From 6f25f0e0f183d5d3ce0c19c341e796a3d8424a05 Mon Sep 17 00:00:00 2001 From: luccul Date: Tue, 6 Jul 2010 04:11:19 +0000 Subject: latex issues --- Problem Set 2.page | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/Problem Set 2.page b/Problem Set 2.page index adde677..a1e0f95 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -99,9 +99,9 @@ Proofs: -$\mathbb{Q}=\mathbf{N}$ by combining the bijections from $\mathbf{N}$ to $\mathbf{Z}$ and from $\mathbf{N}$ to $\mathbb{N} \times \mathbb{N}$. This provides a bijection from $\mathbf{N}$ to $\mathbb{Z} \times \mathbb{Z}$, and since every element of $\mathbb{Q}$ can be represented as the ratio of the two components of an ordered pair of integers, we have a bijection from $\mathbb{Z} \times \mathbb{Z}$ to $\mathbb{Q}$. --Josh explained Cantor's proof of the uncountability of the real numbers on the 28th; Wikipedia provides a good description thereof: 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. +-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: $n \mapsto \tan{\pi n - \pi/2}$. -$2^{\mathbb{N}} = \mathbf{R}$ 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