User:RJGray/Sandboxcantor2

Let the smaller of the two numbers ωκ 1, ωκ 2 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 ωμ 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 ωκ 3 and ωκ 4 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 ωκ 3, ωκ 4 be denoted by α  , the larger by β .

Then one has: α' < α  < β  < β', κ2 < κ3 < κ4; and one sees that all numbers ωμ 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 ωκ 2ν - 1 and ωκ 2ν ) 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 ωμ, for which μ ≤ κ2ν, definitely do not lie in its interior. Since obviously: κ1 < κ2 < κ3 <. . ., ωκ 2ν – 2 < ωκ 2ν – 1 < ωκ 2ν,. .. and these numbers, as indices, are whole numbers, so: κ2ν ≥ 2ν, and hence: ν < κ2ν; thus, we can certainly say (and this is sufficient for the following):

''That if ν is an arbitrary whole number, the [real] quantity ων 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 (an, bn), 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 an + 1 and bn + 1; namely, an + 1 is the least of these two numbers and bn + 1 is the greatest. Cantor's 1874 proof demonstrates that for all n: xn ∉ (an, bn). The sequence an is increasing and bounded above by b, so it has a limit a∞, which satisfies an < a∞. The sequence bn is decreasing and bounded below by a, so it has a limit b∞, which satisfies b∞ < bn. Also, an < bn implies a∞ ≤ b∞. Therefore, an < a∞ ≤ b∞ < bn. If a∞ < b∞, then for every n: xn ∉ (a∞, b∞) because xn is not in the larger interval (an, bn). This contradicts P being dense in [a, b]. Therefore, a∞ = b∞. Since for all n: a∞ ∈ (an, bn) but xn ∉ (an, bn), 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
$$\qquad\quad\;\,\textbf{Fraktur}$$ $$\qquad\quad\text{A}\quad\;\;\mathfrak{A}\quad\mathfrak{a}$$ $$\qquad\quad\text{B}\quad\;\;\mathfrak{B}\quad\mathfrak{b}$$ $$\qquad\quad\text{C}\quad\;\;\mathfrak{C}\quad\mathfrak{c}$$ $$\qquad\quad\text{D}\quad\;\;\mathfrak{D}\quad\mathfrak{d}$$ $$\qquad\quad\text{E}\quad\;\;\mathfrak{E}\quad\mathfrak{e}$$ $$\qquad\quad\text{F}\quad\;\;\mathfrak{F}\quad\mathfrak{f}$$ $$\qquad\quad\text{G}\quad\;\;\mathfrak{G}\quad\mathfrak{g}$$ $$\qquad\quad\text{H}\quad\;\;\mathfrak{H}\quad\mathfrak{h}$$ $$\qquad\quad\text{I}\quad\;\;\;\,\mathfrak{I}\quad\;\mathfrak{i}$$ $$\qquad\quad\text{J}\quad\;\;\;\,\mathfrak{J}\quad\;\mathfrak{j}$$ $$\qquad\quad\text{K}\quad\;\;\;\mathfrak{K}\quad\;\mathfrak{k}$$ $$\qquad\quad\text{L}\quad\;\;\;\,\mathfrak{L}\quad\;\mathfrak{l}$$ $$\qquad\quad\text{M}\quad\;\;\mathfrak{M}\quad\mathfrak{m}$$ $$\qquad\quad\text{N}\quad\;\;\mathfrak{N}\quad\mathfrak{n}$$ $$\qquad\quad\text{O}\quad\;\;\mathfrak{O}\quad\mathfrak{o}$$ $$\qquad\quad\text{P}\quad\;\;\mathfrak{P}\quad\mathfrak{p}$$ $$\qquad\quad\text{Q}\quad\;\;\mathfrak{Q}\quad\mathfrak{q}$$ $$\qquad\quad\text{R}\quad\;\;\mathfrak{R}\quad\mathfrak{r}$$ $$\qquad\quad\text{S}\quad\;\;\mathfrak{S}\quad\mathfrak{s}$$ $$\qquad\quad\text{T}\quad\;\;\mathfrak{T}\quad\mathfrak{t}$$ $$\qquad\quad\text{U}\quad\;\;\mathfrak{U}\quad\mathfrak{u}$$ $$\qquad\quad\text{V}\quad\;\;\mathfrak{V}\quad\mathfrak{v}$$ $$\qquad\quad\text{W}\quad\;\;\!\mathfrak{W}\quad\mathfrak{w}$$ $$\qquad\quad\text{X}\quad\;\;\mathfrak{X}\quad\mathfrak{x}$$ $$\qquad\quad\text{Y}\quad\;\;\mathfrak{Y}\quad\mathfrak{y}$$ $$\qquad\quad\text{Z}\quad\;\;\mathfrak{Z}\quad\;\mathfrak{z}$$