User:RJGray/Cantor draft

End of Cantor's 1879 proof

The sequence a_n is increasing and bounded above by b, so the limit a_∞ = lim_n → ∞ a_n exists. Similarly, the limit b_∞ = lim_n → ∞ b_n exists because the sequence b_n is decreasing and bounded below by a. Also, a_n < b_n for all n implies a_∞ ≤ b_∞. However, if a_∞ < b_∞, then x_n ∉ (a_∞, b_∞) for all n since x_n is not in the larger interval (a_n, b_n). This contradicts P being dense in [a, b]. Hence, a_∞ = b_∞. So for all n, a_∞ ∈ (a_n, b_n) but x_n ∉ (a_n, b_n). Therefore, a_∞ is a number in [a, b] that is not contained in P.^{[proof 1]}

Cantor's 1879 uncountability proof

Everywhere dense

In 1879, Cantor published a new uncountability proof that modifies his 1874 proof. He first defines the topological notion of a point set P being "everywhere dense in an interval":^[A]

If P lies partially or completely in the interval [α, β], then the remarkable case can happen that every interval [γ, δ] contained in [α, β], no matter how small, contains points of P. In such a case, we will say that P is everywhere dense in the interval [α, β].^[B]

In this discussion of Cantor's proof: a, b, c, d are used instead of α, β, γ, δ. Also, Cantor only uses the interval notation [a, b] when a < b.

Since the discussion of Cantor's 1874 proof was simplified by using open intervals rather than closed intervals, the same simplification is used here. This requires an equivalent definition of everywhere dense: A set P is everywhere dense in the interval [a, b] if and only if every open subinterval (c, d) of [a, b] contains at least one point of P.^[1]

Cantor did not specify how many points of P an open subinterval (c, d) must contain. He did not need to specify this because the assumption that every open subinterval contains at least one point of P implies that every open subinterval contains infinitely many points of P. This is proved by generating a sequence of points belonging to both P and (c, d). Since P is dense in [a, b], the subinterval (c, d) contains at least one point x₁ of P. By assumption, the subinterval (x₁, d) contains at least one point x₂ of P and x₂ > x₁ since x₂ belongs to this subinterval. In general, after generating x_n, the subinterval (x_n, d) is used to generate a point x_n + 1 satisfying x_n + 1 > x_n. The infinitely many points x_n belong to both P and (c, d).

Cantor's 1879 proof

Cantor modified his 1874 proof with a new proof of its second theorem: Given any sequence P of real numbers x₁, x₂, x₃, ... and any interval [a, b], there is a number in [a, b] that is not contained in P. Cantor's new proof has two cases (his 1874 proof has three). First, it handles the case of 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 more difficult to handle, but it also reveals the important role denseness plays in the proof. ** TEXT OF CANTOR'S 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 is 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.^{[proof 1]} This case handles case 1 and case 3 of Cantor's 1874 proof.

In the second case, which handles case 2 of Cantor's 1874 proof, P is dense in [a, b]. The denseness of sequence P is used to recursively define a nested sequence of intervals that excludes all of the numbers in P. The base case starts with the interval (a, b). Since P is dense in [a, b], there are infinitely many numbers of P in (a, b). Let x_k₁ be the number with the least index and x_k₂ be the number with the next larger index, and let a₁ be the smaller and b₁ be the larger of these two numbers. Then, k₁ < k₂, a < a₁ < b₁ < b, and (a₁, b₁) is a proper subinterval of (a, b). Also, x_m ∉ (a₁, b₁) for m ≤ k₂ since these x_m are the endpoints of (a₁, b₁). The base case repeats the above proof with the interval (a₁, b₁) to obtain x_k₃, x_k₄, a₂, b₂ such that k₁ < k₂ < k₃ < k₄ and a < a₁ < a₂ < b₂ < b₁ < b and x_m ∉ (a₂, b₂) for m ≤ k₄.^{[proof 1]}

The recursive step starts with the interval (a_n–1, b_n–1), the inequalities k₁ < k₂ < . . . < k_2n–2 < k_2n–1 and a < a₁ < . . . < a_n–1 < b_n–1 . . . < b₁ < b, and the statement that the interval (a_n–1, b_n–1) excludes the first 2n –2 members of the sequence P — that is, x_m ∉ (a_n–1, b_n–1) for m ≤ k_2n–2. Since P is dense in [a, b], there are infinitely many numbers of P in (a_n–1, b_n–1). Let x_{k_2n –1} be the number with the least index and x_{k_2n} be the number with the next larger index, and let a_n be the smaller and b_n be the larger of these two numbers. Then, k_2n –1 < k_2n, a_n–1 < a_n < b_n < b_n–1, and (a_n, b_n) is a proper subinterval of (a_n–1, b_n–1). Combining these inequalities with the ones for step n –1 of the recursion produces k₁ < k₂ < . . . < k_2n–1 < k_2n and a < a₁ < . . . < a_n < b_n . . . < b₁ < b. Also, x_m ∉ (a_n, b_n) for m = k_2n–1 and m = k_2n since these x_m are the endpoints of (a_n, b_n). This together with the statement that (a_n–1, b_n–1) excludes the first 2n –2 members of the sequence P implies that the interval (a_n, b_n) excludes the first 2n members of the sequence P — that is, x_m ∉ (a_n, b_n) for m ≤ k_2n. Therefore, for all n, x_n ∉ (a_n, b_n) since n ≤ k_2n.^{[proof 1]}

The sequence a_n is increasing and bounded by b, so the limit a_∞ = lim_n → ∞ a_n exists. Similarly, the limit b_∞ = lim_n → ∞ b_n exists since the sequence b_n is decreasing and bounded by a. Also, a_n < b_n implies a_∞ ≤ b_∞. 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]. Hence, a_∞ = b_∞. So for all n, a_∞ ∈ (a_n, b_n) but x_n ∉ (a_n, b_n). Therefore, a_∞ is a number in [a, b] that is not contained in P.^{[proof 1]}

Cantor's 1874 & 1879 proof

To find a number in [a, b] that is not contained in the given sequence, construct two sequences of real numbers as follows: Find the first two numbers of the given sequence that are in the open interval (a, b). Denote the smaller of these two numbers by a₁ and the larger by b₁. Similarly, find the first two numbers of the given sequence that are in (a₁, b₁). Denote the smaller by a₂ and the larger by b₂. Continuing this procedure generates a sequence of intervals (a₁, b₁), (a₂, b₂), (a₃, b₃), ... such that each interval in the sequence contains all succeeding intervals — that is, it generates a sequence of nested intervals. This implies that the sequence a₁, a₂, a₃, ... is increasing and the sequence b₁, b₂, b₃, ... is decreasing.^[2]

