December 13th, 2016, 04:10 AM  #1 
Senior Member Joined: Jun 2014 From: USA Posts: 363 Thanks: 26  Continuum Hypothesis
1) Let $q_1, q_2, q_3,$ … be an enumeration of $A = \mathbb{Q} \cap \mathbb{R}(0, 1)$. 2) For any $r \in \mathbb{R}(0, 1)$, let $B_r = \mathbb{Q} \cap \mathbb{R}(r, 1r)$. 3) For each $r \in \mathbb{R}(0,1)$, there exists one and only one $k_r \in A$ such that $f:A \rightarrow B_r$ is a bijection where $f(a) = a  k_r$. Note that $z:\mathbb{R}(0, 1) \rightarrow \mathbb{R}(r, 1r)$ is a bijection where $z(x) = x – r$. If $q \in A \land r \notin A$, then $z(q) = q – r \notin B_r$ because the difference between a rational number and an irrational number is not a rational number. Therefore, $f(q) \neq z(q) \Rightarrow k_r – r = t \land t \neq 0$. 4) If $r, s \in \mathbb{R}(0,1)$, then $r \neq s \Rightarrow k_r \neq k_s$. 5) If $r, s \in \mathbb{R}(0,1)$, then $k_r \neq k_s \Rightarrow r \neq s$. 6) Therefore, $r, s \in \mathbb{R}(0,1) \land r \neq s \iff k_r \neq k_s$. Let $g(k_r) = r$ for any $r \in \mathbb{R}(0, 1)$. 7) Let $C = g(q_1), g(q_2), g(q_3),$ … = $\mathbb{R}(0,1)$. The Continuum Hypothesis states “there is no set whose cardinality is strictly between that of the integers and the real numbers.” Where $C = \mathbb{R}(0, 1)$, this statement is true because: $\mathbb{R}(0,1)$ = $\mathbb{R}$ = $\mathbb{N}$ = $\mathbb{Z}$. 
December 14th, 2016, 09:05 AM  #2  
Senior Member Joined: Jun 2014 From: USA Posts: 363 Thanks: 26  Quote:
 

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