From 0c19db3ab7fcce14941d69fa589881a6a8dfafb4 Mon Sep 17 00:00:00 2001 From: luccul Date: Tue, 6 Jul 2010 04:12:13 +0000 Subject: more latex --- Problem Set 2.page | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/Problem Set 2.page b/Problem Set 2.page index a1e0f95..3801c8e 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -101,7 +101,7 @@ Proofs: -Josh explained Cantor's proof of the uncountability of the real numbers on the 28th; Wikipedia provides a good description thereof: [external](http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument). In a nutshell, you assume you can have an ordered list of the reals (i.e., a bijection to the naturals), then construct a real number not on that list by having its nth digit be different from the nth digit of the nth number on the list. --$(0,1)=\mathbf{R}$ and $[0,1]=\mathbf{R}$ under the same bijection: $n \mapsto \tan{\pi n - \pi/2}$. +-$(0,1)=\mathbf{R}$ and $[0,1]=\mathbf{R}$ under the same bijection: $n \mapsto \tan\left (\pi n - \pi/2 \right)$. -$2^{\mathbb{N}} = \mathbf{R}$ by writing a given real number $r$ as a (possibly infinite) set of natural numbers. For example, write $pi$ as the set of rational numbers $\{3,0.1,0.04,0.001,0.005,...\}$, then replace each number with the natural number it would map to under the bijection $\mathbb{Q}=\mathbf{N}$. -- cgit v1.2.3