로그인하고 있지 않습니다. 편집하면 당신의 IP 주소가 공개적으로 기록됩니다. 계정을 만들고 로그인하면 편집 시 사용자 이름만 보이며, 위키 이용에 여러 가지 편의가 주어집니다.스팸 방지 검사입니다. 이것을 입력하지 마세요!==선택 공리의 직관적/반직관적인 결과== {{인용문|The Axiom of Choice is obviously true; the Well Ordering principle is obviously false; and who can tell about Zorn’s lemma?|Jerry Lloyd Bona<ref>{{서적 인용 |저자= Schechter, Eric |제목= Handbook of analysis and its foundations |쪽= 145 |출판사= Academic Press |날짜= 1997 |doi= 10.1016/B978-0-12-622760-4.50033-9 |Zbl= 0943.26001 }}</ref>. }} <!-- AC 없이 생길 수 있는 일>> There can be a nonempty tree T, with no leaves, but which has no infinite path. That is, every finite path in the tree can be extended one more step, but there is no path that goes forever. A real number can be in the closure of a set X\subset\mathbb{R}, but not the limit of any sequence from X. A function f:\mathbb{R}\to\mathbb{R} can be continuous in the sense that x_n\to x\Rightarrow f(x_n)\to f(x), but not in the \epsilon\ \delta sense. A set can be infinite, but have no countably infinite subset. Thus, it can be incorrect to say that \aleph_0 is the smallest infinite cardinality, since there can be infinite sets of incomparable size with \aleph_0. (see this MO answer.) There can be an equivalence relation on \mathbb{R}, such that the number of equivalence classes is strictly greater than the size of \mathbb{R}. (See François's excellent answer.) This is a consequence of AD, and thus relatively consistent with DC and countable AC. There can be a field with no algebraic closure. The rational field \mathbb{Q} can have different nonisomorphic algebraic closures (due to Läuchli, see Timothy Chow's comment below). Indeed, \mathbb{Q} can have an uncountable algebraic closure, as well as a countable one. There can be a vector space with no basis. There can be a vector space with bases of different cardinalities. The reals can be a countable union of countable sets. Consequently, the theory of Lebesgue measure can fail totally. The first uncountable ordinal \omega_1 can be singular. More generally, it can be that every uncountable \aleph_\alpha is singular. Hence, there are no infinite regular uncountable well-ordered cardinals. 하지만 선택공리를 가정하게 되면, 실수 <math>\mathbb R</math>의 르벡 비가측인 집합이 생기게 The fact that there exist non-measurable sets is highly counter-intuitive; the reason we don't find it so is that we've all been conditioned from day 1 to do measure theory very carefully, and define Borel sets, measurable sets, etc, so we all know that non-measurable sets exist because what would be the point of doing it all so carefully otherwise. At high school we were all taught that the probability of an event occurring was "do it a million times, count how often it happened, divide by a million, and now let a million tend to infinity". And no-one thought to ask "what if this process doesn't tend to a limit?". I bet if anyone asked their teacher they'd say "well it always tends to a limit, that's intuitively clear". But am I right in thinking the following: if we take a subset XX of [0,1] with inner measure 0 and outer measure 1, and we keep choosing random reals uniformly in [0,1] and asking whether they land in XX, and keep a careful table of the result, then the number of times we land in XX divided by the number of times we tried just oscillates around between 0 and 1 without converging? That is fundamentally counterintuitive and in some sense completely goes against the informal (non-rigorous) training that we all got in probability at high school. [if I've got this right!] There can be graphs all of whose cycles have even length and whose chromatic number is greater than two. In fact, let GG be the graph whose vertices are the real numbers, with xx and yy adjacent if |x−y|=2–√+r|x−y|=2+r, where rr is rational. Then GG has only even