By Stone-Weierstrass theorem, must
vanish at some point . Thus,
Note that the subalgebra
is closed in . Since
, we have
; thus, in order to show that
, we just need to show
.
Consider
. If
and
are elements in
, where
and
, then
Since is an
algebra, we have
; therefore,
This shows that
is closed under multiplication. Since the other defining
conditions of a subalgebra in are immediate
for
, we conclude that it is a subalgebra in . Since
and separates
points, so does . Since
and vanishes
nowhere, we conclude that
is nowhere vanishing subalgebra of that separates
points, By Stone-Weierstrass we know that
. But we know that
is closed (because is
one-dimensional and
is closed), and since
, we conclude
showing that
.
If possible, suppose that
; then there exists such
that but
. Since
, we know that there is some non-zero constant
such that
. Since
, this implies
. But implies
, a
contradiction. Hence, .
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