Cantor's 1879 proof

The sequence a_n is increasing and bounded above by b, so the limit a_∞ = lim_n → ∞ a_n exists. Similarly, the limit b_∞ = lim_n → ∞ b_n exists because the sequence b_n is decreasing and bounded below by a. Also, a_n < b_n for all n implies a_∞ ≤ b_∞. However, if a_∞ < b_∞, then x_n ∉ (a_∞, b_∞) for all n since x_n is not in the larger interval (a_n, b_n). This contradicts P being dense in [a, b], so a_∞ = b_∞. Now, for every n, a_∞ ∈ (a_n, b_n) but x_n ∉ (a_n, b_n). Therefore, a_∞ is a number in [a, b] that is not contained in P.^{[proof 1]}

ADD to end of Cantor's 1879 proof

In the Example of Cantor's construction, each successive nested interval excludes rational numbers for two different reasons. It will exclude the finitely many rationals visited in the search for the first two rationals within the interval (these two rationals will have the least indices). These rationals are then used to form an interval that excludes the rationals visited in the search along with infinitely many more rationals. However, it still contains infinitely many rationals since our sequence of rationals is dense in [0, 1]. Forming this interval from the two rationals with the least indices guarantees that this interval excludes an initial segment of our sequence that contains at least two more members than the preceding initial segment. Since the denseness of our sequence guarantees that this process never ends, all rationals will be excluded.^{[proof 1]} Because of the ordering of the rationals in our sequence, the intersection of the nested intervals is the set {√2 − 1}.^[C]

Notes

^ Cantor was not the first to define "everywhere dense" but his terminology was adopted with or without the "everywhere" (everywhere dense: Arkhangel'skii & Fedorchuk 1990, p. 15; dense: Kelley 1991, p. 49). In 1870, Hermann Hankel had defined this concept using different terminology: "a multitude of points … fill the segment if no interval, however small, can be given within the segment in which one does not find at least one point of that multitude" (Ferreirós 2007, p. 155). Hankel was building on Peter Gustav Lejeune Dirichlet's 1829 article that contains the Dirichlet function, a non-(Riemann) integrable function whose value is 0 for rational numbers and 1 for irrational numbers. (Ferreirós 2007, p. 149.)
^ The original German text from Cantor 1879, p. 2 (Cantor's closed set notation (α . . . β) is translated to [α, β]): Liegt P theilweise oder ganz im Intervalle (α . . . β), so kann der bemerkenswerthe Fall eintreten, dass jedes noch so kleine in (α . . . β) enthaltene Intervall (γ . . . δ) Punkte von P enthält. In einem solchen Falle wollen wir sagen, dass P im Intervalle (α . . . β) überall-dicht sei.
^ Different orderings can produce nested intervals whose intersection is {x} for any x: Let x_n be the decimal approximation of x to n-digits. Starting with the interval [x₀, x₀ + 1] and continuing with the intervals [x_n, x_n + 1/10ⁿ ] for n ≥ 1 produces a sequence of intervals whose intersection is {x}.

NEW PROOF

Second theorem

Only the first part of Cantor's second theorem needs to be proved. It states: Given any sequence of real numbers x₁, x₂, x₃, ... and any interval [a, b], there is a number in [a, b] that is not contained in the given sequence.^[A]

To find a number in [a, b] that is not contained in the given sequence, construct two sequences of real numbers as follows: Find the first two numbers of the given sequence that are in the open interval (a, b). Denote the smaller of these two numbers by a₁ and the larger by b₁. Similarly, find the first two numbers of the given sequence that are in (a₁, b₁). Denote the smaller by a₂ and the larger by b₂. Continuing this procedure generates a sequence of intervals (a₁, b₁), (a₂, b₂), (a₃, b₃), ... such that each interval in the sequence contains all succeeding intervals — that is, it generates a sequence of nested intervals. This implies that the sequence a₁, a₂, a₃, ... is increasing and the sequence b₁, b₂, b₃, ... is decreasing.^[3]

Either the number of intervals generated is finite or infinite. If finite, let (a_L, b_L) be the last interval. If infinite, take the limits a_∞ = lim_n → ∞ a_n and b_∞ = lim_n → ∞ b_n. Since a_n < b_n for all n, either a_∞ = b_∞ or a_∞ < b_∞. Thus, there are three cases to consider:

Case 1: There is a last interval (a_L, b_L). Since at most one x_n can be in this interval, every y in this interval except x_n (if it exists) is not contained in the given sequence.

Case 2: a_∞ = b_∞. Then a_∞ is not contained in the given sequence since for all n : a_∞ belongs to the interval (a_n, b_n) but x_n does not belong to (a_n, b_n). In symbols, a_∞ ∈ (a_n, b_n) but x_n ∉ (a_n, b_n).

Proof that for all n : x_n ∉ (a_n, b_n)
This is implied by the stronger result: For all n, (a_n, b_n) excludes x₁, ... , x_2n, which is proved by induction. Basis step: If a₁ = x_j, b₁ = x_k, and E₁ = max(j, k), then (a₁, b₁) excludes x₁, ... , x_E₁. Also, E₁ ≥ 2 since the interval (a₁, b₁) excludes its two endpoints. Inductive step: Assume that (a_n, b_n) excludes x₁, ... , x_{E_n} and E_n ≥ 2n. If a_n+1 = x_j, b_n+1 = x_k, and E_n+1 = max(j, k), then (a_n+1, b_n+1) excludes x₁, ... , x_{E_n+1}. Also, E_n+1 ≥ E_n + 2 ≥ 2n + 2 = 2(n + 1) since the interval (a_n+1, b_n+1) excludes the numbers that (a_n, b_n) excludes plus the two endpoints a_n+1 and b_n+1. Therefore, for all n : (a_n, b_n) excludes x₁, ... , x_2n. (This proof is similar to a proof that Cantor published in 1879. The main difference is that Cantor's proof is embedded in a larger proof and uses notation from it. Our proof does not depend on this notation.^{[proof 1]})

Case 3: a_∞ < b_∞. Then every y in [a_∞, b_∞] is not contained in the given sequence since for all n : y belongs to (a_n, b_n) but x_n does not.^[4]

The proof is complete since, in all cases, at least one real number in [a, b] has been found that is not contained in the given sequence.^[B]

Cantor's proofs are constructive and have been used to write a computer program that generates the digits of a transcendental number. This program applies Cantor's construction to a sequence containing all the real algebraic numbers between 0 and 1. The article that discusses this program gives some of its output, which shows how the construction generates a transcendental.^[5]

Cantor's 1879 proof is the same as his 1874 proof except for a new proof of the first part of his second theorem: Given any sequence P of real numbers x₁, x₂, x₃, ... and any interval [a, b], there is a number in [a, b] that is not contained in the sequence P. The new proof has only two cases. **Big proof ref goes here!**

Cantor's new proof first handles the 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.^{[proof 1]}

In the first case, P is not dense in [a, b]. By definition, P is dense if and only if for all subsets (c, d) of [a, b], there is an x_n in P such that x_n is in (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 subset (c, d) of [a, b] such that for all x_n in P, x_n is not in (c, d). Therefore, every number in (c, d) is not contained in the sequence P.^{[proof 1]} This case handles case 1 and case 3 of Cantor's 1874 proof. In the diagram for case 1, every x_n in (a_L, y) is not in the sequence. In the diagram for case 3, every x_n in (a_∞, b_∞) is not in the sequence.

