added solution to problem 2 in the cardinality section

1 files changed, 5 insertions, 1 deletions

diff --git a/Problem Set 2.page b/Problem Set 2.page
index 5e41ea2..ee95d23 100644
--- a/Problem Set 2.page
+++ b/Problem Set 2.page
@@ -37,6 +37,7 @@ $$\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$$
 
 # Solutions
 
+##Fourier Series
 2. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
 
 $$\begin{array}{ccl} \sin^4 x &=& \frac{{( e^{ix}-e^{-ix} )}^4}{16} \\
@@ -88,7 +89,7 @@ $$\int_0^{2\pi} \sin^4(x) dx = \sum |a_n|^2 = \frac{3 \pi}{4}$$
 
 ## Cardinality
 
-Cardinality of the natural numbers (countable): $\mathbb{N}$,$\mathbf{Z}$,$\mathbb{N} \times \mathbb{N}$,$\mathbb{Q}$
+1. Cardinality of the natural numbers (countable): $\mathbb{N}$,$\mathbf{Z}$,$\mathbb{N} \times \mathbb{N}$,$\mathbb{Q}$
 
 Cardinality of the real numbers (continuum): $\mathbf{R}$,the open interval $(0,1)$,the closed interval $[0,1]$,$2^{\mathbb{N}}$
 
@@ -105,6 +106,9 @@ 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 doesn't 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. 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 it and all the other elements that point to sets that do not contain themselves. 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$ you all $x \. X must point to sets in 2^X$ that contain themselves, but we know that this isn't true since atleast one element in X should point to the empty set.  
+
 # Comments
 
 Here are some issues with the above solutions.  Feel free to make corrections, even if you weren't the original poster.  However, if you do decide to redo a solution, please don't delete the old one, so that people can see where the old idea went wrong and how it was fixed.