Введение
Доказательство в теории множеств
In set theory, Cantor's diagonal argument, also called the diagonalisation argument, the diagonal slash argument, the anti diagonal argument, the diagonal method, and Cantor's diagonalization proof, was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into one to one correspondence with the infinite set of natural numbers. Such sets are now known as uncountable sets, and the size of infinite sets is now treated by the theory of cardinal numbers which Cantor began. The diagonal argument was not Cantor's first proof of the uncountability of the real numbers, which appeared in 1874. However, it demonstrates a general technique that has since been used in a wide range of proofs, including the first of Gödel's incompleteness theorems and Richard's paradox. He begins with a constructive proof of the following lemma:
If s1, s2, , sn, is any enumeration of elements from T, then an element s of T can be constructed that doesn't correspond to any sn in the enumeration. The proof starts with an enumeration of elements from T, for example
{|
|
| s1 = || (0, || 0, || 0, || 0, || 0, || 0, || 0, || )
|
| s2 = || (1, || 1, || 1, || 1, || 1, || 1, || 1, || )
|
| s3 = || (0, || 1, || 0, || 1, || 0, || 1, || 0, || )
|
| s4 = || (1, || 0, || 1, || 0, || 1, || 0, || 1, || )
|
| s5 = || (1, || 1, || 0, || 1, || 0, || 1, || 1, || )
|
| s6 = || (0, || 0, || 1, || 1, || 0, || 1, || 1, || )
|
| s7 = || (1, || 0, || 0, || 0, || 1, || 0, || 0, || )
|
|
|}
Next, a sequence s is constructed by choosing the 1st digit as complementary to the 1st digit of s1 (swapping 0s for 1s and vice versa), the 2nd digit as complementary to the 2nd digit of s2, the 3rd digit as complementary to the 3rd digit of s3, and generally for every n, the nth digit as complementary to the nth digit of sn. For the example above, this yields
{|
|
| s1 || = || (0, || 0, || 0, || 0, || 0, || 0, || 0, || )
|
| s2 || = || (1, || 1, || 1, || 1, || 1, || 1, || 1, || )
|
| s3 || = || (0, || 1, || 0, || 1, || 0, || 1, || 0, || )
|
| s4 || = || (1, || 0, || 1, || 0, || 1, || 0, || 1, || )
|
| s5 || = || (1, || 1, || 0, || 1, || 0, || 1, || 1, || )
|
| s6 || = || (0, || 0, || 1, || 1, || 0, || 1, || 1, || )
|
| s7 || = || (1, || 0, || 0, || 0, || 1, || 0, || 0, || )
|
|
|
|
|
| s || = || (1, || 0, || 1, || 1, || 1, || 0, || 1, || )
|}
By construction, s is a member of T that differs from each sn, since their nth digits differ (highlighted in the example). Hence, s cannot occur in the enumeration. Based on this lemma, Cantor then uses a proof by contradiction to show that:
The set T is uncountable. The proof starts by assuming that T is countable. Then all its elements can be written in an enumeration s1, s2, , sn, Applying the previous lemma to this enumeration produces a sequence s that is a member of T, but is not in the enumeration. However, if T is enumerated, then every member of T, including this s, is in the enumeration. This contradiction implies that the original assumption is false. Therefore, T is uncountable. Constructing a bijection between T and R is slightly more complicated. Instead of mapping 0111 to the decimal 0.0111 , it can be mapped to the base b number: 0.0111 b. This leads to the family of functions: The functions are injections, except for This function will be modified to produce a bijection between T and R.
Construction of a bijection between T and R
This construction uses a method devised by Cantor that was published in 1878. He used it to construct a bijection between the closed interval [0, 1] and the irrationals in the open interval (0, 1). He first removed a countably infinite subset from each of these sets so that there is a bijection between the remaining uncountable sets. Since there is a bijection between the countably infinite subsets that have been removed, combining the two bijections produces a bijection between the original sets. Cantor's method can be used to modify the function to produce a bijection from T to (0, 1). Because some numbers have two binary expansions, is not even injective. For example, 0.1000 2 = 1/2 and 0.0111 2 = 1/2, so both 1000 and 0111 map to the same number, 1/2. To modify , observe that it is a bijection except for a countably infinite subset of (0, 1) and a countably infinite subset of T. It is not a bijection for the numbers in (0, 1) that have two binary expansions. These are called dyadic numbers and have the form where m is an odd integer and n is a natural number. Put these numbers in the sequence: r = (1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, ). Also, is not a bijection to (0, 1) for the strings in T appearing after the binary point in the binary expansions of 0, 1, and the numbers in sequence r. Put these eventually constant strings in the sequence: s = ( , , 1 , 0 , 01 , 00 , 11 , 10 , ). Define the bijection g(t) from T to (0, 1): If t is the nth string in sequence s, let g(t) be the nth number in sequence r; otherwise, g(t) = 0. t2. To construct a bijection from T to R, start with the tangent function tan(x), which is a bijection from (−π/2, π/2) to R (see the figure shown on the right). Next observe that the linear function h(x) = πx – π/2 is a bijection from (0, 1) to (−π/2, π/2) (see the figure shown on the left). The composite function tan(h(x)) = tan(πx – π/2) is a bijection from (0, 1) to R. Composing this function with g(t) produces the function tan(h(g(t))) = tan(πg(t) – π/2), which is a bijection from T to R.
