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| | Psoriasis often also manifests itself on the nails. Non-Steroid Topical Medication- Various other ointments may be applied topically, which do not include steroids as an active ingredient. In Vanuatu, Tamanu Oil is used by local healers (klevas) in their custom medicine and tropical first aid to treat a wide range of aliments including ALL of the ones listed below and more. And it turns out that these microbes have their favourite spots. An autoimmune disease, it is frequently hereditary, and has been linked to obesity, heart disease and diabetes. <br><br>Even if you are experiencing peeling skin or any of type of psoriasis while taking proteins, vitamins A, vitamins B and vitamins C, it is possible that you can get relief. When many people think about dermatology, they tend to think just about acne. Seawater baths are very beneficial in curing psoriasis. You should avoid all commercial versions of herbal skin care products, because their products are not really natural herbs but instead are artificial scents with little bit of essential oils. Most agree that those with exceptionally high IQs are much more likely to be afflicted with psoriasis. <br><br>Most patients who get shingles do not get it again. Read this article to find out more about healing skin psoriasis with oils. Although, there is no [http://www.dominionradio.info/ cure for Psoriasis], there are natural ways to control it which can keep it from flaring up. Most doctors agree that a small amount of natural sunlight tends to help reduce psoriasis; the problem here, of course, is that exposing the penis to sunlight cannot be done just anywhere. Moisturizers maintain skin supple, add water content to the surface of your skin and support you to maintain a youthful look. <br><br>Dandruff as most of us know is the process of continual shedding of skin cells on our scalps. It works effectively by simply stimulating the body to produce natural proteins that are lost because of skin aging. The active constituents are created to help remove the excess of scales on the scalp. Psoriasis is also considered a chronic condition and, though it can go into remission for long periods of time, it is generally one that people must live with forever. Olive oil itself contains oleocanthal, which is an anti-inflammatory agent that aides in softening lines and wrinkles. <br><br>Every segment of the health sector has benefited from the miraculous World Wide Web. Sadly 50-80% associated with psoriasis victims can get toe nail psoriasis. It is likely that you will still have some skin discoloration especially on your back and chest areas. So is there a psoriasis diet cure that actually works, guaranteed. This happened ten years ago and the psoriasis has never come back. |
| <span id="Lead"></span>
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| [[Image:Diagonal argument 01 svg.svg|right|thumb|250px|An illustration of Cantor's diagonal argument (in base 2) for the existence of [[uncountable set]]s. The sequence at the bottom cannot occur anywhere in the enumeration of sequences above.]]
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| [[File:Aplicación 2 inyectiva sobreyectiva02.svg|thumb|An [[infinite set]] may have the same [[cardinality]] as a proper [[subset]] of itself, as the depicted [[bijection]] ''f''(''x'')=2''x'' from the [[natural numbers|natural]] to the [[even numbers|even]] numbers demonstrates. Nevertheless, infinite sets of different cardinalities exist, as '''Cantor's diagonal argument''' shows.]]
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| In [[set theory]], '''Cantor's diagonal argument''', also called the '''diagonalisation argument''', the '''diagonal slash argument''' or the '''diagonal method''', was published in 1891 by [[Georg Cantor]] as a [[mathematical proof]] that there are [[infinite set]]s which cannot be put into [[bijection|one-to-one correspondence]] with the infinite set of [[natural number]]s.<ref>{{cite journal |author=Georg Cantor |title=Ueber eine elementare Frage der Mannigfaltigkeitslehre |journal=Jahresbericht der Deutsche Mathematiker-Vereinigung 1890-1891 |volume=1 |pages=75-78 (84-87 in pdf file) |year=1892 |url=http://gdz-lucene.tc.sub.uni-goettingen.de/gcs/gcs?action=pdf&metsFile=PPN37721857X_0001&divID=LOG_0001&pagesize=original&pdfTitlePage=http://gdz.sub.uni-goettingen.de/dms/load/pdftitle/?metsFile=PPN37721857X_0001%7C&targetFileName=PPN37721857X_0001_LOG_0001.pdf&}} (in german)</ref> | |
| Such sets are now known as [[uncountable set]]s, and the size of infinite sets is now treated by the theory of [[cardinal number]]s which Cantor began.
