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TakuyaMurata (talk | contribs) →Over a real closed field: alternatively, one can use Tarski's principle. |
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The usual proof of the IFT uses Banach's fixed point theorem, which relies on the Cauchy completeness. That part of the argument is replaced by the use of the [[extreme value theorem]], which does not need completeness. Explicitly, in {{section link||A_proof_using_the_contraction_mapping_principle}}, the Cauchy completeness is used only to establish the inclusion <math>B(0, r/2) \subset f(B(0, r))</math>. Here, we shall directly show <math>B(0, r/4) \subset f(B(0, r))</math> instead (which is enough). Given a point <math>y</math> in <math>B(0, r/4)</math>, consider the function <math>P(x) = |f(x) - y|^2</math> defined on a neighborhood of <math>\overline{B}(0, r)</math>. If <math>P'(x) = 0</math>, then <math>0 = P'(x) = 2[f_1(x) - y_1 \cdots f_n(x) - y_n]f'(x)</math> and so <math>f(x) = y</math>, since <math>f'(x)</math> is invertible. Now, by the extreme value theorem, <math>P</math> admits a minimal at some point <math>x_0</math> on the closed ball <math>\overline{B}(0, r)</math>, which can be shown to lie in <math>B(0, r)</math> using <math>2^{-1}|x| \le |f(x)|</math>. Since <math>P'(x_0) = 0</math>, <math>f(x_0) = y</math>, which proves the claimed inclusion. <math>\square</math>
Alternatively, one can deduce the theorem from the one over real numbers by [[Tarski's principle]].{{fact|date=December 2024}}
==See also==
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