В теории множеств диагональный аргумент Кантора, также называемый аргументом диагонализации, диагональным слэш-аргументом, антидиагональным аргументом, диагональным методом и доказательством диагонализации Кантора, был опубликован в 1891 году Георгом Кантором как математическое доказательство того, что существуют бесконечные множества, которые нельзя привести в соответствие одно к одному с бесконечным множеством натуральных чисел. Такие множества теперь известны как несчётные множества, а размер бесконечных множеств теперь изучается теорией кардинальных чисел, которую начал Кантор. Диагональный аргумент не был первым доказательством Кантора о несчётности действительных чисел, которое появилось в 1874 году. Однако он демонстрирует общий метод, который с тех пор использовался в широком диапазоне доказательств, включая первое из теорем неполноты Гёделя и парадокс Ричарда. Он начинается с конструктивного доказательства следующей леммы:
In set theory, Cantor's diagonal argument, also called the diagonalisation argument, the diagonal slash argument, the anti diagonal argument, the diagonal method, and Cantor's diagonalization proof, was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into one to one correspondence with the infinite set of natural numbers. Such sets are now known as uncountable sets, and the size of infinite sets is now treated by the theory of cardinal numbers which Cantor began. The diagonal argument was not Cantor's first proof of the uncountability of the real numbers, which appeared in 1874. However, it demonstrates a general technique that has since been used in a wide range of proofs, including the first of Gödel's incompleteness theorems and Richard's paradox. He begins with a constructive proof of the following lemma:
If s1, s2, , sn, is any enumeration of elements from T, then an element s of T can be constructed that doesn't correspond to any sn in the enumeration. The proof starts with an enumeration of elements from T, for example
{|
|
| s1 = || (0, || 0, || 0, || 0, || 0, || 0, || 0, || )
|
| s2 = || (1, || 1, || 1, || 1, || 1, || 1, || 1, || )
|
| s3 = || (0, || 1, || 0, || 1, || 0, || 1, || 0, || )
|
| s4 = || (1, || 0, || 1, || 0, || 1, || 0, || 1, || )
|
| s5 = || (1, || 1, || 0, || 1, || 0, || 1, || 1, || )
|
| s6 = || (0, || 0, || 1, || 1, || 0, || 1, || 1, || )
|
| s7 = || (1, || 0, || 0, || 0, || 1, || 0, || 0, || )
|
|
|}
Next, a sequence s is constructed by choosing the 1st digit as complementary to the 1st digit of s1 (swapping 0s for 1s and vice versa), the 2nd digit as complementary to the 2nd digit of s2, the 3rd digit as complementary to the 3rd digit of s3, and generally for every n, the nth digit as complementary to the nth digit of sn. For the example above, this yields
{|
|
| s1 || = || (0, || 0, || 0, || 0, || 0, || 0, || 0, || )
|
| s2 || = || (1, || 1, || 1, || 1, || 1, || 1, || 1, || )
|
| s3 || = || (0, || 1, || 0, || 1, || 0, || 1, || 0, || )
|
| s4 || = || (1, || 0, || 1, || 0, || 1, || 0, || 1, || )
|
| s5 || = || (1, || 1, || 0, || 1, || 0, || 1, || 1, || )
|
| s6 || = || (0, || 0, || 1, || 1, || 0, || 1, || 1, || )
|
| s7 || = || (1, || 0, || 0, || 0, || 1, || 0, || 0, || )
|
|
|
|
|
| s || = || (1, || 0, || 1, || 1, || 1, || 0, || 1, || )
|}
By construction, s is a member of T that differs from each sn, since their nth digits differ (highlighted in the example). Hence, s cannot occur in the enumeration. Based on this lemma, Cantor then uses a proof by contradiction to show that:
The set T is uncountable. The proof starts by assuming that T is countable. Then all its elements can be written in an enumeration s1, s2, , sn, Applying the previous lemma to this enumeration produces a sequence s that is a member of T, but is not in the enumeration. However, if T is enumerated, then every member of T, including this s, is in the enumeration. This contradiction implies that the original assumption is false. Therefore, T is uncountable. Constructing a bijection between T and R is slightly more complicated. Instead of mapping 0111 to the decimal 0.0111 , it can be mapped to the base b number: 0.0111 b. This leads to the family of functions: The functions are injections, except for This function will be modified to produce a bijection between T and R.