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| The diagonal argument was not Cantor's [[Cantor's first uncountability proof|first proof]] of the uncountability of the [[real number]]s; it was actually published much later than his first proof, which appeared in 1874.<ref>{{Citation |surname=Gray|given=Robert|year=1994 |url=http://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/Gray819-832.pdf |title=Georg Cantor and Transcendental Numbers|journal=[[American Mathematical Monthly]]|volume=101|pages=819–832}}</ref>
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| However, it demonstrates a powerful and general technique that has since been used in a wide range of proofs, also known as diagonal arguments by analogy with the argument used in this proof. The most famous examples are perhaps [[Russell's paradox]], the first of [[Gödel's incompleteness theorems]], and Turing's answer to the ''[[Entscheidungsproblem]]''.
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| == An uncountable set ==
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| In his 1891 article, Cantor considered the set ''T'' of all infinite sequences of [[binary digits]] (i.e. consisting only of 0es and 1s).
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| First, he [[constructive proof|constructively]] shows the following theorem:
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| :If ''s''<sub>1</sub>, ''s''<sub>2</sub>, … , ''s''<sub>''n''</sub>, … is any enumeration of elements from ''T'', then there is always an element ''s'' of ''T'' which corresponds to no ''s''<sub>''n''</sub> in the enumeration.
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| To prove this, given an enumeration of arbitrary members from ''T'', like e.g.
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| :{|
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| |-
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| | ''s''<sub>1</sub> = || (0, || 0, || 0, || 0, || 0, || 0, || 0, || ...)
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| |-
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| | ''s''<sub>2</sub> = || (1, || 1, || 1, || 1, || 1, || 1, || 1, || ...)
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| |-
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| | ''s''<sub>3</sub> = || (0, || 1, || 0, || 1, || 0, || 1, || 0, || ...)
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| |-
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| | ''s''<sub>4</sub> = || (1, || 0, || 1, || 0, || 1, || 0, || 1, || ...)
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| |-
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| | ''s''<sub>5</sub> = || (1, || 1, || 0, || 1, || 0, || 1, || 1, || ...)
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| |-
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| | ''s''<sub>6</sub> = || (0, || 0, || 1, || 1, || 0, || 1, || 1, || ...)
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| |-
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| | ''s''<sub>7</sub> = || (1, || 0, || 0, || 0, || 1, || 0, || 0, || ...)
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| |-
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| | ...
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| |}
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| he constructs the sequence ''s'' by choosing its ''n''<sup>th</sup> digit as complementary to the ''n''<sup>th</sup> digit of ''s''<sub>''n''</sub>, for every ''n''. In the example, this yields:
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| <!---
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| For each ''m'' and ''n'' let ''s''<sub>''m'',''n''</sub> be the ''n''<sup>th</sup> element of the ''m''<sup>th</sup> sequence on the list. So, for instance, ''s''<sub>2,1</sub> is the first element of the second sequence.
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| It is possible to build a sequence ''s''<sub>0</sub> in such a way that its first element is different from the first element of the first sequence in the list, its second element is different from the second element of the second sequence in the list, and, in general, its ''n''<sup>th</sup> element is different from the ''n''<sup>th</sup> element of the ''n''<sup>th</sup> sequence in the list. That is to say, if ''s''<sub>''n'',''n''</sub> is 1, then ''s''<sub>0,''n''</sub> is 0, otherwise ''s''<sub>0,''n''</sub> is 1. For instance:
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| --->
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| :{|
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| | ''s''<sub>1</sub> || = || (<u>'''0'''</u>, || 0, || 0, || 0, || 0, || 0, || 0, || ...)
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| |-
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| | ''s''<sub>2</sub> || = || (1, || <u>'''1'''</u>, || 1, || 1, || 1, || 1, || 1, || ...)
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| |-
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| | ''s''<sub>3</sub> || = || (0, || 1, || <u>'''0'''</u>, || 1, || 0, || 1, || 0, || ...)
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| |-
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| | ''s''<sub>4</sub> || = || (1, || 0, || 1, || <u>'''0'''</u>, || 1, || 0, || 1, || ...)
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| |-
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| | ''s''<sub>5</sub> || = || (1, || 1, || 0, || 1, || <u>'''0'''</u>, || 1, || 1, || ...)
