Essays北极甜虾 · No. 13

SET THEORY · IV数到无穷之后
第四讲 · 连续统假设

一个答不出的问题

连续统假设:是,也不是
实数有多少个?第二讲说是 ,第三讲说无穷基数排成 。那么 是其中哪一个?Cantor 猜是 ,Hilbert 把它列为二十世纪的第一个问题。答案在 1963 年揭晓:用通常的公理,既证不了,也否定不了。这一讲讲清楚“证不了”是怎么被证明的。七条定理,都带证明或证明要点。
Ⅰ可数性
Ⅱ比大小
Ⅲ序数
Ⅳ连续统
Ⅴ大基数
上一讲

良序就是没有无穷下降;序数是良序的长度。Goodstein 序列涨到天文数字后必落回 0,可 Peano 算术证不出来。所有可数序数排在一起是 ,它的基数是 ,第一个比 大的无穷。

先交作业:467

从 5 出发的 Goodstein 序列,第三项是 255,在底数 4 下写成 。把 4 换成 5 得 ,减 1,第四项是 。

1
THE QUESTION

问题本身

ZFC 只排除了少数几个位置;其余每一个,都和 ZFC 相容
图 4·1 落在 序列的哪一格?本讲的结论是: 不行,其余大多数格子都和通常的公理相容。
连续统假设CH

。换句话说:实数的每个无穷子集,要么可数,要么和全体实数一样多,中间没有别的大小。

两种说法等价: 是比 大的第一个基数,所以“中间没有别的大小”就是“ 恰好是 ”(这里用到选择公理,保证 是某个 )。

Cantor 1878 年就这样猜了,此后几十年一直想证明它,没有成功。1900 年,Hilbert 在巴黎的国际数学家大会上列了一串问题,第一个就是它。同一个问题里,Hilbert 还提到 Cantor 的另一个断言:实数能排成良序(第三讲),并请人给出直接的证明。后一半四年后由 Zermelo 用选择公理解决;前一半,就是这一讲的故事。

直觉上 CH 很可信:我们能具体写出来的实数集合,好像总是要么可数,要么和实数一样多。下面这条定理说,至少对闭集,这个直觉是对的。

定理 1Cantor–Bendixson(1883–84)

实数轴上的每个闭集,都是一个“完美集”和一个可数集的并。完美集要么是空集,要么恰好有 个点。所以每个闭集要么可数,要么有 个点:闭集满足 CH。

证明要点完美集是没有孤立点的闭集。对闭集 ,叫一个点 为“凝聚点”,如果它的任何邻域里都有 的不可数个点。非凝聚点都落在某个以有理数为端点、只含可数个 中点的区间里,这样的区间只有可数个,所以非凝聚点只有可数个。凝聚点组成的集合 是闭的、没有孤立点,就是完美集。

非空完美集 至少有 个点:在 里取两个不交的闭区间,各含 的点(没有孤立点保证能取);每个再分成两个,以此类推,每一步区间长度减半以下。每条“左右左右……”的无穷路径对应一串套着的闭区间,交出 里唯一一个点。不同路径给出不同的点,于是 里有一份 Cantor 集,至少 个点。∎

闭集
去掉孤立点
再去掉
剩下的“完美核”里没有孤立点,它恰好有 个点
蓝色:完美部分 金色:孤立点,一共只有可数个(示意)
图 4·2剥掉孤立点,剥一次可能又露出新的孤立点;一直剥下去(这正是第三讲里 Cantor 发明序数的那个过程),剩下的是完美核。

这条定理的历史有点曲折:Cantor 1883 年先不加证明地写下一个版本,同年 Bendixson 在给他的信里指出其中一处不对,改正后给了证明;“完美集恰好有 个点”是 Cantor 1884 年证的。1916 年,Alexandrov 和 Hausdorff 各自把结论推广到了更复杂的集合(Borel 集),它们也满足 CH。问题在于:实数的子集远不止这些“看得见”的集合。

2
WHAT ZFC CAN SAY

公理能说的一点点

关于 到底是多大,通常的公理系统 ZFC 能证明的东西少得可怜。最重要的一条来自 Julius König,背后还有一段插曲:1904 年他在国际数学家大会上宣布,实数根本不能排成良序,CH 因此没有意义。论证里的一个引理很快被发现有漏洞,第二年他在论文里明确收回了这个结论,只留下下面这条站得住的部分。

定理 2König(1905)

实数集不是可数个“比它小”的集合的并:如果 每个都满足 ,那么它们的并不可能是整个 。

证明第二讲证过 ,所以不妨把 换成全体实数列 。设 ,。把 投影到第 个坐标,得到 ,它的大小不超过 ,所以有实数 。令 。对每个 , 的第 个坐标不在 里,所以 。于是 不在并里,矛盾。∎

