From c91e30ff2cedd33ab6a83114167c28e696436130 Mon Sep 17 00:00:00 2001 From: arathir Date: Fri, 9 Jul 2010 04:50:35 +0000 Subject: fix typo --- 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 5ba091b..5d5cd25 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -107,7 +107,7 @@ Proofs: -$2^{\mathbb{N}} = \mathbf{R}$ 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 $\mathbb{Q}=\mathbf{N}$. 2. So we assume that there is some bijection $f$ between all the elements in $X$ and $2^X$. This means that we can map every element in $X$ to an element in $2^X$. For all the $x \in X$ atleast one $x$ must map to a set in $2^X$ that does not contain itself. This is true since, we know that the empty set is a member of $2^X$, so one of the elements in $X$ must map to the empty set.So lets look at all the $x \in X$ that map to sets in $2^X$ that do not contain the element x. All of these elements that map to an element in $2^X$ that do not contain themselves must be contained in another set that is a member of $2^X$. Since there is a bijection between $X$ and $2^X$, there must be some other -$x \in X$ that points to this set we just created. However, if some other $x \in X$ points to this set, it also points to a set that does not contain itself. Therefore there must be some other set that contains just this element and an $x \in X$ must point to this set. Each time you try to create this set, you create another element that does not point to itself. Therefore, in order to have a bijection between $X$ and $2^X$ all $x \ in X$ must point to sets in $2^X$ that contain themselves, but we know that this isn't true since atleast one element in $X$ should point to the empty set. +$x \in X$ that points to this set we just created. However, if some other $x \in X$ points to this set, it also points to a set that does not contain itself. Therefore there must be some other set that contains just this element and an $x \in X$ must point to this set. Each time you try to create this set, you create another element that does not point to itself. Therefore, in order to have a bijection between $X$ and $2^X$ all $x \in X$ must point to sets in $2^X$ that contain $x$, but we know that this isn't true since atleast one element in $X$ should point to the empty set. # Comments -- cgit v1.2.3