Construction of a bijection between T and R
This construction uses a method devised by Cantor that was published in 1878. He used it to construct a bijection between the closed interval [0, 1] and the irrationals in the open interval (0, 1). He first removed a countably infinite subset from each of these sets so that there is a bijection between the remaining uncountable sets. Since there is a bijection between the countably infinite subsets that have been removed, combining the two bijections produces a bijection between the original sets. Cantor's method can be used to modify the function to produce a bijection from T to (0, 1). Because some numbers have two binary expansions, is not even injective. For example, 0.1000 2 = 1/2 and 0.0111 2 = 1/2, so both 1000 and 0111 map to the same number, 1/2. To modify , observe that it is a bijection except for a countably infinite subset of (0, 1) and a countably infinite subset of T. It is not a bijection for the numbers in (0, 1) that have two binary expansions. These are called dyadic numbers and have the form where m is an odd integer and n is a natural number. Put these numbers in the sequence: r = (1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, ). Also, is not a bijection to (0, 1) for the strings in T appearing after the binary point in the binary expansions of 0, 1, and the numbers in sequence r. Put these eventually constant strings in the sequence: s = ( , , 1 , 0 , 01 , 00 , 11 , 10 , ). Define the bijection g(t) from T to (0, 1): If t is the nth string in sequence s, let g(t) be the nth number in sequence r; otherwise, g(t) = 0. t2. To construct a bijection from T to R, start with the tangent function tan(x), which is a bijection from (−π/2, π/2) to R (see the figure shown on the right). Next observe that the linear function h(x) = πx – π/2 is a bijection from (0, 1) to (−π/2, π/2) (see the figure shown on the left). The composite function tan(h(x)) = tan(πx – π/2) is a bijection from (0, 1) to R. Composing this function with g(t) produces the function tan(h(g(t))) = tan(πg(t) – π/2), which is a bijection from T to R.
Если s1, s2, ..., sn – любое перечисление элементов из T, то можно построить элемент s из T, который не соответствует ни одному sn в перечислении. Следовательно, s не может встречаться в перечислении. Основываясь на этой лемме, Кантор затем использует доказательство от противного, чтобы показать, что: множество T несчётно. Доказательство начинается с предположения, что T счётно. Тогда все его элементы можно записать в перечислении s1, s2, ..., sn. Применение предыдущей леммы к этому перечислению даёт последовательность s, которая является элементом T, но отсутствует в перечислении. Однако, если T перечислено, то каждый элемент T, включая s, должен быть в перечислении. Это противоречие подразумевает, что первоначальное предположение ложно. Поэтому T несчётно. Построение биекции между T и R немного сложнее. Вместо отображения 0111 на десятичную дробь 0,0111, его можно отобразить на число в системе счисления b: 0,0111b. Это приводит к семейству функций: эти функции являются инъекциями, за исключением той, которую необходимо модифицировать для получения биекции между T и R.
In set theory, Cantor's diagonal argument, also called the diagonalisation argument, the diagonal slash argument, the anti diagonal argument, the diagonal method, and Cantor's diagonalization proof, was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into one to one correspondence with the infinite set of natural numbers. Such sets are now known as uncountable sets, and the size of infinite sets is now treated by the theory of cardinal numbers which Cantor began. The diagonal argument was not Cantor's first proof of the uncountability of the real numbers, which appeared in 1874. However, it demonstrates a general technique that has since been used in a wide range of proofs, including the first of Gödel's incompleteness theorems and Richard's paradox. He begins with a constructive proof of the following lemma:
If s1, s2, , sn, is any enumeration of elements from T, then an element s of T can be constructed that doesn't correspond to any sn in the enumeration. The proof starts with an enumeration of elements from T, for example
{|
|
| s1 = || (0, || 0, || 0, || 0, || 0, || 0, || 0, || )
|