In the second case, P is dense in [a, b]. The denseness of P is used to recursively define a nested sequence of intervals is used to prove that this nested sequence excludes all members of P. The base case starts with the observation that since P is dense in [a, b], there are infinitely many real numbers in P that belong to (a, b). Let x_k₁ be the real number with the least index k₁ and let x_k₂ be the real number with the next larger index k₂. Let a₁ be the smaller of these two reals and let b₁ be the larger. Then, a < a₁ < b₁ < b and k₁ < k₂. Since x_k₁ and x_k₂ are the reals in P with the least indexes, for all j ≤ k₂, **MAY NEED MORE** the numbers x_j are not in (a₁, b₁). **MAY NEED MORE** May want to prove this! If j < k2, then if xj is in (a1, b1), it canoot be xj in interval or xj would have been chosen instead of xk+1 or xk+2!!

The recursive part of the definition starts with the interval (a_n, b_n) and with the inequalities a < a₁ < ... < a_n < b_n < ... < b₁ < b and k₁ < k₂ < ... < k_2n − 1 < k_2n such that for all j ≤ k_2n, the numbers x_j are not in (a_n, b_n). Since P is dense in [a, b], there are infinitely many real numbers in P that belong to (a_n, b_n). Let x_{k_2n + 1} be the real number with the least index k_2n + 2 and let x_2n + 2 be the real number with the next larger index k_2n + 2. Let a_n + 1 be the smaller of these two reals and let b_n + 1 be the larger. Then, a_n < a_n + 1 < b_n + 1 < b_n and k_2n + 1 < k_2n + 2. Since x_k₁ and x_k₂ are the reals in P with the least indexes **MAY NEED MORE** for all j ≤ k_2n + 2, the numbers x_j are not in (a_n + 1, b_n + 1). **MAY NEED MORE HERE!**

The sequence a_n is increasing and bounded above by b, so it has a limit a_∞, which satisfies a_n ≤ a_∞ for all n. The sequence b_n is decreasing and bounded below by a, so it has a limit b_∞, which satisfies b_∞ ≤ b_n for all 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.^{[proof 1]} **MAY NEED MORE HERE!** This case handles case 2 of Cantor's 1874 proof.

Second theorem

Only the first part of Cantor's second theorem needs to be proved. It states: Given any sequence of real numbers x₁, x₂, x₃, ... and any interval [a, b], there is a number in [a, b] that is not contained in the given sequence.^[C]

To find a number in [a, b] that is not contained in the given sequence, construct two sequences of real numbers as follows: Find the first two numbers of the given sequence that are in the open interval (a, b). Denote the smaller of these two numbers by a₁ and the larger by b₁. Similarly, find the first two numbers of the given sequence that are in (a₁, b₁). Denote the smaller by a₂ and the larger by b₂. Continuing this procedure generates a sequence of intervals (a₁, b₁), (a₂, b₂), (a₃, b₃), ... such that each interval in the sequence contains all succeeding intervals—that is, it generates a sequence of nested intervals. This implies that the sequence a₁, a₂, a₃, ... is increasing and the sequence b₁, b₂, b₃, ... is decreasing.^[6]

Either the number of intervals generated is finite or infinite. If finite, let (a_L, b_L) be the last interval. If infinite, take the limits a_∞ = lim_n → ∞ a_n and b_∞ = lim_n → ∞ b_n. Since a_n < b_n for all n, either a_∞ = b_∞ or a_∞ < b_∞. Thus, there are three cases to consider:

Case 1: There is a last interval (a_L, b_L). Since at most one x_n can be in this interval, every y in this interval except x_n (if it exists) is not contained in the given sequence.

Case 2: a_∞ = b_∞. Then a_∞ is not contained in the given sequence since for all n : a_∞ belongs to the interval (a_n, b_n), but as Cantor observes, x_n does not.

Proof that for all n : x_n ∉ (a_n, b_n)}}
This is implied by the stronger result: For all n, (a_n, b_n) excludes x₁, ... , x_2n, which is proved by induction. Basis step: If a₁ = x_j, b₁ = x_k, and E₁ = max(j, k), then (a₁, b₁) excludes x₁, ... , x_E₁. Also, E₁ ≥ 2 since the interval (a₁, b₁) excludes its two endpoints. Inductive step: Assume that (a_n, b_n) excludes x₁, ... , x_{E_n} and E_n ≥ 2n. If a_n+1 = x_j, b_n+1 = x_k, and E_n+1 = max(j, k), then (a_n+1, b_n+1) excludes x₁, ... , x_{E_n+1}. Also, E_n+1 ≥ E_n + 2 ≥ 2n + 2 = 2(n + 1) since the interval (a_n+1, b_n+1) excludes the numbers that (a_n, b_n) excludes plus the two endpoints a_n+1 and b_n+1. Therefore, for all n : (a_n, b_n) excludes x₁, ... , x_2n. (This proof is similar to a proof that Cantor published in 1879. The main difference is that Cantor's proof is embedded in a larger proof and uses notation from it. Our proof does not depend on this notation.^{[proof 1]})

Case 3: a_∞ < b_∞. Then every y in [a_∞, b_∞] is not contained in the given sequence since for all n : y belongs to (a_n, b_n) but x_n does not.^[4]

In the first case, P is not dense in [a, b]. By definition, P is dense if and only if for all (c, d) ⊆ [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 (c, d) ⊆ [a, b] such that for all x ∈ P, we have x ∉ (c, d). Thus, every number in (c, d) is not contained in the sequence P.^{[proof 1]} 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 c, which satisfies a_n < c. The sequence b_n is decreasing and bounded below by a, so it has a limit d, which satisfies d < b_n. Also, a_n < b_n implies c ≤ d. Therefore, a_n < c ≤ d < b_n. If c < d, then for every n: x_n ∉ (c, d) because x_n is not in the larger interval (a_n, b_n). This contradicts P being dense in [a, b]. Therefore, c = d. Since for all n: c ∈ (a_n, b_n) but x_n ∉ (a_n, b_n), the limit c is a real number that is not contained in the sequence P.^{[proof 1]} This case handles case 2 of Cantor's 1874 proof.

