User:RJGray/Cantor draft

End of Cantor's 1879 proof
The sequence an is increasing and bounded above by b, so the limit a∞ = limn → ∞ an exists. Similarly, the limit b∞ = limn → ∞ bn exists because the sequence bn is decreasing and bounded below by a. Also, an < bn for all n implies a∞ ≤ b∞. However, if a∞ < b∞, then xn ∉ (a∞, b∞) for all n since xn is not in the larger interval (an, bn). This contradicts P being dense in [a, b]. Hence, a∞ = b∞. So for all n, a∞ ∈ (an, bn) but xn ∉ (an, bn). Therefore, a∞ is a number in [a, b] that is not contained in P.

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":


 * 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].

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.

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 xn, the subinterval (xn, d) is used to generate a point xn + 1 satisfying xn + 1 > xn. The infinitely many points xn 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. 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 xk 1 be the number with the least index and xk 2 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, xm ∉ (a1, b1) for m ≤ k2 since these xm are the endpoints of (a1, b1). The base case repeats the above proof with the interval (a1, b1) to obtain xk 3, xk 4 , a2, b2 such that k1 < k2 < k3 < k4 and a < a1 < a2 < b2 < b1 < b and xm ∉ (a2, b2) for m ≤ k4.

The recursive step starts with the interval (an–1, bn–1), the inequalities k1 < k2 <. . . < k2n–2 < k2n–1 and a < a1 <. . . < an–1 < bn–1. . . < b1 < b, and the statement that the interval (an–1, bn–1) excludes the first 2n–2 members of the sequence P—that is, xm ∉ (an–1, bn–1) for m ≤ k2n–2. Since P is dense in [a, b], there are infinitely many numbers of P in (an–1, bn–1). Let xk 2n–1 be the number with the least index and xk 2n be the number with the next larger index, and let an be the smaller and bn be the larger of these two numbers. Then, k2n–1 < k2n, an–1 < an < bn < bn–1, and (an, bn) is a proper subinterval of (an–1, bn–1). Combining these inequalities with the ones for step n–1 of the recursion produces k1 < k2 <. . . < k2n–1 < k2n and a < a1 <. . . < an < bn. . . < b1 < b. Also, xm ∉ (an, bn) for m = k2n–1 and m = k2n since these xm are the endpoints of (an, bn). This together with the statement that (an–1, bn–1) excludes the first 2n–2 members of the sequence P implies that the interval (an, bn) excludes the first 2n members of the sequence P—that is, xm ∉ (an, bn) for m ≤ k2n. Therefore, for all n, xn ∉ (an, bn) since n ≤ k2n.

The sequence an is increasing and bounded by b, so the limit a∞ = limn → ∞ an exists. Similarly, the limit b∞ = limn → ∞ bn exists since the sequence bn is decreasing and bounded by a. Also, an < bn implies a∞ ≤ b∞. 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]. Hence, a∞ = b∞. So for all n, a∞ ∈ (an, bn) but xn ∉ (an, bn). Therefore, a∞ is a number in [a, b] that is not contained in P.

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.

Cantor's 1879 proof
The sequence an is increasing and bounded above by b, so the limit a∞ = limn → ∞ an exists. Similarly, the limit b∞ = limn → ∞ bn exists because the sequence bn is decreasing and bounded below by a. Also, an < bn for all n implies a∞ ≤ b∞. However, if a∞ < b∞, then xn ∉ (a∞, b∞) for all n since xn is not in the larger interval (an, bn). This contradicts P being dense in [a, b], so a∞ = b∞. Now, for every n, a∞ ∈ (an, bn) but xn ∉ (an, bn). Therefore, a∞ is a number in [a, b] that is not contained in P.

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. Because of the ordering of the rationals in our sequence, the intersection of the nested intervals is the set {{{sqrt|2}} &minus; 1}.

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.

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.

Either the number of intervals generated is finite or infinite. If finite, let (aL, bL) be the last interval. If infinite, take the limits a∞ = limn → ∞ an and b∞ = limn → ∞ bn. Since an < bn for all n, either a∞ = b∞ or a∞ < b∞. Thus, there are three cases to consider:
 * Case 1: There is a last interval (aL, bL). Since at most one xn can be in this interval, every y in this interval except xn (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 (an, bn) but xn does not belong to (an, bn). In symbols, a∞ ∈ (an, bn) but xn ∉ (an, bn).