| s2 = || (1, || 1, || 1, || 1, || 1, || 1, || 1, || )
|
| s3 = || (0, || 1, || 0, || 1, || 0, || 1, || 0, || )
|
| s4 = || (1, || 0, || 1, || 0, || 1, || 0, || 1, || )
|
| s5 = || (1, || 1, || 0, || 1, || 0, || 1, || 1, || )
|
| s6 = || (0, || 0, || 1, || 1, || 0, || 1, || 1, || )
|
| s7 = || (1, || 0, || 0, || 0, || 1, || 0, || 0, || )
|
|
|}
Next, a sequence s is constructed by choosing the 1st digit as complementary to the 1st digit of s1 (swapping 0s for 1s and vice versa), the 2nd digit as complementary to the 2nd digit of s2, the 3rd digit as complementary to the 3rd digit of s3, and generally for every n, the nth digit as complementary to the nth digit of sn. For the example above, this yields
{|
|
| s1 || = || (0, || 0, || 0, || 0, || 0, || 0, || 0, || )
|
| s2 || = || (1, || 1, || 1, || 1, || 1, || 1, || 1, || )
|
| s3 || = || (0, || 1, || 0, || 1, || 0, || 1, || 0, || )
|
| s4 || = || (1, || 0, || 1, || 0, || 1, || 0, || 1, || )
|
| s5 || = || (1, || 1, || 0, || 1, || 0, || 1, || 1, || )
|
| s6 || = || (0, || 0, || 1, || 1, || 0, || 1, || 1, || )
|
| s7 || = || (1, || 0, || 0, || 0, || 1, || 0, || 0, || )
|
|
|
|
|
| s || = || (1, || 0, || 1, || 1, || 1, || 0, || 1, || )
|}
By construction, s is a member of T that differs from each sn, since their nth digits differ (highlighted in the example). Hence, s cannot occur in the enumeration. Based on this lemma, Cantor then uses a proof by contradiction to show that:
The set T is uncountable. The proof starts by assuming that T is countable. Then all its elements can be written in an enumeration s1, s2, , sn, Applying the previous lemma to this enumeration produces a sequence s that is a member of T, but is not in the enumeration. However, if T is enumerated, then every member of T, including this s, is in the enumeration. This contradiction implies that the original assumption is false. Therefore, T is uncountable. Constructing a bijection between T and R is slightly more complicated. Instead of mapping 0111 to the decimal 0.0111 , it can be mapped to the base b number: 0.0111 b. This leads to the family of functions: The functions are injections, except for This function will be modified to produce a bijection between T and R.
Construction of a bijection between T and R
This construction uses a method devised by Cantor that was published in 1878. He used it to construct a bijection between the closed interval [0, 1] and the irrationals in the open interval (0, 1). He first removed a countably infinite subset from each of these sets so that there is a bijection between the remaining uncountable sets. Since there is a bijection between the countably infinite subsets that have been removed, combining the two bijections produces a bijection between the original sets. Cantor's method can be used to modify the function to produce a bijection from T to (0, 1). Because some numbers have two binary expansions, is not even injective. For example, 0.1000 2 = 1/2 and 0.0111 2 = 1/2, so both 1000 and 0111 map to the same number, 1/2. To modify , observe that it is a bijection except for a countably infinite subset of (0, 1) and a countably infinite subset of T. It is not a bijection for the numbers in (0, 1) that have two binary expansions. These are called dyadic numbers and have the form where m is an odd integer and n is a natural number. Put these numbers in the sequence: r = (1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, ). Also, is not a bijection to (0, 1) for the strings in T appearing after the binary point in the binary expansions of 0, 1, and the numbers in sequence r. Put these eventually constant strings in the sequence: s = ( , , 1 , 0 , 01 , 00 , 11 , 10 , ). Define the bijection g(t) from T to (0, 1): If t is the nth string in sequence s, let g(t) be the nth number in sequence r; otherwise, g(t) = 0. t2. To construct a bijection from T to R, start with the tangent function tan(x), which is a bijection from (−π/2, π/2) to R (see the figure shown on the right). Next observe that the linear function h(x) = πx – π/2 is a bijection from (0, 1) to (−π/2, π/2) (see the figure shown on the left). The composite function tan(h(x)) = tan(πx – π/2) is a bijection from (0, 1) to R. Composing this function with g(t) produces the function tan(h(g(t))) = tan(πg(t) – π/2), which is a bijection from T to R.