Notes

^ This implies the rest of the theorem — namely, there are infinitely many numbers in [a, b] that are not contained in the given sequence. For example, let $[0,1]$ be the interval and consider its subintervals $[0,{\tfrac {1}{2}}],$ $[{\tfrac {3}{4}},{\tfrac {7}{8}}],$ $[{\tfrac {15}{16}},{\tfrac {31}{32}}],$ $\ldots .$ Since these subintervals are pairwise disjoint, applying the first part of the theorem to each subinterval produces infinitely many numbers in $[0,1]$ that are not contained in the given sequence. In general, for the interval $[a,b],$ apply the first part of the theorem to the subintervals $[a,a+{\tfrac {1}{2}}(b-a)],$ $[a+{\tfrac {3}{4}}(b-a),a+{\tfrac {7}{8}}(b-a)],$ $[a+{\tfrac {15}{16}}(b-a),a+{\tfrac {31}{32}}(b-a)],$ $\ldots .$
^ The main difference between Cantor's proof and the above proof is that he generates the sequence of closed intervals [a_n, b_n]. To find a_n + 1 and b_n + 1, he uses the interior of the interval [a_n, b_n], which is the open interval (a_n, b_n). Generating open intervals combines Cantor's use of closed intervals and their interiors, which allows the case diagrams to depict all the details of the proof.
^ This implies the rest of the theorem — namely, there are infinitely many numbers in [a, b] that are not contained in the given sequence. For example, let $[0,1]$ be the interval and consider its subintervals $[0,{\tfrac {1}{2}}],$ $[{\tfrac {3}{4}},{\tfrac {7}{8}}],$ $[{\tfrac {15}{16}},{\tfrac {31}{32}}],$ $\ldots .$ Since these subintervals are pairwise disjoint, applying the first part of the theorem to each subinterval produces infinitely many numbers in $[0,1]$ that are not contained in the given sequence. In general, for the interval $[a,b],$ apply the first part of the theorem to the subintervals $[a,a+{\tfrac {1}{2}}(b-a)],$ $[a+{\tfrac {3}{4}}(b-a),a+{\tfrac {7}{8}}(b-a)],$ $[a+{\tfrac {15}{16}}(b-a),a+{\tfrac {31}{32}}(b-a)],$ $\ldots .$

Accessibility

Accessibility. For screen readers: used Template:lang-de to enclose German sentences in notes.

Deutsch

Seite 5: Dem widerspricht aber ein sehr allgemeiner Satz, welchen wir in Borchardt's Journal, Bd. 77, pag. 260, mit aller Strenge bewiesen haben, nämlich der folgende Satz:

Francais

French: mit

Seite 5: Dem widerspricht aber ein sehr allgemeiner Satz, welchen wir in Borchardt's Journal, Bd. 77, pag. 260, mit aller Strenge bewiesen haben, nämlich der folgende Satz:

Liegt P theilweise oder ganz im Intervalle (α . . . β), so kann der bemerkenswerthe Fall eintreten, dass jedes noch so kleine in (α . . . β) enthaltene Intervall (γ . . . δ) Punkte von P enthält. In einem solchen Falle wollen wir sagen, dass P im Intervalle (α . . . β) überall-dicht sei.}}

"Hat man eine einfach unendliche Reihe
ω₁, ω₂, . . . , ω_ν, . . .
von reellen, ungleichen Zahlen, die nach irgend einem Gesetz fortschreiten, so lässt sich in jedem vorgegebenen, Intervalle (α . . . β) eine Zahl η (und folglich lassen sich deren unendlich viele) angeben, welche nicht in jener Reihe (als Glied derselben) vorkommt."

If P lies partially or completely in the interval [α, β], then the remarkable case can happen that every intervals [γ, δ] contained in [α, β], no matter how small, contains points of P. In such a case, we will say that P is everywhere dense in the interval [α, β].^[A]

Notes

^ The original German text from Cantor 1879, p. 2 (Cantor's closed set notation (α . . . β) is translated to [α, β]): Liegt P theilweise oder ganz im Intervalle (α . . . β), so kann der bemerkenswerthe Fall eintreten, dass jedes noch so kleine in (α . . . β) enthaltene Intervall (γ . . . δ) Punkte von P enthält. In einem solchen Falle wollen wir sagen, dass P im Intervalle (α . . . β) überall-dicht sei.

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Cantor's 1874 & 1879 proof

To find a number in [a, b] that is not contained in the given sequence, construct two sequences of real numbers as follows: Find the first two numbers of the given sequence that are in the open interval (a, b). Denote the smaller of these two numbers by a₁ and the larger by b₁. Similarly, find the first two numbers of the given sequence that are in (a₁, b₁). Denote the smaller by a₂ and the larger by b₂. Continuing this procedure generates a sequence of intervals (a₁, b₁), (a₂, b₂), (a₃, b₃), ... such that each interval in the sequence contains all succeeding intervals — that is, it generates a sequence of nested intervals. This implies that the sequence a₁, a₂, a₃, ... is increasing and the sequence b₁, b₂, b₃, ... is decreasing.^[7]

Cantor's 1879 proof

Cantor's 1879 proof is the same as his 1874 proof except for a new proof of the first part of his second theorem: Given any sequence P of real numbers x₁, x₂, x₃, ... and any interval [a, b], there is a number in [a, b] that is not contained in the sequence P. Cantor's new proof has only two cases: It first handles the 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 more difficult to handle, but it also reveals the important role denseness plays in the proof.^{[proof 1]}

In the first case, P is not dense in [a, b]. By definition, P is dense 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.^{[proof 1]} This case handles case 1 and case 3 of Cantor's 1874.

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. (Cantor's definition of this sequence is same as his 1874 definition and he uses essentially his same α, β notation for the nested sequence of intervals.^[A]) The base case of the recursion is (a₁, b₁) = (a, b). The recursive step starts with the interval (a_n, b_n) which contains infinitely many elements of P since P is dense in [a, b]. WORKING HERE! Then the construction of the next interval follows: The two members of P with the smallest indices are used to define a_n + 1 and b_n + 1 — namely, a_n + 1 is the least of these two members 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.^{[proof 1]} This case handles case 2 of Cantor's 1874 proof.

In the Example of Cantor's construction, the sequence P successive nested interval excludes rational numbers for two different reasons. It will exclude the finitely many rationals visited in the search for the first two rationals within the interval (these two rationals will have the least indices). These rationals are then used to form an interval that excludes the rationals visited in the search along with infinitely many more rationals. However, it still contains infinitely many rationals since our sequence of rationals is dense in [0, 1]. Forming this interval from the two rationals with the least indices guarantees that this interval excludes an initial segment of our sequence that contains at least two more members than the preceding initial segment. Since the denseness of our sequence guarantees that this process never ends, all rationals will be excluded.^{[proof 1]} Because of the ordering of the rationals in our sequence, the intersection of the nested intervals is the set {√2 − 1}.