坐标
第 个坐标 避开 ,于是 不在任何 里
图 4·3又是一条对角线:第 个坐标专门对付第 个集合。

推论马上就来:。因为 恰好是 这可数个更小基数的并。同样的理由排除 、 等所有“能由可数步逼近”的 。König 本人只证了这种可数步的情形;它的一般形式(无穷多个基数之和小于对应的积)是 Zermelo 1908 年写下的。

除此之外,ZFC 对 几乎什么都说不出来。后面会看到,这不是我们不够聪明。

3
GÖDEL

Gödel:否定不了

“证明一个命题证不出来”听起来像绕口令。办法其实很朴素:找一个满足所有公理、却让这个命题为假的“世界”(模型)。如果命题能从公理推出来,它在所有满足公理的世界里都得成立。

1938 年 Gödel 造出了一个让 CH 成立的世界,叫“可构造宇宙” 。

:全体集合
里的实数,都在 之前出现
每一层只收前一层里“能用公式定义出来”的子集
图 4·4全体集合 是一个越往上越宽的锥; 是它里面一个瘦长的子宇宙(示意)。

一层一层搭起来: 是空集;每一层 只收 里“能用一句一阶公式定义出来”的子集;到了极限序数就把前面的并起来。和全体集合比, 非常“吝啬”:它只放进能被说出来的集合。

定理 3Gödel(1938)

满足 ZFC 的全部公理,而且在 里 CH 成立(广义连续统假设也成立)。因此,如果 ZF 不矛盾,ZFC 加上 CH 也不矛盾:ZFC 否定不了 CH。

为什么 CH 在 里成立关键一步叫“凝聚引理”: 里每个实数(也就是 的子集),在某个可数层 ()里就已经出现了。原因是,描述这个实数只需要可数多的“原料”,把用到的原料收拢起来,会塌缩成某一个可数层。每个可数层只有可数个元素,一共 层,所以 里的实数至多 个。(这里只讲思路,不证)

4
COHEN

Cohen:也证明不了

另一半更难:造一个 CH 不成立的世界。 是把宇宙“削瘦”,Cohen 要做的是“加胖”,往宇宙里塞进新的实数。可塞什么?宇宙里已经有了“所有”实数,新实数从哪来?

Cohen 1963 年的办法叫力迫法(forcing)。先打个比方:在有理数里加进 ,你不需要知道 “到底是什么”,只需要知道它满足什么方程,就能把扩张后的世界算清楚。力迫法是一样的思路:只用“有限的信息”去描述一个新实数,再证明这些信息能拼成一个和原来所有实数都不同的实数。

说得具体一点。从一个满足 ZFC 的可数模型 出发。一个“条件” 是一个有限的 0/1 串,表示新实数的前几位;更长的串是更强的条件。 里的一个集合 叫“稠密”,如果任何条件都能延长到 里。

定理 4泛型实数(Rasiowa–Sikorski 引理的特例)

设 可数。则存在一个无穷 0/1 串 ,它的某个前段落在 里的每一个稠密集里。这样的 不等于 里的任何实数。

证明 可数,所以它的稠密集只有可数个,排成 。从空串出发,第 步把当前的串延长到 里(稠密保证做得到)。这些串一个比一个长,合起来就是 ,它的前段依次落进了每个 。

对 里的任何实数 ,令 是所有“在某一位上和 不同”的有限串。任何有限串都能在后面添一位和 相反的数字,所以 稠密,而且它在 里。于是 的某个前段在 里, 在那一位和 不同,。∎

金色:泛型实数 的前几位
蓝圈: 里的几个实数, 都在某一位上和它们分开
图 4·5 沿二叉树一路往下,每遇到一个稠密集就走进去,于是躲开了 里的每一个实数。仔细看,这又是第一讲那条对角线,只是对手换成了一个可数模型。

把 加进 ,再把能由它造出来的集合都放进去,就得到更大的模型 。力迫法的核心定理说: 仍然满足 ZFC,而且 里什么成立,可以在 里面用“条件”算出来。

定理 5Cohen(1963)

如果 ZF 不矛盾,那么 ZFC 加上 也不矛盾。因此 ZFC 证明不了 CH。

证明要点不是加一个新实数,而是一次加 个(条件换成“对 个新实数,各给有限多位”)。剩下的难点是:加完以后,原来的 、 会不会塌缩成可数,使得“ 个”名不副实?Cohen 证明不会:这些条件满足一种叫“可数链条件”的组合性质,任何两两不相容的条件组都只有可数个,这保证了基数不塌缩。(这里只讲思路,不证)

Cohen 把证明寄给 Gödel 检查;Gödel 读完写信称赞,并亲自把它推荐给《美国国家科学院院刊》发表。Gödel 加 Cohen:CH 和 ZFC 独立,既证不出来,也否定不了。Cohen 因此获得 1966 年的菲尔兹奖。2019 年,Han 和 van Doorn 在证明助手 Lean 里把这整套论证形式化了一遍。

5
ALMOST ANYTHING GOES

几乎什么都可以

Cohen 的方法不只能让 。换一换加入的新实数的个数,就能让它等于各种各样的 。到底哪些可以?