Конструкция биекции между T и R
In set theory, Cantor's diagonal argument, also called the diagonalisation argument, the diagonal slash argument, the anti diagonal argument, the diagonal method, and Cantor's diagonalization proof, was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into one to one correspondence with the infinite set of natural numbers. Such sets are now known as uncountable sets, and the size of infinite sets is now treated by the theory of cardinal numbers which Cantor began. The diagonal argument was not Cantor's first proof of the uncountability of the real numbers, which appeared in 1874. However, it demonstrates a general technique that has since been used in a wide range of proofs, including the first of Gödel's incompleteness theorems and Richard's paradox. He begins with a constructive proof of the following lemma:
If s1, s2, , sn, is any enumeration of elements from T, then an element s of T can be constructed that doesn't correspond to any sn in the enumeration. The proof starts with an enumeration of elements from T, for example
{|
|
| s1 = || (0, || 0, || 0, || 0, || 0, || 0, || 0, || )
|
| s2 = || (1, || 1, || 1, || 1, || 1, || 1, || 1, || )
|
| s3 = || (0, || 1, || 0, || 1, || 0, || 1, || 0, || )
|
| s4 = || (1, || 0, || 1, || 0, || 1, || 0, || 1, || )
|
| s5 = || (1, || 1, || 0, || 1, || 0, || 1, || 1, || )
|
| s6 = || (0, || 0, || 1, || 1, || 0, || 1, || 1, || )
|
| s7 = || (1, || 0, || 0, || 0, || 1, || 0, || 0, || )
|
|
|}
Next, a sequence s is constructed by choosing the 1st digit as complementary to the 1st digit of s1 (swapping 0s for 1s and vice versa), the 2nd digit as complementary to the 2nd digit of s2, the 3rd digit as complementary to the 3rd digit of s3, and generally for every n, the nth digit as complementary to the nth digit of sn. For the example above, this yields
{|
|
| s1 || = || (0, || 0, || 0, || 0, || 0, || 0, || 0, || )
|
| s2 || = || (1, || 1, || 1, || 1, || 1, || 1, || 1, || )
|
| s3 || = || (0, || 1, || 0, || 1, || 0, || 1, || 0, || )
|
| s4 || = || (1, || 0, || 1, || 0, || 1, || 0, || 1, || )
|
| s5 || = || (1, || 1, || 0, || 1, || 0, || 1, || 1, || )
|
| s6 || = || (0, || 0, || 1, || 1, || 0, || 1, || 1, || )
|
| s7 || = || (1, || 0, || 0, || 0, || 1, || 0, || 0, || )
|
|
|
|
|
| s || = || (1, || 0, || 1, || 1, || 1, || 0, || 1, || )
|}
By construction, s is a member of T that differs from each sn, since their nth digits differ (highlighted in the example). Hence, s cannot occur in the enumeration. Based on this lemma, Cantor then uses a proof by contradiction to show that:
The set T is uncountable. The proof starts by assuming that T is countable. Then all its elements can be written in an enumeration s1, s2, , sn, Applying the previous lemma to this enumeration produces a sequence s that is a member of T, but is not in the enumeration. However, if T is enumerated, then every member of T, including this s, is in the enumeration. This contradiction implies that the original assumption is false. Therefore, T is uncountable. Constructing a bijection between T and R is slightly more complicated. Instead of mapping 0111 to the decimal 0.0111 , it can be mapped to the base b number: 0.0111 b. This leads to the family of functions: The functions are injections, except for This function will be modified to produce a bijection between T and R.
Construction of a bijection between T and R
This construction uses a method devised by Cantor that was published in 1878. He used it to construct a bijection between the closed interval [0, 1] and the irrationals in the open interval (0, 1). He first removed a countably infinite subset from each of these sets so that there is a bijection between the remaining uncountable sets. Since there is a bijection between the countably infinite subsets that have been removed, combining the two bijections produces a bijection between the original sets. Cantor's method can be used to modify the function to produce a bijection from T to (0, 1). Because some numbers have two binary expansions, is not even injective. For example, 0.1000 2 = 1/2 and 0.0111 2 = 1/2, so both 1000 and 0111 map to the same number, 1/2. To modify , observe that it is a bijection except for a countably infinite subset of (0, 1) and a countably infinite subset of T. It is not a bijection for the numbers in (0, 1) that have two binary expansions. These are called dyadic numbers and have the form where m is an odd integer and n is a natural number. Put these numbers in the sequence: r = (1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, ). Also, is not a bijection to (0, 1) for the strings in T appearing after the binary point in the binary expansions of 0, 1, and the numbers in sequence r. Put these eventually constant strings in the sequence: s = ( , , 1 , 0 , 01 , 00 , 11 , 10 , ). Define the bijection g(t) from T to (0, 1): If t is the nth string in sequence s, let g(t) be the nth number in sequence r; otherwise, g(t) = 0. t2. To construct a bijection from T to R, start with the tangent function tan(x), which is a bijection from (−π/2, π/2) to R (see the figure shown on the right). Next observe that the linear function h(x) = πx – π/2 is a bijection from (0, 1) to (−π/2, π/2) (see the figure shown on the left). The composite function tan(h(x)) = tan(πx – π/2) is a bijection from (0, 1) to R. Composing this function with g(t) produces the function tan(h(g(t))) = tan(πg(t) – π/2), which is a bijection from T to R.