 * {| class="wikitable collapsible collapsed"

! style="background: f5f5f5;" |Proof that for all n: xn ∉ (an, bn)
 * - style="text-align: left; vertical-align: top; background: white"
 * style="padding-left: 1em; padding-right: 1em" | This is implied by the stronger result: For all n, (an, bn) excludes x1, ..., x2n, which is proved by induction. Basis step: If a1 = xj, b1 = xk, and E1 = max(j, k), then (a1, b1) excludes x1, ... , xE 1 . Also, E1 ≥ 2 since the interval (a1, b1) excludes its two endpoints. Inductive step: Assume that (an, bn) excludes x1, ... , xE n and En ≥ 2n. If an+1 = xj, bn+1 = xk, and En+1 = max(j, k), then (an+1, bn+1) excludes x1, ... , xE n+1 . Also, En+1 ≥ En + 2 ≥ 2n + 2 = 2(n + 1) since the interval (an+1, bn+1) excludes the numbers that (an, bn) excludes plus the two endpoints an+1 and bn+1. Therefore, for all n: (an, bn) excludes x1, ... , x2n. (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. )
 * }
 * Case 3: a∞ < b∞. Then every y in [a∞, b∞] is not contained in the given sequence since for all n: y belongs to (an, bn) but xn does not.

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.

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.

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.

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 xn in P such that xn 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 xn in P, xn is not in (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 diagram for case 1, every xn in (aL, y) is not in the sequence. In the diagram for case 3, every xn 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 xk 1 be the real number with the least index k1 and let xk 2 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 xk 1 and xk 2 are the reals in P with the least indexes, for all j ≤ k2, **MAY NEED MORE** the numbers xj 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 (an, bn) and with the inequalities a < a1 < ... < an < bn < ... < b1 < b and k1 < k2 < ... < k2n−1 < k2n such that for all j ≤ k2n, the numbers xj are not in (an, bn). Since P is dense in [a, b], there are infinitely many real numbers in P that belong to (an, bn). Let xk 2n+1 be the real number with the least index k2n+2 and let x2n+2 be the real number with the next larger index k2n+2. Let an+1 be the smaller of these two reals and let bn+1 be the larger. Then, an < an+1 < bn+1 < bn and k2n+1 < k2n+2. Since xk 1 and xk 2 are the reals in P with the least indexes **MAY NEED MORE** for all j ≤ k2n+2, the numbers xj are not in (an+1, bn+1). **MAY NEED MORE HERE!**

The sequence an is increasing and bounded above by b, so it has a limit a∞, which satisfies an ≤ a∞ for all n. The sequence bn is decreasing and bounded below by a, so it has a limit b∞, which satisfies b∞ ≤ bn for all n. 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. **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.

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.

Either the number of intervals generated is finite or infinite. If finite, let (aL, bL) be the last interval. If infinite, take the limits a∞ = limn → ∞ an and b∞ = limn → ∞ bn. Since an < bn for all n, either a∞ = b∞ or a∞ < b∞. Thus, there are three cases to consider:


 * Case 1: There is a last interval (aL, bL). Since at most one xn can be in this interval, every y in this interval except xn (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 (an, bn), but as Cantor observes, xn does not.


 * {| class="wikitable collapsible collapsed"

! style="background: f5f5f5;" |Proof that for all n: xn ∉ (an, bn)}}
 * - style="text-align: left; vertical-align: top; background: white"
 * style="padding-left: 1em; padding-right: 1em" | This is implied by the stronger result: For all n, (an, bn) excludes x1, ..., x2n, which is proved by induction. Basis step: If a1 = xj, b1 = xk, and E1 = max(j, k), then (a1, b1) excludes x1, ... , xE 1 . Also, E1 ≥ 2 since the interval (a1, b1) excludes its two endpoints. Inductive step: Assume that (an, bn) excludes x1, ... , xE n and En ≥ 2n. If an+1 = xj, bn+1 = xk, and En+1 = max(j, k), then (an+1, bn+1) excludes x1, ... , xE n+1 . Also, En+1 ≥ En + 2 ≥ 2n + 2 = 2(n + 1) since the interval (an+1, bn+1) excludes the numbers that (an, bn) excludes plus the two endpoints an+1 and bn+1. Therefore, for all n: (an, bn) excludes x1, ... , x2n. (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. )
 * }
 * Case 3: a∞ < b∞. Then every y in [a∞, b∞] is not contained in the given sequence since for all n: y belongs to (an, bn) but xn does not.

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. 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 c, which satisfies an < c. The sequence bn is decreasing and bounded below by a, so it has a limit d, which satisfies d < bn. Also, an < bn implies c ≤ d. Therefore, an < c ≤ d < bn. If c < d, then for every n: xn ∉ (c, d) because xn is not in the larger interval (an, bn). This contradicts P being dense in [a, b]. Therefore, c = d. Since for all n: c ∈ (an, bn) but xn ∉ (an, bn), the limit c is a real number that is not contained in the sequence P. This case handles case 2 of Cantor's 1874 proof.