length cycles. Assuming that every subset of RR is measurable (which is consistent with ZF), then the chromatic number of GG is uncountable. This is a result of Shelah and Soifer. If we assume the Axiom of Choice, then the chromatic number of GG is two. -----------이하 AC일 때의 문제 Banach Tarski paradox There are of course reasons to care about the axiom of choice because there are categories in which epimorphisms do not split. However if one sticks to the category of sets my position could be (provocatively) described as follows: The axiom of choice is obviously false but that doesn't stop me from using it. Before I go on to explain why I think it is false let me make a general remark. Set theory is a mathematical model for mathematics it isn't mathematics itself. We all know that all models usually manage to model some part of what they model but they almost never correctly model everything. Things are a little bit more complicated in the case of set theory but it is also supposed to be the common language of mathematics. However, it really works as such only as a protocol for conflict resolution; in case we disagree over a proof we are supposed to work our way down to formal set theory where there couldn't possibly be any conflicts. However, most mathematicians would rather, I believe, voluntarily submit to extended flagellation than actually work with formal set theory. Luckily, in practice all disagreements are resolved long before one reaches that level. Furthermore, most working mathematicians show a cavalier towards set theory. It is quite common to speak of the free group on isomorphism classes of objects in some large category which is not possible in formal set theory as the the isomorphism classes are themselves proper classes and hence can not be members of some class. Of course, when pressed a mathematician using such a phrase would probably modify it by speaking of skeleta but I have once been criticised when using a slightly different formulation that avoided the problem without speaking of skeleta as being wrong for set-theoretic reasons. (This is not meant as a criticism of the person in question, a working mathematician should have the right to ignore the horny parts of set theory, at their own peril of course.) Now, the reason why I believe that the axiom of choice is obviously false is that gives us an embedding of the field of pp-adic numbers into CC which seems fishy as they are constructed in such different ways. In fact if you try to pin down such an embedding by asking for it to fulfil more conditions then it doesn't exist. This is true if you ask that it be measurable or take definable numbers to definable numbers and so on. My own feeling is that its existence is so counter-intuitive that it couldn't possibly existst. On the other hand such an embedding is used over and over again in say the theory of ℓℓ-adic cohomology. It is true that in that case at least it can be avoided (Deligne seems to share some of my disbelief as in his second paper on the Weil conjecture he starts with a short discussion on how to avoid it but still uses it as it cuts down on uninteresting arguments). My feeling about the axiom of choice is pragmatic; it is useful and doesn't seem to get us in trouble so I have no qualms using it (even though I don't believe in it fully). I have also a picture of sets which could be used to justify this contradictory (I am not trying to formalise it so it should not be considered a competing model of mathematics). To me all elements of a set are not on equal footing. Taking my cue from algebraic geometry, there are closed points which are "real" elements but also non-closed points. Hence, the set of embeddings of the pp-adic numbers in CC is under the axiom of choice a non-empty set but in my