Эта конструкция использует метод, разработанный Кантором и опубликованный в 1878 году. Он использовал его для построения биекции между замкнутым интервалом [0, 1] и иррациональными числами в открытом интервале (0, 1). Сначала он удалил счётное бесконечное подмножество из каждого из этих множеств, чтобы между оставшимися несчётными множествами существовала биекция. Поскольку существует биекция между удалёнными счётными бесконечными подмножествами, объединение этих двух биекций даёт биекцию между исходными множествами. Метод Кантора можно использовать для модификации функции, чтобы получить биекцию из T в (0, 1). Поскольку некоторые числа имеют два двоичных представления, исходная функция даже не является инъективной. Например, 0,10002 = 1/2 и 0,01112 = 1/2, поэтому и 1000, и 0111 отображаются в одно и то же число, 1/2. Чтобы модифицировать функцию, заметим, что она является биекцией, за исключением счётного бесконечного подмножества (0, 1) и счётного бесконечного подмножества T. Она не является биекцией для чисел в (0, 1), имеющих два двоичных представления. Эти числа называются диадными и имеют вид m/2n, где m – нечётное целое число, а n – натуральное число. Поместим эти числа в последовательность: r = (1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, ...). Также исходная функция не является биекцией на (0, 1) для строк в T, следующих за двоичной точкой в двоичных представлениях 0, 1 и чисел в последовательности r. Поместим эти в конечном итоге постоянные строки в последовательность: s = ( , , 1, 0, 01, 00, 11, 10, ...). Определим биекцию g(t) из T в (0, 1): если t – n-я строка в последовательности s, пусть g(t) будет n-м числом в последовательности r; в противном случае g(t) = 0,t2. Чтобы построить биекцию из T в R, начнём с тангенциальной функции tan(x), которая является биекцией из (−π/2, π/2) в R (см. рисунок справа). Далее заметим, что линейная функция h(x) = πx – π/2 является биекцией из (0, 1) в (−π/2, π/2) (см. рисунок слева). Композитная функция tan(h(x)) = tan(πx – π/2) является биекцией из (0, 1) в R. Композиция этой функции с g(t) даёт функцию tan(h(g(t))) = tan(πg(t) – π/2), которая является биекцией из T в R.
In set theory, Cantor's diagonal argument, also called the diagonalisation argument, the diagonal slash argument, the anti diagonal argument, the diagonal method, and Cantor's diagonalization proof, was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into one to one correspondence with the infinite set of natural numbers. Such sets are now known as uncountable sets, and the size of infinite sets is now treated by the theory of cardinal numbers which Cantor began. The diagonal argument was not Cantor's first proof of the uncountability of the real numbers, which appeared in 1874. However, it demonstrates a general technique that has since been used in a wide range of proofs, including the first of Gödel's incompleteness theorems and Richard's paradox. He begins with a constructive proof of the following lemma:
If s1, s2, , sn, is any enumeration of elements from T, then an element s of T can be constructed that doesn't correspond to any sn in the enumeration. The proof starts with an enumeration of elements from T, for example
{|
|
| s1 = || (0, || 0, || 0, || 0, || 0, || 0, || 0, || )
|
| s2 = || (1, || 1, || 1, || 1, || 1, || 1, || 1, || )
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| s3 = || (0, || 1, || 0, || 1, || 0, || 1, || 0, || )
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| s4 = || (1, || 0, || 1, || 0, || 1, || 0, || 1, || )
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| s5 = || (1, || 1, || 0, || 1, || 0, || 1, || 1, || )
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| s6 = || (0, || 0, || 1, || 1, || 0, || 1, || 1, || )
|
| s7 = || (1, || 0, || 0, || 0, || 1, || 0, || 0, || )
|
|
|}
Next, a sequence s is constructed by choosing the 1st digit as complementary to the 1st digit of s1 (swapping 0s for 1s and vice versa), the 2nd digit as complementary to the 2nd digit of s2, the 3rd digit as complementary to the 3rd digit of s3, and generally for every n, the nth digit as complementary to the nth digit of sn. For the example above, this yields
{|
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| s1 || = || (0, || 0, || 0, || 0, || 0, || 0, || 0, || )
|
| s2 || = || (1, || 1, || 1, || 1, || 1, || 1, || 1, || )
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| s3 || = || (0, || 1, || 0, || 1, || 0, || 1, || 0, || )
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| s4 || = || (1, || 0, || 1, || 0, || 1, || 0, || 1, || )
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| s5 || = || (1, || 1, || 0, || 1, || 0, || 1, || 1, || )
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| s6 || = || (0, || 0, || 1, || 1, || 0, || 1, || 1, || )
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| s7 || = || (1, || 0, || 0, || 0, || 1, || 0, || 0, || )
|
|
|
|
|
| s || = || (1, || 0, || 1, || 1, || 1, || 0, || 1, || )
|}
By construction, s is a member of T that differs from each sn, since their nth digits differ (highlighted in the example). Hence, s cannot occur in the enumeration. Based on this lemma, Cantor then uses a proof by contradiction to show that:
The set T is uncountable. The proof starts by assuming that T is countable. Then all its elements can be written in an enumeration s1, s2, , sn, Applying the previous lemma to this enumeration produces a sequence s that is a member of T, but is not in the enumeration. However, if T is enumerated, then every member of T, including this s, is in the enumeration. This contradiction implies that the original assumption is false. Therefore, T is uncountable. Constructing a bijection between T and R is slightly more complicated. Instead of mapping 0111 to the decimal 0.0111 , it can be mapped to the base b number: 0.0111 b. This leads to the family of functions: The functions are injections, except for This function will be modified to produce a bijection between T and R.