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
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, . . ., ων, . . . 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].

Example

 * File: "Cantor's first set theory article"
 * Click on Cantor photo
 * Click on More details
 * Click on Edit (at top of page)
 * Change or add to Permission = Template:PD-old or Template:PD-old|PD-other

Cantor

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 * Date          =circa 1870
 * Permission    =PD-old}}
 * }}
 * }}

Perron

 * Source=https://opc.mfo.de/detail?photo_id=3261
 * Author=Jacobs, Konrad
 * Date=1948
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 * DEFAULTSORT:Perron, Oskar}}
 * Category:1948 in Munich]]
 * Category:Pictures from Oberwolfach Photo Collection]]
 * Category:Oskar Perron (mathematician)]]

Fraenkel

 * int:filedesc}} ==
 * Information
 * 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])
 * Author=unknown|author}}
 * Date=between 1939 and 1949
 * Permission=
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}}

int:license-header}} == PD-Israel}}


 * Category:Abraham Fraenkel]]
 * Category:Photographs from the Central Zionist Archives]]
 * Notation:PHG\1110747
 * Original number:110747
 * Name of photographer/institution:Unknown
 * Copyright:By the archives

Weierstrass

 * int:filedesc}} ==
 * Source: from de:Image:Karl_Weierstrass.jpg,
 * from http://www.sil.si.edu/digitalcollections/hst/scientific-identity/explore.htm


 * Category:Karl Weierstraß]]
 * Category:Rectors of Humboldt-Universität zu Berlin]]


 * int:license-header}} ==
 * PD-Art|PD-old-70}}


 * Category:Artworks missing infobox template]]

Kronecker

 * int:filedesc}} ==
 * Information
 * Description=Leopold Kronecker
 * Source=http://www.britannica.com/EBchecked/media/28346/Kronecker-1865
 * Author=author
 * Date=1865
 * Permission=PD-old-70}}
 * other_versions=
 * }}


 * Category:Leopold Kronecker]]

Dedekind

 * int:filedesc}} ==

Category:Richard Dedekind

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.

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. Since Cantor's proof has not been published in English, an English translation is given alongside the original German text, which is from. The translation starts one sentence before the proof because this sentence mentions Cantor's 1874 proof. Cantor states it was printed in Borchardt's Journal. Crelle’s Journal was also called Borchardt’s Journal from 1856-1880 when Carl Wilhelm Borchardt edited the journal. Square brackets are used to identify this mention of Cantor's earlier proof, to clarify the translation, and to provide page numbers. Also, "Mannichfaltigkeit" (manifold) has been translated to "set" and Cantor's notation for closed sets (α . . . β) has been translated to [α, β]. Cantor changed his terminology from Mannichfaltigkeit to Menge (set) in his 1883 article, which introduced sets of ordinal numbers. Currently in mathematics, a manifold is type of topological space.

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. 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.) The base case of the recursion is (a1, b1) = (a, b). The recursive step starts with the interval (an, bn) 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 an + 1 and bn + 1—namely, an + 1 is the least of these two members 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.

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. Because of the ordering of the rationals in our sequence, the intersection of the nested intervals is the set {{{sqrt|2}} &minus; 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. Because of the ordering of the rationals in our sequence, the intersection of the nested intervals is the set {{{sqrt|2}} &minus; 1}.

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.

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.

Either the number of intervals generated is finite or infinite. If finite, let (aL, bL) be the last interval. If infinite, take the limits a∞ = limn → ∞ an and b∞ = limn → ∞ bn. Since an < bn for all n, either a∞ = b∞ or a∞ < b∞. Thus, there are three cases to consider:
 * Case 1: There is a last interval (aL, bL). Since at most one xn can be in this interval, every y in this interval except xn (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 (an, bn) but xn does not belong to (an, bn). In symbols, a∞ ∈ (an, bn) but xn ∉ (an, bn).


 * {| class="wikitable collapsible collapsed"

! style="background: f5f5f5;" |Proof that for all n: xn ∉ (an, bn)
 * - style="text-align: left; vertical-align: top; background: white"
 * style="padding-left: 1em; padding-right: 1em" | This is implied by the stronger result: For all n, (an, bn) excludes x1, ..., x2n, which is proved by induction. Basis step: If a1 = xj, b1 = xk, and E1 = max(j, k), then (a1, b1) excludes x1, ... , xE 1 . Also, E1 ≥ 2 since the interval (a1, b1) excludes its two endpoints. Inductive step: Assume that (an, bn) excludes x1, ... , xE n and En ≥ 2n. If an+1 = xj, bn+1 = xk, and En+1 = max(j, k), then (an+1, bn+1) excludes x1, ... , xE n+1 . Also, En+1 ≥ En + 2 ≥ 2n + 2 = 2(n + 1) since the interval (an+1, bn+1) excludes the numbers that (an, bn) excludes plus the two endpoints an+1 and bn+1. Therefore, for all n: (an, bn) excludes x1, ... , x2n. (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. )
 * }
 * Case 3: a∞ < b∞. Then every y in [a∞, b∞] is not contained in the given sequence since for all n: y belongs to (an, bn) but xn does not.