ADD to end of Cantor's 1879 proof

In the Example of Cantor's construction, each successive nested interval excludes rational numbers for two different reasons. It will exclude the finitely many rationals visited in the search for the first two rationals within the interval (these two rationals will have the least indices). These rationals are then used to form an interval that excludes the rationals visited in the search along with infinitely many more rationals. However, it still contains infinitely many rationals since our sequence of rationals is dense in [0, 1]. Forming this interval from the two rationals with the least indices guarantees that this interval excludes an initial segment of our sequence that contains at least two more members than the preceding initial segment. Since the denseness of our sequence guarantees that this process never ends, all rationals will be excluded.^{[proof 1]} Because of the ordering of the rationals in our sequence, the intersection of the nested intervals is the set {√2 − 1}.^[B]

Notes

^ Cantor's notation only differs by the position of the dots. 1874 notation: (α · · · β), (α' · · · β'), (α'' · · · β''), · · · (Cantor 1874, p. 261). 1879 notation: (α . . . β), (α' . . . β'), (α'' . . . β''), . . . (Cantor 1879, p. 4).
^ Different orderings can produce nested intervals whose intersection is {x} for any x: Let x_n be the decimal approximation of x to n-digits. Starting with the interval [x₀, x₀ + 1] and continuing with the intervals [x_n, x_n + 1/10ⁿ ] for n ≥ 1 produces a sequence of intervals whose intersection is {x}.

NEW PROOF

Second theorem

Only the first part of Cantor's second theorem needs to be proved. It states: Given any sequence of real numbers x₁, x₂, x₃, ... and any interval [a, b], there is a number in [a, b] that is not contained in the given sequence.^[A]

To find a number in [a, b] that is not contained in the given sequence, construct two sequences of real numbers as follows: Find the first two numbers of the given sequence that are in the open interval (a, b). Denote the smaller of these two numbers by a₁ and the larger by b₁. Similarly, find the first two numbers of the given sequence that are in (a₁, b₁). Denote the smaller by a₂ and the larger by b₂. Continuing this procedure generates a sequence of intervals (a₁, b₁), (a₂, b₂), (a₃, b₃), ... such that each interval in the sequence contains all succeeding intervals — that is, it generates a sequence of nested intervals. This implies that the sequence a₁, a₂, a₃, ... is increasing and the sequence b₁, b₂, b₃, ... is decreasing.^[8]

Either the number of intervals generated is finite or infinite. If finite, let (a_L, b_L) be the last interval. If infinite, take the limits a_∞ = lim_n → ∞ a_n and b_∞ = lim_n → ∞ b_n. Since a_n < b_n for all n, either a_∞ = b_∞ or a_∞ < b_∞. Thus, there are three cases to consider:

Case 1: There is a last interval (a_L, b_L). Since at most one x_n can be in this interval, every y in this interval except x_n (if it exists) is not contained in the given sequence.

Case 2: a_∞ = b_∞. Then a_∞ is not contained in the given sequence since for all n : a_∞ belongs to the interval (a_n, b_n) but x_n does not belong to (a_n, b_n). In symbols, a_∞ ∈ (a_n, b_n) but x_n ∉ (a_n, b_n).

Proof that for all n : x_n ∉ (a_n, b_n)
This is implied by the stronger result: For all n, (a_n, b_n) excludes x₁, ... , x_2n, which is proved by induction. Basis step: If a₁ = x_j, b₁ = x_k, and E₁ = max(j, k), then (a₁, b₁) excludes x₁, ... , x_E₁. Also, E₁ ≥ 2 since the interval (a₁, b₁) excludes its two endpoints. Inductive step: Assume that (a_n, b_n) excludes x₁, ... , x_{E_n} and E_n ≥ 2n. If a_n+1 = x_j, b_n+1 = x_k, and E_n+1 = max(j, k), then (a_n+1, b_n+1) excludes x₁, ... , x_{E_n+1}. Also, E_n+1 ≥ E_n + 2 ≥ 2n + 2 = 2(n + 1) since the interval (a_n+1, b_n+1) excludes the numbers that (a_n, b_n) excludes plus the two endpoints a_n+1 and b_n+1. Therefore, for all n : (a_n, b_n) excludes x₁, ... , x_2n. (This proof is similar to a proof that Cantor published in 1879. The main difference is that Cantor's proof is embedded in a larger proof and uses notation from it. Our proof does not depend on this notation.^{[proof 1]})

Case 3: a_∞ < b_∞. Then every y in [a_∞, b_∞] is not contained in the given sequence since for all n : y belongs to (a_n, b_n) but x_n does not.^[4]

The proof is complete since, in all cases, at least one real number in [a, b] has been found that is not contained in the given sequence.^[B]

Cantor's proofs are constructive and have been used to write a computer program that generates the digits of a transcendental number. This program applies Cantor's construction to a sequence containing all the real algebraic numbers between 0 and 1. The article that discusses this program gives some of its output, which shows how the construction generates a transcendental.^[9]

Cantor's 1879 proof is the same as his 1874 proof except for a new proof of the first part of his second theorem: Given any sequence P of real numbers x₁, x₂, x₃, ... and any interval [a, b], there is a number in [a, b] that is not contained in the sequence P. The new proof has only two cases. **Big proof ref goes here!**

Cantor's new proof first handles the 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.^{[proof 1]}

In the first case, P is not dense in [a, b]. By definition, P is dense if and only if for all subsets (c, d) of [a, b], there is an x_n in P such that x_n is in (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 subset (c, d) of [a, b] such that for all x_n in P, x_n is not in (c, d). Therefore, every number in (c, d) is not contained in the sequence P.^{[proof 1]} This case handles case 1 and case 3 of Cantor's 1874 proof. In the diagram for case 1, every x_n in (a_L, y) is not in the sequence. In the diagram for case 3, every x_n in (a_∞, b_∞) is not in the sequence.

In the second case, P is dense in [a, b]. The denseness of P is used to recursively define a nested sequence of intervals is used to prove that this nested sequence excludes all members of P. The base case starts with the observation that since P is dense in [a, b], there are infinitely many real numbers in P that belong to (a, b). Let x_k₁ be the real number with the least index k₁ and let x_k₂ be the real number with the next larger index k₂. Let a₁ be the smaller of these two reals and let b₁ be the larger. Then, a < a₁ < b₁ < b and k₁ < k₂. Since x_k₁ and x_k₂ are the reals in P with the least indexes, for all j ≤ k₂, **MAY NEED MORE** the numbers x_j are not in (a₁, b₁). **MAY NEED MORE** May want to prove this! If j < k2, then if xj is in (a1, b1), it canoot be xj in interval or xj would have been chosen instead of xk+1 or xk+2!!