Construction of a bijection between T and R
This construction uses a method devised by Cantor that was published in 1878. He used it to construct a bijection between the closed interval [0, 1] and the irrationals in the open interval (0, 1). He first removed a countably infinite subset from each of these sets so that there is a bijection between the remaining uncountable sets. Since there is a bijection between the countably infinite subsets that have been removed, combining the two bijections produces a bijection between the original sets. Cantor's method can be used to modify the function to produce a bijection from T to (0, 1). Because some numbers have two binary expansions, is not even injective. For example, 0.1000 2 = 1/2 and 0.0111 2 = 1/2, so both 1000 and 0111 map to the same number, 1/2. To modify , observe that it is a bijection except for a countably infinite subset of (0, 1) and a countably infinite subset of T. It is not a bijection for the numbers in (0, 1) that have two binary expansions. These are called dyadic numbers and have the form where m is an odd integer and n is a natural number. Put these numbers in the sequence: r = (1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, ). Also, is not a bijection to (0, 1) for the strings in T appearing after the binary point in the binary expansions of 0, 1, and the numbers in sequence r. Put these eventually constant strings in the sequence: s = ( , , 1 , 0 , 01 , 00 , 11 , 10 , ). Define the bijection g(t) from T to (0, 1): If t is the nth string in sequence s, let g(t) be the nth number in sequence r; otherwise, g(t) = 0. t2. To construct a bijection from T to R, start with the tangent function tan(x), which is a bijection from (−π/2, π/2) to R (see the figure shown on the right). Next observe that the linear function h(x) = πx – π/2 is a bijection from (0, 1) to (−π/2, π/2) (see the figure shown on the left). The composite function tan(h(x)) = tan(πx – π/2) is a bijection from (0, 1) to R. Composing this function with g(t) produces the function tan(h(g(t))) = tan(πg(t) – π/2), which is a bijection from T to R.
Общие наборы
Кантор использовал обобщенную форму диагонального аргумента для доказательства теоремы Кантора: для любого множества S, множество степеней S, то есть множество всех подмножеств S (обозначаемое здесь как P(S)), не может находиться во взаимно однозначном соответствии с самим S. Доказательство строится следующим образом:
Пусть f — произвольная функция из S в P(S). Достаточно доказать, что f не может быть сюръективной. Это означает, что существует подмножество T множества P(S), то есть некоторое подмножество S, которое не принадлежит области значений f. В качестве кандидата рассмотрим множество:
T = { s ∈ S: s ∉ f(s) }. Для любого s из S, либо s принадлежит T, либо не принадлежит. Если s принадлежит T, то по определению T, s не принадлежит f(s), следовательно, T не равно f(s). С другой стороны, если s не принадлежит T, то по определению T, s принадлежит f(s), следовательно, опять же T не равно f(s); см. иллюстрацию. Более подробное изложение этого доказательства можно найти в статье о теореме Кантора.
Орден кардиналов
С равенством, определенным как существование биекции между их базовыми множествами, Кантор также определяет бинарный предикат кардинальностей и в терминах существования инъекций между и . Он обладает свойствами предпорядка и здесь обозначается "". Можно вложить натуральные числа в двоичные последовательности, тем самым явно доказывая различные утверждения о существовании инъекций, так что в этом смысле , где обозначает пространство функций. Однако, следуя аргументам предыдущих разделов, сюръекции не существует, а следовательно, и биекции, то есть множество несчетно. Для этого можно записать , где "" понимается как существование инъекции вместе с доказанным отсутствием биекции (в отличие от альтернатив, таких как отрицание предпорядка Кантора или определение через назначенные ординалы). Также в этом смысле, как было показано, и одновременно выполняется , для всех множеств.
Принимая закон исключённого третьего, характеристические функции отображаются на булеаны, а затем , так что несчётное также не перечислимо и его также можно отобразить на . Классически, теорема Шредера — Бернштейна верна и утверждает, что любые два множества, находящиеся в инъективном образе друг друга, находятся и в биективном образе. Здесь каждое неограниченное подмножество находится в биекции с самим собой, а каждое счётное подмножество (свойство, выраженное через сюръекции) уже счётно, то есть находится в сюръективном образе . В этом контексте все возможности исчерпаны, что делает "" нестрогим частичным порядком, или даже полным порядком при допущении аксиомы выбора. Таким образом, диагональный аргумент устанавливает, что, хотя оба рассматриваемых множества бесконечны, бесконечных последовательностей единиц и нулей на самом деле больше, чем натуральных чисел. Результат Кантора также подразумевает, что понятие множества всех множеств противоречиво: если бы было множеством всех множеств, то было бы одновременно больше, чем и подмножеством .