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.

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.

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.

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 xn in P such that xn 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 xn in P, xn is not in (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 diagram for case 1, every xn in (aL, y) is not in the sequence. In the diagram for case 3, every xn 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 xk 1 be the real number with the least index k1 and let xk 2 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 xk 1 and xk 2 are the reals in P with the least indexes, for all j ≤ k2, **MAY NEED MORE** the numbers xj 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 (an, bn) and with the inequalities a < a1 < ... < an < bn < ... < b1 < b and k1 < k2 < ... < k2n−1 < k2n such that for all j ≤ k2n, the numbers xj are not in (an, bn). Since P is dense in [a, b], there are infinitely many real numbers in P that belong to (an, bn). Let xk 2n+1 be the real number with the least index k2n+2 and let x2n+2 be the real number with the next larger index k2n+2. Let an+1 be the smaller of these two reals and let bn+1 be the larger. Then, an < an+1 < bn+1 < bn and k2n+1 < k2n+2. Since xk 1 and xk 2 are the reals in P with the least indexes **MAY NEED MORE** for all j ≤ k2n+2, the numbers xj are not in (an+1, bn+1). **MAY NEED MORE HERE!**

The sequence an is increasing and bounded above by b, so it has a limit a∞, which satisfies an ≤ a∞ for all n. The sequence bn is decreasing and bounded below by a, so it has a limit b∞, which satisfies b∞ ≤ bn for all n. 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. **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.

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.

Either the number of intervals generated is finite or infinite. If finite, let (aL, bL) be the last interval. If infinite, take the limits a∞ = limn → ∞ an and b∞ = limn → ∞ bn. Since an < bn for all n, either a∞ = b∞ or a∞ < b∞. Thus, there are three cases to consider:


 * Case 1: There is a last interval (aL, bL). Since at most one xn can be in this interval, every y in this interval except xn (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 (an, bn), but as Cantor observes, xn does not.


 * {| class="wikitable collapsible collapsed"

! style="background: f5f5f5;" |Proof that for all n: xn ∉ (an, bn)
 * - style="text-align: left; vertical-align: top; background: white"
 * style="padding-left: 1em; padding-right: 1em" | This is implied by the stronger result: For all n, (an, bn) excludes x1, ..., x2n, which is proved by induction. Basis step: If a1 = xj, b1 = xk, and E1 = max(j, k), then (a1, b1) excludes x1, ... , xE 1 . Also, E1 ≥ 2 since the interval (a1, b1) excludes its two endpoints. Inductive step: Assume that (an, bn) excludes x1, ... , xE n and En ≥ 2n. If an+1 = xj, bn+1 = xk, and En+1 = max(j, k), then (an+1, bn+1) excludes x1, ... , xE n+1 . Also, En+1 ≥ En + 2 ≥ 2n + 2 = 2(n + 1) since the interval (an+1, bn+1) excludes the numbers that (an, bn) excludes plus the two endpoints an+1 and bn+1. Therefore, for all n: (an, bn) excludes x1, ... , x2n. (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. )
 * }
 * Case 3: a∞ < b∞. Then every y in [a∞, b∞] is not contained in the given sequence since for all n: y belongs to (an, bn) but xn does not.

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. 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 c, which satisfies an < c. The sequence bn is decreasing and bounded below by a, so it has a limit d, which satisfies d < bn. Also, an < bn implies c ≤ d. Therefore, an < c ≤ d < bn. If c < d, then for every n: xn ∉ (c, d) because xn is not in the larger interval (an, bn). This contradicts P being dense in [a, b]. Therefore, c = d. Since for all n: c ∈ (an, bn) but xn ∉ (an, bn), the limit c is a real number that is not contained in the sequence P. This case handles case 2 of Cantor's 1874 proof.

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
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, . . ., ων, . . . 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].

Example

 * File: "Cantor's first set theory article"
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Cantor

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 * Date          =circa 1870
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Perron

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Fraenkel

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 * Information
 * 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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}}

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Weierstrass

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Kronecker

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 * Description=Leopold Kronecker
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 * Date=1865
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Dedekind

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Category:Richard Dedekind