The recursive part of the definition starts with the interval (a_n, b_n) and with the inequalities a < a₁ < ... < a_n < b_n < ... < b₁ < b and k₁ < k₂ < ... < k_2n − 1 < k_2n such that for all j ≤ k_2n, the numbers x_j are not in (a_n, b_n). Since P is dense in [a, b], there are infinitely many real numbers in P that belong to (a_n, b_n). Let x_{k_2n + 1} be the real number with the least index k_2n + 2 and let x_2n + 2 be the real number with the next larger index k_2n + 2. Let a_n + 1 be the smaller of these two reals and let b_n + 1 be the larger. Then, a_n < a_n + 1 < b_n + 1 < b_n and k_2n + 1 < k_2n + 2. Since x_k₁ and x_k₂ are the reals in P with the least indexes **MAY NEED MORE** for all j ≤ k_2n + 2, the numbers x_j are not in (a_n + 1, b_n + 1). **MAY NEED MORE HERE!**

The sequence a_n is increasing and bounded above by b, so it has a limit a_∞, which satisfies a_n ≤ a_∞ for all n. The sequence b_n is decreasing and bounded below by a, so it has a limit b_∞, which satisfies b_∞ ≤ b_n for all 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.^{[proof 1]} **MAY NEED MORE HERE!** This case handles case 2 of Cantor's 1874 proof.

Second theorem

Only the first part of Cantor's second theorem needs to be proved. It states: Given any sequence of real numbers x₁, x₂, x₃, ... and any interval [a, b], there is a number in [a, b] that is not contained in the given sequence.^[C]

To find a number in [a, b] that is not contained in the given sequence, construct two sequences of real numbers as follows: Find the first two numbers of the given sequence that are in the open interval (a, b). Denote the smaller of these two numbers by a₁ and the larger by b₁. Similarly, find the first two numbers of the given sequence that are in (a₁, b₁). Denote the smaller by a₂ and the larger by b₂. Continuing this procedure generates a sequence of intervals (a₁, b₁), (a₂, b₂), (a₃, b₃), ... such that each interval in the sequence contains all succeeding intervals—that is, it generates a sequence of nested intervals. This implies that the sequence a₁, a₂, a₃, ... is increasing and the sequence b₁, b₂, b₃, ... is decreasing.^[10]

Either the number of intervals generated is finite or infinite. If finite, let (a_L, b_L) be the last interval. If infinite, take the limits a_∞ = lim_n → ∞ a_n and b_∞ = lim_n → ∞ b_n. Since a_n < b_n for all n, either a_∞ = b_∞ or a_∞ < b_∞. Thus, there are three cases to consider:

Case 1: There is a last interval (a_L, b_L). Since at most one x_n can be in this interval, every y in this interval except x_n (if it exists) is not contained in the given sequence.

Case 2: a_∞ = b_∞. Then a_∞ is not contained in the given sequence since for all n : a_∞ belongs to the interval (a_n, b_n), but as Cantor observes, x_n does not.

Proof that for all n : x_n ∉ (a_n, b_n)
This is implied by the stronger result: For all n, (a_n, b_n) excludes x₁, ... , x_2n, which is proved by induction. Basis step: If a₁ = x_j, b₁ = x_k, and E₁ = max(j, k), then (a₁, b₁) excludes x₁, ... , x_E₁. Also, E₁ ≥ 2 since the interval (a₁, b₁) excludes its two endpoints. Inductive step: Assume that (a_n, b_n) excludes x₁, ... , x_{E_n} and E_n ≥ 2n. If a_n+1 = x_j, b_n+1 = x_k, and E_n+1 = max(j, k), then (a_n+1, b_n+1) excludes x₁, ... , x_{E_n+1}. Also, E_n+1 ≥ E_n + 2 ≥ 2n + 2 = 2(n + 1) since the interval (a_n+1, b_n+1) excludes the numbers that (a_n, b_n) excludes plus the two endpoints a_n+1 and b_n+1. Therefore, for all n : (a_n, b_n) excludes x₁, ... , x_2n. (This proof is similar to a proof that Cantor published in 1879. The main difference is that Cantor's proof is embedded in a larger proof and uses notation from it. Our proof does not depend on this notation.^{[proof 1]})

Case 3: a_∞ < b_∞. Then every y in [a_∞, b_∞] is not contained in the given sequence since for all n : y belongs to (a_n, b_n) but x_n does not.^[4]

In the first case, P is not dense in [a, b]. By definition, P is dense if and only if for all (c, d) ⊆ [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 (c, d) ⊆ [a, b] such that for all x ∈ P, we have x ∉ (c, d). Thus, every number in (c, d) is not contained in the sequence P.^{[proof 1]} 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 c, which satisfies a_n < c. The sequence b_n is decreasing and bounded below by a, so it has a limit d, which satisfies d < b_n. Also, a_n < b_n implies c ≤ d. Therefore, a_n < c ≤ d < b_n. If c < d, then for every n: x_n ∉ (c, d) because x_n is not in the larger interval (a_n, b_n). This contradicts P being dense in [a, b]. Therefore, c = d. Since for all n: c ∈ (a_n, b_n) but x_n ∉ (a_n, b_n), the limit c is a real number that is not contained in the sequence P.^{[proof 1]} This case handles case 2 of Cantor's 1874 proof.

Notes

^ This implies the rest of the theorem — namely, there are infinitely many numbers in [a, b] that are not contained in the given sequence. For example, let $[0,1]$ be the interval and consider its subintervals $[0,{\tfrac {1}{2}}],$ $[{\tfrac {3}{4}},{\tfrac {7}{8}}],$ $[{\tfrac {15}{16}},{\tfrac {31}{32}}],$ $\ldots .$ Since these subintervals are pairwise disjoint, applying the first part of the theorem to each subinterval produces infinitely many numbers in $[0,1]$ that are not contained in the given sequence. In general, for the interval $[a,b],$ apply the first part of the theorem to the subintervals $[a,a+{\tfrac {1}{2}}(b-a)],$ $[a+{\tfrac {3}{4}}(b-a),a+{\tfrac {7}{8}}(b-a)],$ $[a+{\tfrac {15}{16}}(b-a),a+{\tfrac {31}{32}}(b-a)],$ $\ldots .$
^ The main difference between Cantor's proof and the above proof is that he generates the sequence of closed intervals [a_n, b_n]. To find a_n + 1 and b_n + 1, he uses the interior of the interval [a_n, b_n], which is the open interval (a_n, b_n). Generating open intervals combines Cantor's use of closed intervals and their interiors, which allows the case diagrams to depict all the details of the proof.
^ This implies the rest of the theorem — namely, there are infinitely many numbers in [a, b] that are not contained in the given sequence. For example, let $[0,1]$ be the interval and consider its subintervals $[0,{\tfrac {1}{2}}],$ $[{\tfrac {3}{4}},{\tfrac {7}{8}}],$ $[{\tfrac {15}{16}},{\tfrac {31}{32}}],$ $\ldots .$ Since these subintervals are pairwise disjoint, applying the first part of the theorem to each subinterval produces infinitely many numbers in $[0,1]$ that are not contained in the given sequence. In general, for the interval $[a,b],$ apply the first part of the theorem to the subintervals $[a,a+{\tfrac {1}{2}}(b-a)],$ $[a+{\tfrac {3}{4}}(b-a),a+{\tfrac {7}{8}}(b-a)],$ $[a+{\tfrac {15}{16}}(b-a),a+{\tfrac {31}{32}}(b-a)],$ $\ldots .$

