Help With a Couple Short Proofs

[AplanisTophet]

Please read and check the following two short proofs, which I presume to be inaccurate given what is proven, as I am looking to know what is wrong as the case may be. Thank you in advance!:

 

1) Let [latex]q_1, q_2, q_3,[/latex] … be an enumeration of [latex]A = \mathbb{Q} \cap (0,1)[/latex].

 

2) Select a real number [latex]r \in (0,1)[/latex] uniformly at random. Note: A common way of doing this cited by mathematicians is the theoretical flipping of a coin infinitely many times followed by some trivial 'clean-up' given that there are two binary expansions of each dyadic rational, e.g. 0.1 = 0.01111111...

 

3) Let [latex]B = \mathbb{Q} \cap (-r, 1-r)[/latex]. Note that [latex]r \in (0,1) \Rightarrow 0 \in B[/latex]. Also note that the length of [latex](-r, 0)[/latex] is completely dependent upon [latex]r[/latex] which was selected uniformly at random, so the length has likewise been selected uniformly at random.

 

4) Let [latex]f:A \rightarrow B[/latex] be a bijection where a constant [latex]k \in A[/latex] such that [latex]f(a) = a - k[/latex].* This step is to relate some [latex]q_i[/latex] to the element [latex]0 \in B[/latex] in a way that allows any [latex]i \in \mathbb{N}[/latex] a chance of being selected uniformly at random, just as [latex]r[/latex] and the length of [latex](-r, 0)[/latex] were. There will be one and only one possible constant [latex]k[/latex] for any [latex]r \in \mathbb{R}(0, 1)[/latex] such that [latex]f[/latex] is a bijection given [latex]f(a) = a - k[/latex].

 

5) Let [latex]C = \{ r + q : q \in B \}[/latex]. Note that [latex]0 \in B \Rightarrow r \in C[/latex].

 

6) Let [latex]g:B \rightarrow C[/latex] be a bijective function: [latex]g(b) = b + r[/latex]. Note that [latex]0 \in B \Rightarrow \exists g(0) = r[/latex].

 

7) [latex]\exists i \in \mathbb{N} : f(q_i) = 0 \Rightarrow g(f(q_i)) = g(0) = r[/latex]. Therefore, where [latex]r[/latex] was selected uniformly at random, so was [latex]i[/latex].

 

 

* The cardinality of A is equal to the cardinality of B, so a bijection can exist. There is something interesting here, however. Note that [latex]z:\mathbb{R}(0,1) \rightarrow \mathbb{R}(-r, 1-r)[/latex] is a bijection where [latex]z(x) = x - r[/latex], but [latex]q \in A \Rightarrow z(q) \notin B[/latex] if [latex]r \notin \mathbb{Q}[/latex] because any rational less an irrational is not rational. I therefore suggest that [latex]\exists k \in A[/latex] such that [latex]f:A \rightarrow B[/latex] is a bijection where [latex]f(a) = a - k[/latex]. This poses the question, what is the result [latex]t[/latex] of the equation: [latex](1/3 - 0) - (f(1/3) - z(0)) = 1/3 - ((1/3 - k) - (0 - r)) = 1/3 - 1/3 + k - r = k - r = t[/latex]? What is unusual is that [latex]k - r = t[/latex] must be all but infinitesimal… It is the smallest difference between two disjoint “shifted copies” of the rational numbers in [latex]\mathbb{R}[/latex] \ [latex]\mathbb{Q}[/latex].

 

In keeping with the above definitions, we can continue:

 

1) For each [latex]r \in \mathbb{R}(0,1)[/latex], there will exist a constant [latex]k \in A[/latex] such that [latex]f:A \rightarrow B[/latex] is a bijection where [latex]f(a) = a - k[/latex].

 

2) A Vitali set [latex]V[/latex] would produce a different [latex]k[/latex] for each [latex]v \in V[/latex]. More specifically, the length of [latex](0, k) = k[/latex] will be unique for any element from each disjoint “shifted copy” of the rationals in the quotient group [latex]\mathbb{R} / \mathbb{Q}[/latex] (where each element of the quotient group is of a form [latex]\mathbb{Q} + r = \{ r + q : q \in \mathbb{Q} \}[/latex] for some [latex]r \in \mathbb{R}[/latex]), and there exists one and only one [latex]v \in V[/latex] for each element of [latex]\mathbb{R} / \mathbb{Q}[/latex]. Note: Vitali sets are usually restricted to [0, 1], not (0, 1), be we can trivially assume such a restriction to (0,1) here. I use them here for simplicity despite their creation relying on the axiom of choice, noting this proof could also be carried out without the use of Vitali sets and the axiom of choice.

 

3) For each [latex]v \in V[/latex], let [latex]v \sim k \Rightarrow f(a) = a - k[/latex] is a bijection from [latex]A[/latex] onto [latex]B = \mathbb{Q} \cap (-v, 1-v)[/latex].

 

4) Where each possible [latex]k \in A[/latex], the set [latex]D = \{ k : \exists v \in V[/latex] where [latex]v \sim k \}[/latex] has a cardinality equal to the cardinality of [latex]\mathbb{N}[/latex].

 

5) Let [latex]d_1, d_2, d_3,[/latex] ... be an enumeration of [latex]D[/latex].

 

6) Let [latex]\mathbb{R}(0,1) / \mathbb{Q} = \{ x \cap \mathbb{R}(0, 1) : x \in \mathbb{R} / \mathbb{Q} \}[/latex]. The elements of [latex]\mathbb{R}(0,1) / \mathbb{Q}[/latex] therefore partition [latex]\mathbb{R}(0,1)[/latex] by definition (just as [latex]\mathbb{R} / \mathbb{Q}[/latex] is a partition of [latex]\mathbb{R}[/latex]) and each element of [latex]\mathbb{R}(0,1) / \mathbb{Q}[/latex] is enumerable because it is merely a shifted copy of [latex]A[/latex] equal to [latex]v + B = C = \{ v + q : q \in B \}[/latex] for some [latex]v \in V[/latex].

 

7) Let [latex]E = \bigcup_{i = 1}^\infty (C_{d_i})[/latex], where [latex]C_{d_i} = \{ v + q : v \sim d_i \land q \in B = \mathbb{Q} \cap (-v, 1-v) \}[/latex].

 

8 ) Each [latex]C_{d_i}[/latex] is enumerable by definition so the union [latex]E[/latex] is a countable union of countable sets that is also enumerable.

 

9) The union [latex]E[/latex] is also equal to [latex]\mathbb{R}(0,1)[/latex] because [latex]\{ C_{d_i} : i \in \mathbb{N} \} = \mathbb{R}(0,1) / \mathbb{Q}[/latex] is a partition of [latex]\mathbb{R}(0,1)[/latex] by definition.

 

10) The fact that [latex]E[/latex] is enumerable and [latex]E = \mathbb{R}(0,1)[/latex] by definition [latex]\Rightarrow \mathbb{R}(0,1)[/latex] is enumerable.

 

Ok, I’ve now contradicted Cantor’s Theorem, so unless math is broken (which I doubt), please help me find what’s wrong with the above!