“real: chapter_10” — 2011/5/22 — 23:26 — page 12 — #12
10-12 Real Analysis
Theorem 10.2.13 Let
∞
n=0
a
n
x
n
and
∞
n=0
b
n
x
n
converge in
S ={x/|x| <δ, δ>0}. Let E ={x ∈ S/
∞
n=0
a
n
x
n
=
∞
n=0
b
n
x
n
}.
If E has a limit point in S, then a
n
= b
n
for all n and hence E = S.
Proof Let f (x) =
∞
n=0
c
n
x
n
, where c
n
= a
n
− b
n
so that f (x) = 0 for
x ∈ E. Let A be the set of all limit points of E in S and B = S \ A.
Since A is closed (note that A is the derived set of E), B is open. If
only we can prove that A is open, then S = A ∪ B gives a separation
for the connected interval S. This shows that either A =∅or B =∅.
However by hypothesis A =∅and hence B =∅or that S = A. We now
observe that the continuity of f on S (note that f is even differentiable
in S) implies that S = A ⊂ E (note ...