opinion it does not contain any closed points. (In fact all its elements are probably generic, i.e., their closure is the whole set.) As long as you only deal with "concrete" objects (which should be closed in any set in which they are contained but maybe should be something more) the conclusions about them that are obtained by using the axiom of choice should be OK. The Axiom of Determinacy (AD) fails. What that means: Partition the set ωω into two sets S and T, and think of this partition as a game (S, T) with two players. To play, player 1 picks a natural number a0, then player 2 picks b0 (as a function of a0), then player 1 picks a1 (as a function of b0), then player 2 picks b1 (as a function of a0 and a1), and so on until an and bn are selected for all n ∈ ω. Then the sequence a0, b0, a1, b1, … is either in S (in which case player 1 wins), or in T (in which case player 2 wins). The game (S, T) is determined if either player 1 or player 2 has a winning strategy, i.e., if there are functions fn: nω → ω where choosing an = fn( b0, …, bn–1 ) guaranteed player 1 victory, or similarly for player 2. (We can't have both.) AD is just the statement that every such game is determined, which is false in ZFC. As with most of the weird examples, the undetermined game is constructed with a well-ordering of R. What makes this so unintuitive to me is that both AC and AD are generalizations of statements that are easily seen for finite objects. (Any finite game, or even any game with finite depth, is determined, by an easy induction on the depth.) There are apparently many set theorists that agree with this assessment, since they try to rescue AD as relativized to L(R). That the relative consistency strength of this statement is equivalent to that of large cardinals is considered good evidence that those large cardinals are, in fact, consistent. More precisely, ZF + AD is consistent iff ZFC + "there are infinitely many Woodin cardinals" is consistent, and ADL(R) is outright provable in ZFC + "there is a measurable cardinal which is greater than infinitely many Woodin cardinals". 공사중 --> 요약: 리브레 위키에서의 모든 기여는 크리에이티브 커먼즈 저작자표시-동일조건변경허락 3.0 라이선스로 배포됩니다(자세한 내용에 대해서는 리브레 위키:저작권 문서를 읽어주세요). 만약 여기에 동의하지 않는다면 문서를 저장하지 말아 주세요. 글이 직접 작성되었거나 호환되는 라이선스인지 확인해주세요. 리그베다 위키, 나무위키, 오리위키, 구스위키, 디시위키 및 CCL 미적용 사이트 등에서 글을 가져오실 때는 본인이 문서의 유일한 기여자여야 하고, 만약 본인이 문서의 유일한 기여자라는 증거가 없다면 그 문서는 불시에 삭제될 수 있습니다. 취소 편집 도움말 (새 창에서 열림) | () [] [[]] {{}} {{{}}} · <!-- --> · [[분류:]] · [[파일:]] · [[미디어:]] · #넘겨주기 [[]] · {{ㅊ|}} · <onlyinclude></onlyinclude> · <includeonly></includeonly> · <noinclude></noinclude> · <br /> · <ref></ref> · {{각주}} · {|class="wikitable" · |- · rowspan=""| · colspan=""| · |} {{lang|}} · {{llang||}} · {{인용문|}} · {{인용문2|}} · {{유튜브|}} · {{다음팟|}} · {{니코|}} · {{토막글}} {{삭제|}} · {{특정판삭제|}}(이유를 적지 않을 경우 기각될 가능성이 높습니다. 반드시 이유를 적어주세요.) {{#expr:}} · {{#if:}} · {{#ifeq:}} · {{#iferror:}} · {{#ifexist:}} · {{#switch:}} · {{#time:}} · {{#timel:}} · {{#titleparts:}} __NOTOC__ · __FORCETOC__ · __TOC__ · {{PAGENAME}} · {{SITENAME}} · {{localurl:}} · {{fullurl:}} · {{ns:}} –(대시) ‘’(작은따옴표) “”(큰따옴표) ·(가운뎃점) …(말줄임표) ‽(물음느낌표) 〈〉(홑화살괄호) 《》(겹화살괄호) ± − × ÷ ≈ ≠ ∓ ≤ ≥ ∞ ¬ ¹ ² ³ ⁿ ¼ ½ ¾ § € £ ₩ ¥ ¢ † ‡ • ← → ↔ ‰ °C µ(마이크로) Å °(도) ′(분) ″(초) Α α Β β Γ γ Δ δ Ε ε Ζ ζ Η η Θ θ Ι ι Κ κ Λ λ Μ μ(뮤) Ν ν Ξ ξ Ο ο Π π Ρ ρ Σ σ ς Τ τ Υ υ Φ φ Χ χ Ψ ψ Ω ω · Ά ά Έ έ Ή ή Ί ί Ό ό Ύ ύ Ώ ώ · Ϊ ϊ Ϋ ϋ · ΐ ΰ Æ æ Đ(D with stroke) đ Ð(eth) ð ı Ł ł Ø ø Œ œ ß Þ þ · Á á Ć ć É é Í í Ĺ ĺ Ḿ ḿ Ń ń Ó ó Ŕ ŕ Ś ś Ú ú Ý ý Ź ź · À à È è Ì ì Ǹ ǹ Ò ò Ù ù · İ Ż ż ·  â Ĉ ĉ Ê ê Ĝ ĝ Ĥ ĥ Î î Ĵ ĵ Ô ô Ŝ ŝ Û û · Ä ä Ë ë Ï ï Ö ö Ü ü Ÿ ÿ · ǘ ǜ ǚ ǖ · caron/háček: Ǎ ǎ Č č Ď ď Ě ě Ǐ ǐ Ľ ľ Ň ň Ǒ ǒ Ř ř Š š Ť ť Ǔ ǔ Ž ž · breve: Ă ă Ğ ğ Ŏ ŏ Ŭ ŭ · Ā ā Ē ē Ī ī Ō ō Ū ū · à ã Ñ ñ Õ õ · Å å Ů ů · Ą ą Ę ę · Ç ç Ş ş Ţ ţ · Ő ő Ű ű · Ș ș Ț ț