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| |-
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| | ''s''<sub>6</sub> || = || (0, || 0, || 1, || 1, || 0, || <u>'''1'''</u>, || 1, || ...)
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| |-
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| | ''s''<sub>7</sub> || = || (1, || 0, || 0, || 0, || 1, || 0, || <u>'''0'''</u>, || ...)
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| |-
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| | ...
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| |-
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| |-
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| | ''s'' || = || (<u>'''1'''</u>, || <u>'''0'''</u>, || <u>'''1'''</u>, || <u>'''1'''</u>, || <u>'''1'''</u>, || <u>'''0'''</u>, || <u>'''1'''</u>, || ...)
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| |}
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| By construction, ''s'' differs from each ''s''<sub>''n''</sub>, since their ''n''<sup>th</sup> digits differ (highlighted in the example).
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| Hence, ''s'' cannot occur in the enumeration.
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| Based on this theorem, Cantor then uses an [[indirect argument]] to show that:
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| :The set ''T'' is uncountable.
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| He assumes for contradiction that ''T'' was countable.
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| Then (all) its elements could be written as an enumeration ''s''<sub>1</sub>, ''s''<sub>2</sub>, … , ''s''<sub>''n''</sub>, … .
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| Applying the above theorem to this enumeration would produce a sequence ''s'' not belonging to the enumeration.
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| This contradicts the assumption, so ''T'' must be uncountable.
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| <!---
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| The elements ''s''<sub>1,1</sub>, ''s''<sub>2,2</sub>, ''s''<sub>3,3</sub>, and so on, are here highlighted, showing the origin of the name "diagonal argument". Each element in ''s''<sub>0</sub> is, by definition, different from the highlighted element in the corresponding column of the table above it. In short, ''s''<sub>0,n</sub> ≠ ''s''<sub>n,n</sub>.
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| Therefore this new sequence ''s''<sub>0</sub> is distinct from all the sequences in the list. This follows from the fact that if it were identical to, say, the 10th sequence in the list, then we would have ''s''<sub>0,10</sub> = ''s''<sub>10,10</sub>. In general, we would have ''s''<sub>0,''n''</sub> = ''s''<sub>''n'',''n''</sup>, which, due to the construction of ''s''<sub>0</sub>, is impossible. In short, by its definition ''s''<sub>0</sub> is not contained in the countable sequence ''S''.
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| Let ''T'' be a set consisting of all infinite sequences of 0s and 1s. By its definition, this set must contain not only the sequences included in ''S'', but also ''s''<sub>0</sub>, which is just another sequence of 0s and 1s. However, ''s''<sub>0</sub> does not appear anywhere in ''S''. Hence, ''T'' cannot coincide with ''S''.
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| Because this argument applies to any countable set ''S'' of sequences of 0s and 1s, it follows that ''T'' cannot be equal to any such set. If ''T'' was countable, there would be some ''S'' that did have exactly the same members as ''T'', namely, any ''S'' that simply enumerates all the elements of ''T''. But the preceding proof shows no ''S'' contains all the elements of ''T''. Thus ''T'' is uncountable: it cannot be placed in one-to-one correspondence with the set of natural numbers <math>\mathbb{N}</math>.
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| --->
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| === Interpretation ===
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| The interpretation of Cantor's result will depend upon one's view of mathematics. To [[constructivism (mathematics)|constructivists]], the argument shows no more than that there is no [[bijection]] between the natural numbers and ''T''. It does not rule out the possibility that the latter are [[subcountable]]. In the context of [[classical mathematics]], this is impossible, and the diagonal argument establishes that, although both sets are infinite, there are actually ''more'' infinite sequences of ones and zeros than there are natural numbers.
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| === Real numbers ===
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| {| style="float:right"
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| | [[File:Tangent one period.svg|thumb|100px|The function tan: (−π/2,π/2) → '''R''']]
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| |}
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| {| style="float:right"
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| | [[File:Linear transformation svg.svg|thumb|100px|The function ''h'': (0,1) → (−π/2,π/2)]]
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| |}
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| The uncountability of the [[real number]]s was already established by [[Cantor's first uncountability proof]], but it also follows from the above result. To see this, we will build a one-to-one correspondence between the set ''T'' of infinite binary strings and a [[subset]] of '''R''' (the set of real numbers). Since ''T'' is uncountable, this subset of '''R''' must be uncountable. Hence '''R''' is uncountable.
