Added some comments about the solutions

1 files changed, 15 insertions, 6 deletions

diff --git a/Problem Set 2.page b/Problem Set 2.page
index 23109d8..a7ddc60 100644
--- a/Problem Set 2.page
+++ b/Problem Set 2.page
@@ -7,8 +7,8 @@
     - $\mathbb{N} \times \mathbb{N}$
     - $\mathbb{Q}$ = Set of all fractions $\frac{n}{m}$ where $n,m \in \mathbb{Z}$
     - $\mathbb{R}$
-    - The open interval $(0,1)$
-    - The closed interval $[0,1]$
+    - The open interval $(0,1) = \{ x \in \mathbb{R} : 1 < x < 1 \}$
+    - The closed interval $[0,1] = \{ x \in \mathbb{R} : 0 \leq x \leq 1 \}$
     - $2^{\mathbb{N}}$ = Set of all subsets of $\mathbb{N}$.
     - $2^{\mathbb{R}}$ = Set of all subsets of $\mathbb{R}$.
     - $\mathbb{R}^{\mathbb{R}}$ = Set of all functions from $\mathbb{R}$ to itself.
@@ -18,7 +18,6 @@ Cook up other examples and post them on the wiki!
 2. Let $X$ be any set.  Show that the cardinality of $2^{X}$ is larger than the cardinality of $X$.  
 (Hint: Let $f: X \to 2^X$ be a bijection.  Consider the set of all elements $x \in X$ such that $x$ is not an element of $f(x)$.)
 
-
 ## Fourier Series
 
 
@@ -89,7 +88,7 @@ $$\int_0^{2\pi} \sin^4(x) dx = \sum |a_n|^2 = \frac{3 \pi}{4}$$
 
 ## Cardinality
 
-Cardinality of the natural numbers (countable): $\mathbf{N}$,$\mathbf{Z}$,$\mathbb{N} \times \mathbb{N}$,$\mathbb{Q}$
+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 $[0,1]$,$2^{\mathbb{N}}$
 
@@ -102,6 +101,16 @@ Proofs:
 
 -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.
 
--$(0,1)=\mathbf{R}$ 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 $\mathbb{Q}=\mathbf{N}$.
+
+# Comments
+
+Here are some issues with the above solutions.  Feel free to make corrections, even if they 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.
+
+- The map from $\mathbb{Z} \times \mathbb{Z}$ to $\mathbb{Q}$ sending a pair $(a,b)$ to a ratio $\frac{a}{b}$ is right in principle, but not exactly right.  One minor problem with it is that the map is not defined when $b = 0$.  Another more serious issue is that multiple pairs can map to the same rational.  For example the pairs $(2,1)$ and $(4,2)$ each map to the rational number $2$.  So, while it's true that every rational number is a ratio of some pair, it's not true that there's a \emph{unique} such pair.  To make it unique, you might demand that the pair be in lowest terms (and there's also an issue with negative numbers that you'd have to incorporate into the idea of lowest terms) but then you would have a bijection between some \emph{subset} of $\mathbb{Z} \times \mathbb{Z}$ and $\mathbb{Q}$.  So a little bit more is needed to make this work.
+
+- The bijection from $[0,1]$ to $\mathbb{R}$ doesn't work because the function given would send $0$ to $- \infty$ and $1$ to $+ \infty$.  Somehow there are two extra points you have to sneak in.  It might be helpful to consider the bijection from $\{0,1,2,3, \dots \}$ and $\mathbb{N} = \{1,2,3,\dots \}$ given by sending $n$ to $n+1$.  How could you modify this idea to put $[0,1)$ in bijection with $(0,1)$?
 
--$2^{\mathbb{N}}\mathbf{R}$ by writing a given real number $r$ as a (possibly infinite) set of natural numbers. For example, write $pi$ as the set of natural 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 $\mathbb{Q}=\mathbf{N}$.
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+- The map from $\mathbf{R}$ to $2^{\mathbb{Q}}$ is not a bijection, since not every subset of the natural numbers is obtained in this way.  For example, the number $\frac{1}{3}$ is never contained in any of the subsets, since it has a repeating decimal expansion instead of a terminating one.  So for example the subset containing just the element $\frac{1}{3}$ is never hit by the map.  A hint for this problem is to think about binary expansions.
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