定理 6Solovay、Cohen(1963–65);Easton(1970)

对任何一个 ,只要它不能写成可数个更小基数的“极限”(即共尾性大于 ),“”就和 ZFC 相容。Easton 进一步证明,对所有正则基数 , 的取值几乎可以随意指定,只需满足单调性和 König 那条限制。

可以
可以
可以
不行
可以
规则: 可以是任何“不能写成可数个更小基数之极限”的
图 4·6König 的定理排除的,恰好是全部能被排除的;其余每个位置都有一个满足 ZFC 的世界。

换句话说:除了“它比 大”,ZFC 对 的全部了解,就是定理 2 那一条。

6
IS IT TRUE?

那它到底对不对

“独立”回答了“能不能证”,没有回答“对不对”。如果你相信集合有一个确定的宇宙,那 CH 在里面要么真、要么假,只是 ZFC 的公理不够多,看不出来。Gödel 就这么想。1947 年他写道,CH 最可能是独立的,而他期待新的公理会否定它;晚年的一份手稿里,他甚至猜实数的个数是 。

有一个漂亮的等价形式,能让人对 CH 产生“几何直觉”。

定理 7Sierpiński(1919)

CH 成立,当且仅当平面 能分成两部分 、:每条水平线和 只交于可数个点,每条竖直线和 只交于可数个点。

证明(⇒)在 CH 下,把 排成 型的良序 ,于是每个实数前面只有可数个实数。令 ,。固定 ,水平线上属于 的点是 ,可数;固定 ,竖直线上属于 的点是 ,可数。

(⇐)假设 CH 不成立,有这样的分解。取 的一个大小为 的子集 。每条竖直线 ()上属于 的点只有可数个,它们的纵坐标合起来至多 个,比 少,所以有一个 不在其中。这条水平线上,所有 ()都不在 里,只能在 里,于是这条水平线和 交于 个点,不可数,矛盾。∎

(按 型的良序排)
竖线上的 段、横线上的 段,都只是一个可数的前段
图 4·7把横轴、纵轴都按 型的良序排,对角线一刀切开平面。(示意:真实的良序和实数的大小顺序毫无关系)

1986 年 Freiling 把这个等价形式变成了一个思想实验。往实数轴上随机掷两支飞镖,得到 、。如果 CH 成立,按上面的分解,要么 落在“ 那条竖线上可数个坏点”里,要么 落在“ 那条横线上可数个坏点”里。可一个随机点落进一个事先给定的可数集,概率是 0。按“掷镖的直觉”,这两件事都不该发生,所以 CH 应该是假的。

这个论证的漏洞在于:那些“可数集”是由不可测的分解给出的,概率论在这里并不适用。所以它是直觉,不是证明,信不信由你。

今天的集合论学者在这个问题上分成几派。有人在找新的、大家都能接受的公理来决定 CH:一类强力的“力迫公理”推出 ,2021 年 Asperó 和 Schindler 把其中两条主要的候选公理连了起来。另一条路线希望找到一个“终极的 ”,它会让 CH 成立。还有人认为根本没有唯一的宇宙,CH 在有的宇宙里真、有的宇宙里假,就像平行公设在不同的几何里一样。这是下一讲的话题。

7
CODA

写在最后

这一讲的主角是一个答不出的问题。Cantor 猜 ;闭集都满足它,这让猜想显得很可信;König 只能排除 这类位置;Gödel 造了一个瘦的宇宙让它成立,Cohen 造了一个胖的宇宙让它不成立。证明“证不出来”的那把钥匙,竟然又是第一讲那条对角线。