Assuming the law of excluded middle, characteristic functions surject onto powersets, and then So the uncountable is also not enumerable and it can also be mapped onto Classically, the Schröder–Bernstein theorem is valid and says that any two sets which are in the injective image of one another are in bijection as well. Here, every unbounded subset of is then in bijection with itself, and every subcountable set (a property in terms of surjections) is then already countable, i. e. in the surjective image of In this context the possibilities are then exhausted, making "" a non strict partial order, or even a total order when assuming choice. The diagonal argument thus establishes that, although both sets under consideration are infinite, there are actually more infinite sequences of ones and zeros than there are natural numbers. Cantor's result then also implies that the notion of the set of all sets is inconsistent: If were the set of all sets, then would at the same time be bigger than and a subset of .
В отсутствие исключенных средних
Также в конструктивной математике нет сюръекции из полного домена на пространство функций или на коллекцию подмножеств, то есть эти две коллекции несчётны. Снова используя "" для доказанного существования инъекции в сочетании с отсутствием биекции, получаем и далее, , как было отмечено ранее. Аналогично, , и, конечно, , также в конструктивной теории множеств. Однако упорядочить ординалы и кардиналы конструктивно сложнее или невозможно. Например, теорема Шрёдера — Бернштейна требует закона исключённого третьего. Фактически, стандартный порядок на действительных числах, расширяющий порядок рациональных чисел, также не обязательно является разрешимым. Большинство свойств интересных классов функций также не разрешимы по теореме Райса, то есть множество счётных чисел для подчётных множеств может не быть рекурсивным и, следовательно, не быть счётным. Сложная коллекция подмножеств множества конструктивно не взаимозаменяема с коллекцией его характеристических функций. В конструктивном контексте (в котором закон исключённого третьего не принимается как аксиома), непротиворечиво принять неклассические аксиомы, противоречащие следствиям закона исключённого третьего. Несчётные множества, такие как или , могут быть заявлены как подчётные. Это понятие размера избыточно в классическом контексте, но в противном случае не обязательно подразумевает счётность. Существование инъекций из несчётного или в также возможно. Таким образом, кардинальное отношение не является антисимметричным. Следовательно, даже в присутствии множеств функциональных пространств, которые даже классически несчётны, интуиционисты не принимают это отношение как иерархию трансфинитных размеров. Когда аксиома степени не принимается, в конструктивной структуре даже подчётность всех множеств является непротиворечивой. Всё это говорит о том, что несуществование множества всех множеств в обычной теории множеств уже следует из предикативного разделения. В теории множеств моделируются теории математики. Более слабые логические аксиомы означают меньше ограничений и, следовательно, позволяют создавать более богатый класс моделей. Множество может быть идентифицировано как модель поля действительных чисел, если оно удовлетворяет некоторым аксиомам действительных чисел или их конструктивной переформулировке. Изучались различные модели, такие как действительные числа Коши или действительные числа Дедекинда, среди прочих. Первые связаны с частными от деления последовательностей, в то время как последние — это хорошо определённые сечения, взятые из степени множества, если они существуют. При наличии исключённого третьего все они изоморфны и несчётны. В противном случае варианты действительных чисел Дедекинда могут быть счётными или вкладываться в натуральные числа, но не одновременно. При предположении о счётном выборе конструктивные действительные числа Коши, даже без явного модуля сходимости, являются коши-полными, а действительные числа Дедекинда упрощаются, становясь изоморфными им. Действительно, здесь выбор также помогает диагональным построениям, и при его предположении коши-полные модели действительных чисел несчётны.
Открытые вопросы
Вдохновлённые осознанием того, что множество вещественных чисел "мощнее", чем множество натуральных чисел, возникает вопрос, существует ли множество, чья мощность находится "между" мощностью целых и вещественных чисел. Этот вопрос приводит к знаменитой гипотезе континуума. Аналогично, вопрос о том, существует ли множество, чья мощность находится между |S| и |P(S)| для некоторого бесконечного множества S, приводит к обобщённой гипотезе континуума.
Диагонализация в более широком контексте
Парадокс Рассела показал, что наивная теория множеств, основанная на неограниченной схеме выделения, противоречива. Обратите внимание на сходство между построением T и множеством в парадоксе Рассела. Следовательно, в зависимости от того, как мы модифицируем аксиоматическую схему выделения, чтобы избежать парадокса Рассела, аргументы, такие как несуществование множества всех множеств, могут оставаться или не оставаться обоснованными. Аналоги аргумента по диагонали широко используются в математике для доказательства существования или несуществования определенных объектов. Например, стандартное доказательство неразрешимости проблемы останова по сути является аргументом по диагонали. Также диагонализация первоначально использовалась для демонстрации существования классов сложности произвольной сложности и сыграла ключевую роль в ранних попытках доказать, что P не равно NP.