2+c_1a_2+c_2a_1)+c_1c_2$

Since ${\bf a}$ is an algebra, we have $a_1a_2+c_1a_2+c_2a_1\in {\bf a}$ ; therefore,

$b_1b_2=(a_1a_2+c_1a_2+c_2a_1)+c_1c_2\in {\bf a}+\mathbb R1$

This shows that $B={\bf a}+\mathbb R1$ is closed under multiplication. Since the other defining conditions of a subalgebra in
C(X)
are immediate for $B={\bf a}+\mathbb R1$ , we conclude that it is a subalgebra in
C(X)
. Since ${\bf a}\subset B$ and ${\bf a}$ separates points, so does . Since $\mathbb R1\subset B$ and
R1
vanishes nowhere, we conclude that $B={\bf a}+\mathbb R1$ is nowhere vanishing subalgebra of
C(X)
that separates points, By Stone-Weierstrass we know that
B = C(X)
. But we know that ${\overline{\bf a}}+\mathbb R1$ is closed (because
R1
is one-dimensional and ${\overline{\bf a}}$ is closed), and since $B\subseteq {\overline{\bf a}}+\mathbb R1$ , we conclude

$C(X)={\overline B}\subseteq {\overline{\bf a}}+\mathbb R1$

showing that $C(X)={\overline{\bf a}}+\mathbb R1$ .

If possible, suppose that ${\overline{\bf a}}\subsetneq \{f\in C(X):f(x_0)=0\}$ ; then there exists
feC(x
such that
T0 1o
but $f\notin {\overline {\bf a}}$ . Since $f\in C(X)={\overline{\bf a}}+\mathbb R1$ , we know that there is some non-zero constant $c\in\mathbb R$ such that $f-c\in{\overline{\bf a}}$ . Since ${\overline{\bf a}}\subsetneq \{f\in C(X):f(x_0)=0\}$ , this implies $f(x_0)-c=0~\Rightarrow~f(x_0)=c$ . But
T0 1o
implies $0=f(x_0)=c$, a contradiction. Hence,  ${\overline{\bf a}}= \{f\in C(X):f(x_0)=0\}$.