User:RJGray/Sandboxcantor2

     Let the smaller of the two numbers ω_κ1, ω_κ2 be denoted by α', the larger by β'. (Their equality is impossible because we assumed that our sequence consists of nothing but unequal numbers.)

     Then according to the definition:
            α < α' < β' < β ,
furthermore:
                 κ₁ < κ₂ ;
and all numbers ω_μ of our sequence, for which μ ≤ κ₂, do not lie in the interior of the interval [α', β'], as is immediately clear from the definition of the numbers κ₁, κ₂. Similarly, let ω_κ3 and ω_κ4 be the two numbers of our sequence with smallest indices that fall in the interior of the interval [α', β'] and let the smaller of the numbers ω_κ3, ω_κ4 be denoted by α'', the larger by β''.

     Then one has:
            α' < α'' < β'' < β' ,
              κ₂ < κ₃ < κ₄ ;
and one sees that all numbers ω_μ of our sequence, for which μ ≤ κ₄, do not fall into the interior of the interval [α'', β''].

      After one has followed this rule to reach an interval [α^{(ν - 1)}, β^{(ν - 1)}], the next interval is produced by selecting the first two (i. e. with lowest indices) numbers of our sequence (ω) (let them be ω_{κ2ν - 1} and ω_κ2ν) that fall into the interior of [α^{(ν - 1)}, β^{(ν - 1)}]. Let the smaller of these two numbers be denoted by α^(ν), the larger by β^(ν).

     The interval [α^(ν), β^(ν)] then lies in the interior of all preceding intervals and has the specific relation with our sequence (ω) that all numbers ω_μ, for which μ ≤ κ_2ν, definitely do not lie in its interior. Since obviously:
            κ₁ < κ₂ < κ₃ < . . . , ω_{κ2ν – 2} < ω_{κ2ν – 1} < ω_κ2ν , . . .
and these numbers, as indices, are whole numbers, so:
            κ_2ν ≥ 2ν ,
and hence:
            ν < κ_2ν ;
thus, we can certainly say (and this is sufficient for the following):

     That if ν is an arbitrary whole number, the [real] quantity ω_ν lies outside the interval [α^(ν) . . . β^(ν)].

Cantor's new proof first handles the easy case of the sequence P not being dense in the interval. Then it deals with the more difficult case of P being dense. This division into cases not only indicates which sequences are most difficult to handle, but it also reveals the important role denseness plays in the proof.

In the first case, P is not dense in [a, b]. By definition, P is dense in [a, b] if and only if for all subintervals (c, d) of [a, b], there is an x ∈ P such that x ∈ (c, d). Taking the negation of each side of the "if and only if" produces: P is not dense in [a, b] if and only if there exists a subinterval (c, d) of [a, b] such that for all x ∈ P : x ∉ (c, d). Therefore, every number in (c, d) is not contained in the sequence P. This case handles case 1 and case 3 of Cantor's 1874 proof.

In the second case, P is dense in [a, b]. The denseness of P is used to recursively define a nested sequence of intervals that excludes all elements of P. The definition begins with a₁ = a and b₁ = b. The definition's inductive case starts with the interval (a_n, b_n), which because of the denseness of P contains infinitely many elements of P. From these elements of P, the two with smallest indices are used to define a_n + 1 and b_n + 1; namely, a_n + 1 is the least of these two numbers and b_n + 1 is the greatest. Cantor's 1874 proof demonstrates that for all n :  x_n ∉ (a_n, b_n). The sequence a_n is increasing and bounded above by b, so it has a limit a_∞, which satisfies a_n < a_∞. The sequence b_n is decreasing and bounded below by a, so it has a limit b_∞, which satisfies b_∞ < b_n. Also, a_n < b_n implies a_∞ ≤ b_∞. Therefore, a_n < a_∞ ≤ b_∞ < b_n. If a_∞ < b_∞, then for every n: x_n ∉ (a_∞, b_∞) because x_n is not in the larger interval (a_n, b_n). This contradicts P being dense in [a, b]. Therefore, a_∞ = b_∞. Since for all n: a_∞ ∈ (a_n, b_n) but x_n ∉ (a_n, b_n), the limit a_∞ is a real number that is not contained in the sequence P. This case handles case 2 of Cantor's 1874 proof.

Fraktur

$\qquad \quad \;\,{\textbf {Fraktur}}$

$\qquad \quad {\text{A}}\quad \;\;{\mathfrak {A}}\quad {\mathfrak {a}}$

$\qquad \quad {\text{B}}\quad \;\;{\mathfrak {B}}\quad {\mathfrak {b}}$

$\qquad \quad {\text{C}}\quad \;\;{\mathfrak {C}}\quad {\mathfrak {c}}$

$\qquad \quad {\text{D}}\quad \;\;{\mathfrak {D}}\quad {\mathfrak {d}}$

$\qquad \quad {\text{E}}\quad \;\;{\mathfrak {E}}\quad {\mathfrak {e}}$

$\qquad \quad {\text{F}}\quad \;\;{\mathfrak {F}}\quad {\mathfrak {f}}$

$\qquad \quad {\text{G}}\quad \;\;{\mathfrak {G}}\quad {\mathfrak {g}}$

$\qquad \quad {\text{H}}\quad \;\;{\mathfrak {H}}\quad {\mathfrak {h}}$

$\qquad \quad {\text{I}}\quad \;\;\;\,{\mathfrak {I}}\quad \;{\mathfrak {i}}$

$\qquad \quad {\text{J}}\quad \;\;\;\,{\mathfrak {J}}\quad \;{\mathfrak {j}}$

$\qquad \quad {\text{K}}\quad \;\;\;{\mathfrak {K}}\quad \;{\mathfrak {k}}$

$\qquad \quad {\text{L}}\quad \;\;\;\,{\mathfrak {L}}\quad \;{\mathfrak {l}}$

$\qquad \quad {\text{M}}\quad \;\;{\mathfrak {M}}\quad {\mathfrak {m}}$

$\qquad \quad {\text{N}}\quad \;\;{\mathfrak {N}}\quad {\mathfrak {n}}$

$\qquad \quad {\text{O}}\quad \;\;{\mathfrak {O}}\quad {\mathfrak {o}}$

$\qquad \quad {\text{P}}\quad \;\;{\mathfrak {P}}\quad {\mathfrak {p}}$

$\qquad \quad {\text{Q}}\quad \;\;{\mathfrak {Q}}\quad {\mathfrak {q}}$

$\qquad \quad {\text{R}}\quad \;\;{\mathfrak {R}}\quad {\mathfrak {r}}$

$\qquad \quad {\text{S}}\quad \;\;{\mathfrak {S}}\quad {\mathfrak {s}}$

$\qquad \quad {\text{T}}\quad \;\;{\mathfrak {T}}\quad {\mathfrak {t}}$

$\qquad \quad {\text{U}}\quad \;\;{\mathfrak {U}}\quad {\mathfrak {u}}$

$\qquad \quad {\text{V}}\quad \;\;{\mathfrak {V}}\quad {\mathfrak {v}}$

$\qquad \quad {\text{W}}\quad \;\;\!{\mathfrak {W}}\quad {\mathfrak {w}}$

$\qquad \quad {\text{X}}\quad \;\;{\mathfrak {X}}\quad {\mathfrak {x}}$

$\qquad \quad {\text{Y}}\quad \;\;{\mathfrak {Y}}\quad {\mathfrak {y}}$

$\qquad \quad {\text{Z}}\quad \;\;{\mathfrak {Z}}\quad \;{\mathfrak {z}}$