A question about intuitionistic analysis - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T23:29:55Zhttp://mathoverflow.net/feeds/question/115264http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/115264/a-question-about-intuitionistic-analysisA question about intuitionistic analysisSet2012-12-03T08:45:59Z2012-12-03T08:45:59Z
<p>In Michael Dummett's book "Elements of Intuitionism" , the product of real numbers are defined as follow:</p>
<p>$x\cdot y=$ { $\langle r_n\rangle \cdot \langle s_n\rangle$ | $\langle r_n\rangle\in x , \langle s_n\rangle\in y$ } , where $\langle r_n\rangle ,\langle s_n\rangle$ are Cauchy sequences of rational numbers, and $\langle r_n\rangle \cdot \langle s_n\rangle=\langle r_n\cdot s_n\rangle$ .</p>
<p>This definition is valid iff we can prove intuitionistically { $\langle r_n\rangle \cdot \langle s_n\rangle$ | $\langle r_n\rangle\in x , \langle s_n\rangle\in y$ } is indeed a real number, i.e. { $\langle r_n\rangle \cdot \langle s_n\rangle$ | $\langle r_n\rangle\in x , \langle s_n\rangle\in y$ } is closed under "equivalent" (we say that $\langle r_n\rangle$ is equivalent to $\langle s_n\rangle$ if for all natural number k, we can find a natural number n such that $|r_m-s_m|<2^{-k} $ for all $m>n$ ). This is to say, we must prove the following proposition intuitionistically :</p>
<p>Let $\langle r_n\rangle , \langle s_n\rangle , \langle t_n\rangle $ be Cauchy sequences of rational numbers, and $\langle t_n\rangle $ is equivalent to $\langle r_n\rangle \cdot \langle s_n\rangle$. Then we can construct two Cauchy sequences $\langle r_n'\rangle , \langle s_n'\rangle $ of rational numbers such that</p>
<p>(1) $\langle r_n'\rangle$ is equivalent to $\langle r_n\rangle$ ;</p>
<p>(2) $\langle s_n'\rangle$ is equivalent to $\langle s_n\rangle$ ;</p>
<p>(3) $\langle t_n\rangle =\langle r_n'\rangle \cdot \langle s_n'\rangle$.</p>
<p>But Michael Dummett doesn't justify his definition, and I find it's very difficult to prove the above proposition intuitionistically. Could you help me?</p>