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| To build this one-to-one correspondence (or [[bijection]]), observe that the string ''t'' = 0111… appears after the [[binary point]] in the [[binary numeral system#Fractions in binary|binary expansion]] 0.0111…. This suggests defining the function ''f''(''t'') = 0.''t'', where ''t'' is a string in ''T''. Unfortunately, ''f''(1000…) = 0.1000… = 1/2, and ''f''(0111…) = 0.0111… = [[Infinite series|1/4 + 1/8 + 1/16 + …]] = 1/2. So this function is not a bijection since two strings correspond to one number—a number having two binary expansions.
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| However, modifying this function produces a bijection from ''T'' to the [[interval (mathematics)|interval]] (0, 1)—that is, the real numbers > 0 and < 1. The idea is to remove the "problem" elements from ''T'' and (0, 1), and handle them separately. From (0, 1), remove the numbers having two binary expansions. Put these numbers in a sequence: ''a'' = (1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, …). From ''T'', remove the strings appearing after the binary point in the binary expansions of 0, 1, and the numbers in sequence ''a''. Put these eventually-constant strings in a sequence: ''b'' = ({{color|#808080|000}}…, {{color|#808080|111}}…, 1{{color|#808080|000}}…, 0{{color|#808080|111}}…, 01{{color|#808080|000}}…, 11{{color|#808080|000}}…, 00{{color|#808080|111}}…, 10{{color|#808080|111}}…, ...). A bijection ''g''(''t'') from ''T'' to (0, 1) is defined by: If ''t'' is the ''n''<sup>th</sup> string in sequence ''b'', let ''g''(''t'') be the ''n''<sup>th</sup> number in sequence ''a''; otherwise, let ''g''(''t'') = 0.''t''.
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| To build a bijection from ''T'' to '''R''': start with the [[trigonometric functions|tangent function]] tan(''x''), which provides a bijection from (−π/2, π/2) to '''R'''; see right picture. Next observe that the [[linear function]] ''h''(''x'') = π''x'' - π/2 provides a bijection from (0, 1) to (−π/2, π/2); see left picture. The [[function composition|composite function]] tan(''h''(''x'')) = tan(π''x'' - π/2) provides a bijection from (0, 1) to '''R'''. Compose this function with ''g''(''t'') to obtain tan(''h''(''g''(''t''))) = tan(π''g''(''t'') - π/2), which is a bijection from ''T'' to '''R'''. This means that ''T'' and '''R''' have the same [[cardinality]]—this cardinality is called the "[[cardinality of the continuum]]."
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| == General sets ==
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| [[File:Diagonal argument powerset svg.svg|thumb|250px|Illustration of the generalized diagonal argument: The set ''T'' = {''n''∈ℕ: ''n''∉''f''(''n'')} at the bottom cannot occur anywhere in the range of ''f'':[[natural number|ℕ]]→[[power set|'''''P''''']](ℕ). The example mapping ''f'' happens to correspond to the example enumeration ''s'' in the [[#Lead|above]] picture.]]
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| A generalized form of the diagonal argument was used by Cantor to prove [[Cantor's theorem]]: for every [[Set (mathematics)|set]] ''S'' the [[power set]] of ''S'', i.e., the set of all [[subset]]s of ''S'' (here written as '''''P'''''(''S'')), has a larger cardinality than ''S'' itself. This proof proceeds as follows:
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| Let ''f'' be any [[Function (mathematics)|function]] from ''S'' to '''''P'''''(''S''). It suffices to prove ''f'' cannot be [[surjective]]. That means that some member ''T'' of '''''P'''''(''S''), i.e., some subset of ''S'', is not in the [[Image (mathematics)|image]] of ''f''. As a candidate consider the set:
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| :''T'' = { ''s'' ∈ ''S'': ''s'' ∉ ''f''(''s'') }.
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| For every ''s'' in ''S'', either ''s'' is in ''T'' or not. If ''s'' is in ''T'', then by definition of ''T'', ''s'' is not in ''f''(''s''), so ''T'' is not equal to ''f''(''s''). On the other hand, if ''s'' is not in ''T'', then by definition of ''T'', ''s'' is in ''f''(''s''), so again ''T'' is not equal to ''f''(''s''); cf. picture.