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Seite 5: Dem widerspricht aber ein sehr allgemeiner Satz, welchen wir in Borchardt's Journal, Bd. 77, pag. 260, mit aller Strenge bewiesen haben, nämlich der folgende Satz:

Francais

French: mit

Seite 5: Dem widerspricht aber ein sehr allgemeiner Satz, welchen wir in Borchardt's Journal, Bd. 77, pag. 260, mit aller Strenge bewiesen haben, nämlich der folgende Satz:

Liegt P theilweise oder ganz im Intervalle (α . . . β), so kann der bemerkenswerthe Fall eintreten, dass jedes noch so kleine in (α . . . β) enthaltene Intervall (γ . . . δ) Punkte von P enthält. In einem solchen Falle wollen wir sagen, dass P im Intervalle (α . . . β) überall-dicht sei.}}

"Hat man eine einfach unendliche Reihe
ω₁, ω₂, . . . , ω_ν, . . .
von reellen, ungleichen Zahlen, die nach irgend einem Gesetz fortschreiten, so lässt sich in jedem vorgegebenen, Intervalle (α . . . β) eine Zahl η (und folglich lassen sich deren unendlich viele) angeben, welche nicht in jener Reihe (als Glied derselben) vorkommt."

If P lies partially or completely in the interval [α, β], then the remarkable case can happen that every intervals [γ, δ] contained in [α, β], no matter how small, contains points of P. In such a case, we will say that P is everywhere dense in the interval [α, β].^[A]

Notes

^ The original German text from Cantor 1879, p. 2 (Cantor's closed set notation (α . . . β) is translated to [α, β]): Liegt P theilweise oder ganz im Intervalle (α . . . β), so kann der bemerkenswerthe Fall eintreten, dass jedes noch so kleine in (α . . . β) enthaltene Intervall (γ . . . δ) Punkte von P enthält. In einem solchen Falle wollen wir sagen, dass P im Intervalle (α . . . β) überall-dicht sei.

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Fraenkel

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|Description=en|1=Abraham Halevi (Adolf) Fraenkel, German-born Israeli mathematician, first Dean of Mathematics as the Hebrew University of Jerusalem and Rector of the University; 1956, recipient of the Israel Prize for exact sciences (1956); member of the Israel Academy of Sciences and Humanities; Zionist, member of the Jewish National Council and the Jewish Assembly of Representatives under the British mandate, belonged to the "Mizrachi"}}

|Source=The David B. Keidan Collection of Digital Images from the Central Zionist Archives

(via http://id.lib.harvard.edu/images/olvwork485526/catalog Harvard University Library])

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|Date=between 1939 and 1949

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Weierstrass

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Kronecker

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|Description=Leopold Kronecker

|Source=http://www.britannica.com/EBchecked/media/28346/Kronecker-1865

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|Date=1865

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Dedekind

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|Description= Photo de Richard Dedekind vers 1870

|Source=http://dbeveridge.web.wesleyan.edu/wescourses/2001f/chem160/01/Photo_Gallery_Science/Dedekind/FrameSet.htm

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Cite error: There are <ref group=proof> tags on this page, but the references will not show without a {{reflist|group=proof}} template (see the help page).

^ Arkhangel'skii & Fedorchuk 1990, p. 16.
^ Cantor 1874, pp. 260–261. English translation: Ewald 1996, pp. 841–842.
^ Cantor 1874, pp. 260–261. English translation: Ewald 1996, pp. 841–842.
^ ^a ^b ^c ^d Cantor 1874, p. 261. English translation: Ewald 1996, p. 842.
^ Gray 1994, p. 822.
^ Cantor 1874, pp. 260–261. English translation: Ewald 1996, pp. 841–842.
^ Cantor 1874, pp. 260–261. English translation: Ewald 1996, pp. 841–842.
^ Cantor 1874, pp. 260–261. English translation: Ewald 1996, pp. 841–842.
^ Gray 1994, p. 822.
^ Cantor 1874, pp. 260–261. English translation: Ewald 1996, pp. 841–842.

[2] Cantor was not the first to define "everywhere dense" but his terminology was adopted with or without the "everywhere" (everywhere dense: Arkhangel'skii & Fedorchuk 1990, p. 15; dense: Kelley 1991, p. 49). In 1870, Hermann Hankel had defined this concept using different terminology: "a multitude of points … fill the segment if no interval, however small, can be given within the segment in which one does not find at least one point of that multitude" (Ferreirós 2007, p. 155). Hankel was building on Peter Gustav Lejeune Dirichlet's 1829 article that contains the Dirichlet function, a non-(Riemann) integrable function whose value is 0 for rational numbers and 1 for irrational numbers. (Ferreirós 2007, p. 149.)

[3] The original German text from Cantor 1879, p. 2 (Cantor's closed set notation (α . . . β) is translated to [α, β]): Liegt P theilweise oder ganz im Intervalle (α . . . β), so kann der bemerkenswerthe Fall eintreten, dass jedes noch so kleine in (α . . . β) enthaltene Intervall (γ . . . δ) Punkte von P enthält. In einem solchen Falle wollen wir sagen, dass P im Intervalle (α . . . β) überall-dicht sei.

[6] Different orderings can produce nested intervals whose intersection is {x} for any x: Let x_n be the decimal approximation of x to n-digits. Starting with the interval [x₀, x₀ + 1] and continuing with the intervals [x_n, x_n + 1/10ⁿ ] for n ≥ 1 produces a sequence of intervals whose intersection is {x}.

