現代数学の系譜工学物理雑談古典ガロア理論も読む49

[過去ﾛｸﾞ] 現代数学の系譜工学物理雑談古典ガロア理論も読む49 (658ﾚｽ)
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72(1): 現代数学の系譜雑談古典ガロア理論も読む ◆e.a0E5TtKE 2018/01/01(月)17:11 ID:dCRrvhl7(8/27) AA×
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72: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [sage] 2018/01/01(月) 17:11:49.95 ID:dCRrvhl7

>>71 つづき

and
I(n, xn,m) ∩ Kc ⊂ f−1((f(x) − ε, f(x) + ε)).
Hence, f|Kc is continuous at x.

To prove the last part of the theorem, note first that (iii) implies (ii) even
without the restriction that J contains no interval. Now suppose that J contains
no interval and that f,K are as in (ii). Define
(1) G(x) = lim sup t→x,t∈Kc f(t)
and
(2) g(x) = G(x) when G(x) is finite,
　 or = f(x) otherwise.
In particular, it follows from (ii) that f|Kc = g|Kc . Let x ∈ Kc and ε > 0.
According to (ii) there is a δ > 0 such that
(3) |g(y) − g(x)| = |f(y) − f(x)| < ε/2
whenever y ∈ (x − δ, x + δ) ∩Kc. If z ∈ (x − δ, x + δ) ∩K, then the assumption
that K can contain no nonempty open set implies the existence of a sequence
{zn : n ∈ N} ⊂ (x − δ, x + δ) ∩ Kc
such that f(zn) → G(z). Hence, by (3), G(z) is finite, so g(z) = G(z) and
|g(z) − g(x)| ? ε/2 < ε. Therefore, g is continuous at x. QED

つづく

http://rio2016.5ch.net/test/read.cgi/math/1514376850/72

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