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| For a more complete account of this proof, see [[Cantor's theorem]].
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| ===Consequences===
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| This result implies that the notion of the [[set of all sets]] is an inconsistent notion. If ''S'' were the set of all sets then '''''P'''''(''S'') would at the same time be bigger than ''S'' and a subset of ''S''.
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| [[Russell's Paradox]] has shown us that [[naive set theory]], based on an [[unrestricted comprehension]] scheme, is contradictory. Note that there is a similarity between the construction of ''T'' and the set in Russell's paradox. Therefore, depending on how we modify the axiom scheme of comprehension in order to avoid Russell's paradox, arguments such as the non-existence of a set of all sets may or may not remain valid.
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| The diagonal argument shows that the set of real numbers is "bigger" than the set of natural numbers (and therefore, the integers and rationals as well). Therefore, we can ask if there is a set whose [[cardinality]] is "between" that of the integers and that of the reals. This question leads to the famous [[continuum hypothesis]]. Similarly, the question of whether there exists a set whose cardinality is between |''S''| and |'''''P'''''(''S'')| for some infinite ''S'' leads to the [[generalized continuum hypothesis]].
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| Analogues of the diagonal argument are widely used in mathematics to prove the existence or nonexistence of certain objects. For example, the conventional proof of the unsolvability of the [[halting problem]] is essentially a diagonal argument. Also, diagonalization was originally used to show the existence of arbitrarily hard [[complexity classes]] and played a key role in early attempts to prove [[P versus NP|P does not equal NP]]. In 2008, diagonalization was used to "slam the door" on [[Laplace's demon]].<ref>{{cite journal|author=P. Binder|title=Theories of almost everything| journal=Nature|volume=455|year=2008|pages=884–885|url=http://www.astro.uhh.hawaii.edu/documents/Binder_nv-toae.pdf}}</ref>
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| ===Version for Quine's New Foundations===
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| The above proof fails for [[W. V. Quine]]'s "[[New Foundations]]" set theory (NF). In NF, the [[unrestricted comprehension|naive axiom scheme of comprehension]] is modified to avoid the paradoxes by introducing a kind of "local" [[type theory]]. In this axiom scheme,
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| :{ ''s'' ∈ ''S'': ''s'' ∉ ''f''(''s'') }
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| is ''not'' a set — i.e., does not satisfy the axiom scheme. On the other hand, we might try to create a modified diagonal argument by noticing that
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| :{ ''s'' ∈ ''S'': ''s'' ∉ ''f''({''s''}) }
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| ''is'' a set in NF. In which case, if '''''P'''''<sub>1</sub>(''S'') is the set of one-element subsets of ''S'' and ''f'' is a proposed bijection from '''''P'''''<sub>1</sub>(''S'') to '''''P'''''(''S''), one is able to use reductio to prove that |'''''P'''''<sub>1</sub>(''S'')| < |'''''P'''''(''S'')|.
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| The proof follows by the fact that if ''f'' were indeed a map ''onto'' '''''P'''''(''S''), then we could find ''r'' in ''S'', such that ''f''({''r''}) coincides with the modified diagonal set, above. We would conclude that if ''r'' is not in ''f''({''r''}), then ''r'' is in ''f''({''r''}) and vice-versa.
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| It is ''not'' possible to put '''''P'''''<sub>1</sub>(''S'') in a one-to-one relation with ''S'', as the two have different types, and so any function so defined would violate the typing rules for the comprehension scheme.
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| ==See also==
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| *[[Cantor's first uncountability proof]]
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| *[[Controversy over Cantor's theory]]
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| ==References==
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| <references/>
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| == External links ==
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| *[http://www.mathpages.com/home/kmath371.htm Cantor's Diagonal Proof] at MathPages
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| *{{MathWorld |title=Cantor Diagonal Method |id=CantorDiagonalMethod }}
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| {{Set theory}}
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| {{DEFAULTSORT:Cantor's Diagonal Argument}}
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| [[Category:Set theory]]
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| [[Category:Theorems in the foundations of mathematics]]
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| [[Category:Mathematical proofs]]
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| [[Category:Infinity]]
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| [[Category:Arguments]]
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| [[Category:Cardinal numbers]]
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