[7] This implies the rest of the theorem — namely, there are infinitely many numbers in [a, b] that are not contained in the given sequence. For example, let $[0,1]$ be the interval and consider its subintervals $[0,{\tfrac {1}{2}}],$ $[{\tfrac {3}{4}},{\tfrac {7}{8}}],$ $[{\tfrac {15}{16}},{\tfrac {31}{32}}],$ $\ldots .$ Since these subintervals are pairwise disjoint, applying the first part of the theorem to each subinterval produces infinitely many numbers in $[0,1]$ that are not contained in the given sequence. In general, for the interval $[a,b],$ apply the first part of the theorem to the subintervals $[a,a+{\tfrac {1}{2}}(b-a)],$ $[a+{\tfrac {3}{4}}(b-a),a+{\tfrac {7}{8}}(b-a)],$ $[a+{\tfrac {15}{16}}(b-a),a+{\tfrac {31}{32}}(b-a)],$ $\ldots .$

[10] The main difference between Cantor's proof and the above proof is that he generates the sequence of closed intervals [a_n, b_n]. To find a_n + 1 and b_n + 1, he uses the interior of the interval [a_n, b_n], which is the open interval (a_n, b_n). Generating open intervals combines Cantor's use of closed intervals and their interiors, which allows the case diagrams to depict all the details of the proof.

[12] This implies the rest of the theorem — namely, there are infinitely many numbers in [a, b] that are not contained in the given sequence. For example, let $[0,1]$ be the interval and consider its subintervals $[0,{\tfrac {1}{2}}],$ $[{\tfrac {3}{4}},{\tfrac {7}{8}}],$ $[{\tfrac {15}{16}},{\tfrac {31}{32}}],$ $\ldots .$ Since these subintervals are pairwise disjoint, applying the first part of the theorem to each subinterval produces infinitely many numbers in $[0,1]$ that are not contained in the given sequence. In general, for the interval $[a,b],$ apply the first part of the theorem to the subintervals $[a,a+{\tfrac {1}{2}}(b-a)],$ $[a+{\tfrac {3}{4}}(b-a),a+{\tfrac {7}{8}}(b-a)],$ $[a+{\tfrac {15}{16}}(b-a),a+{\tfrac {31}{32}}(b-a)],$ $\ldots .$

[14] The original German text from Cantor 1879, p. 2 (Cantor's closed set notation (α . . . β) is translated to [α, β]): Liegt P theilweise oder ganz im Intervalle (α . . . β), so kann der bemerkenswerthe Fall eintreten, dass jedes noch so kleine in (α . . . β) enthaltene Intervall (γ . . . δ) Punkte von P enthält. In einem solchen Falle wollen wir sagen, dass P im Intervalle (α . . . β) überall-dicht sei.

[16] Cantor's notation only differs by the position of the dots. 1874 notation: (α · · · β), (α' · · · β'), (α'' · · · β''), · · · (Cantor 1874, p. 261). 1879 notation: (α . . . β), (α' . . . β'), (α'' . . . β''), . . . (Cantor 1879, p. 4).

[17] Different orderings can produce nested intervals whose intersection is {x} for any x: Let x_n be the decimal approximation of x to n-digits. Starting with the interval [x₀, x₀ + 1] and continuing with the intervals [x_n, x_n + 1/10ⁿ ] for n ≥ 1 produces a sequence of intervals whose intersection is {x}.

[18] This implies the rest of the theorem — namely, there are infinitely many numbers in [a, b] that are not contained in the given sequence. For example, let $[0,1]$ be the interval and consider its subintervals $[0,{\tfrac {1}{2}}],$ $[{\tfrac {3}{4}},{\tfrac {7}{8}}],$ $[{\tfrac {15}{16}},{\tfrac {31}{32}}],$ $\ldots .$ Since these subintervals are pairwise disjoint, applying the first part of the theorem to each subinterval produces infinitely many numbers in $[0,1]$ that are not contained in the given sequence. In general, for the interval $[a,b],$ apply the first part of the theorem to the subintervals $[a,a+{\tfrac {1}{2}}(b-a)],$ $[a+{\tfrac {3}{4}}(b-a),a+{\tfrac {7}{8}}(b-a)],$ $[a+{\tfrac {15}{16}}(b-a),a+{\tfrac {31}{32}}(b-a)],$ $\ldots .$

[20] The main difference between Cantor's proof and the above proof is that he generates the sequence of closed intervals [a_n, b_n]. To find a_n + 1 and b_n + 1, he uses the interior of the interval [a_n, b_n], which is the open interval (a_n, b_n). Generating open intervals combines Cantor's use of closed intervals and their interiors, which allows the case diagrams to depict all the details of the proof.

[22] This implies the rest of the theorem — namely, there are infinitely many numbers in [a, b] that are not contained in the given sequence. For example, let $[0,1]$ be the interval and consider its subintervals $[0,{\tfrac {1}{2}}],$ $[{\tfrac {3}{4}},{\tfrac {7}{8}}],$ $[{\tfrac {15}{16}},{\tfrac {31}{32}}],$ $\ldots .$ Since these subintervals are pairwise disjoint, applying the first part of the theorem to each subinterval produces infinitely many numbers in $[0,1]$ that are not contained in the given sequence. In general, for the interval $[a,b],$ apply the first part of the theorem to the subintervals $[a,a+{\tfrac {1}{2}}(b-a)],$ $[a+{\tfrac {3}{4}}(b-a),a+{\tfrac {7}{8}}(b-a)],$ $[a+{\tfrac {15}{16}}(b-a),a+{\tfrac {31}{32}}(b-a)],$ $\ldots .$

[24] The original German text from Cantor 1879, p. 2 (Cantor's closed set notation (α . . . β) is translated to [α, β]): Liegt P theilweise oder ganz im Intervalle (α . . . β), so kann der bemerkenswerthe Fall eintreten, dass jedes noch so kleine in (α . . . β) enthaltene Intervall (γ . . . δ) Punkte von P enthält. In einem solchen Falle wollen wir sagen, dass P im Intervalle (α . . . β) überall-dicht sei.

[4] Arkhangel'skii & Fedorchuk 1990, p. 16.

[5] Cantor 1874, pp. 260–261. English translation: Ewald 1996, pp. 841–842.

[8] Cantor 1874, pp. 260–261. English translation: Ewald 1996, pp. 841–842.

[Ewald842-9] Cantor 1874, p. 261. English translation: Ewald 1996, p. 842.

[11] Gray 1994, p. 822.

[13] Cantor 1874, pp. 260–261. English translation: Ewald 1996, pp. 841–842.

[15] Cantor 1874, pp. 260–261. English translation: Ewald 1996, pp. 841–842.

[19] Cantor 1874, pp. 260–261. English translation: Ewald 1996, pp. 841–842.

[21] Gray 1994, p. 822.

[23] Cantor 1874, pp. 260–261. English translation: Ewald 1996, pp. 841–842.

[proof 1]

[A]

[B]

[1]

[2]

[C]

[A]

[3]

[4]

[B]

[5]

[C]

[6]

[A]

[7]

[A]

[B]

[A]

[8]

[B]

[9]

[C]

[10]

[A]