如果公理不够用,要不要加新公理?加什么?“更大的无穷”能不能帮上忙?下一讲,这个系列的最后一讲,往无穷的最高处走。

公理说不出的,
不一定没有答案;
也可能是问题本身有好几个答案。
✎ 留言区小题

下面哪些可能是 (在某个满足 ZFC 的世界里):、、、?提示:看它能不能由可数步逼近。答案下一讲公布。

参考资料
  1. D. Hilbert, “Mathematische Probleme”(1900),第一问题
  2. G. Cantor, Math. Ann. 21 (1883);Acta Math. 4 (1884);I. Bendixson, Acta Math. 2 (1883);P. Alexandroff, C. R. Paris 162 (1916);F. Hausdorff, Math. Ann. 77 (1916)
  3. J. König, “Zum Kontinuum-Problem”, Math. Ann. 60 (1905);E. Zermelo, Math. Ann. 65 (1908)
  4. K. Gödel, PNAS 24 (1938);The Consistency of the Continuum Hypothesis(1940)
  5. P. J. Cohen, “The independence of the continuum hypothesis” I、II, PNAS 50 (1963)、51 (1964)
  6. H. Rasiowa, R. Sikorski, Fund. Math. 37 (1950)
  7. R. M. Solovay, “ can be anything it ought to be”(1965);W. B. Easton, “Powers of regular cardinals”, Ann. Math. Logic 1 (1970)
  8. W. Sierpiński, Bull. Int. Acad. Sci. Cracovie(1919 年 2 月 24 日会议);Hypothèse du continu(1934)
  9. C. Freiling, “Axioms of symmetry: throwing darts at the real number line”, J. Symbolic Logic 51 (1986)
  10. K. Gödel, “What is Cantor’s continuum problem?”, Amer. Math. Monthly 54 (1947)
  11. D. Asperó, R. Schindler, “Martin’s Maximum⁺⁺ implies Woodin’s axiom (*)”, Ann. of Math. 193 (2021)
  12. J. M. Han, F. van Doorn, “A formal proof of the independence of the continuum hypothesis”, CPP(2020)
  13. T. Jech, Set Theory(第三版,2003)
注:文中插图均为本文自制示意图。
犹剑在心
北极甜虾(剑心犹在!)
— 第四讲完 —
Translated from the Chinese original, “一个答不出的问题”.
SET THEORY · IVCounting Past Infinity
Lecture 4 · The Continuum Hypothesis

A Question with No Answer

The continuum hypothesis: yes, and also no
How many real numbers are there? Lecture 2 gave the answer ; Lecture 3 showed that the infinite cardinals line up as . So which of them is ? Cantor guessed , and Hilbert put the question first on his list of problems for the twentieth century. The answer came in 1963: with the usual axioms, it can be neither proved nor refuted. This lecture explains how “it cannot be proved” was itself proved. Seven theorems, each with a proof or a proof sketch.
ⅠCountable
ⅡComparing Sizes
ⅢOrdinals
ⅣContinuum
ⅤLarge Cardinals
Last lecture

A well-ordering is an order with no infinite descent; an ordinal is the length of a well-ordering. Goodstein sequences climb to astronomical heights and yet always fall back to 0, but Peano arithmetic cannot prove it. All the countable ordinals, lined up together, make , whose cardinality is , the first infinity larger than .

Homework first: 467

In the Goodstein sequence starting from 5, the third term is 255, written in base 4 as . Replace 4 by 5 to get , subtract 1, and the fourth term is .

1
THE QUESTION

The question itself

ZFC rules out only a few; every other spot is consistent
Figure 4·1Where in the sequence of does fall? This lecture's conclusion: is impossible, and most of the other slots are consistent with the usual axioms.
The continuum hypothesisCH

. In other words: every infinite set of real numbers is either countable or as large as the set of all real numbers, with no size in between.

The two forms say the same thing: is the first cardinal larger than , so “no size in between” means exactly “ is ” (this uses the axiom of choice, which guarantees that is some ).

Cantor made this conjecture in 1878 and spent decades afterwards trying to prove it, without success. In 1900, at the International Congress of Mathematicians in Paris, Hilbert presented a list of problems, and this was the first. In the same problem Hilbert also mentioned another of Cantor's claims, that the real numbers can be well-ordered (Lecture 3), and asked for a direct proof. Zermelo settled that second half four years later with the axiom of choice; the first half is the story of this lecture.

Intuitively, CH is very believable: every set of real numbers we can actually write down seems to be either countable or as large as the reals. The next theorem says that, for closed sets at least, this intuition is right.

Theorem 1Cantor–Bendixson (1883–84)

Every closed set on the real line is the union of a “perfect set” and a countable set. A perfect set is either empty or has exactly points. So every closed set is either countable or has points: closed sets satisfy CH.

Proof sketchA perfect set is a closed set with no isolated points. For a closed set , call a point a “condensation point” if every neighborhood of it contains uncountably many points of . Every other point lies in some interval with rational endpoints that contains only countably many points of ; there are only countably many such intervals, so there are only countably many points that are not condensation points. The set of condensation points is closed and has no isolated points: it is a perfect set.

A nonempty perfect set has at least points. Inside pick two disjoint closed intervals, each containing points of (having no isolated points makes this possible); split each of them in two again, and so on, with the intervals at least halving in length at every step. Each infinite path “left, right, left, right, …” gives a nested sequence of closed intervals, whose intersection is a single point of . Different paths give different points, so contains a copy of the Cantor set and has at least points.∎

Closed set
Drop isolated points
And again
Perfect kernel left: no isolated points, exactly points
Blue: perfect part; gold: isolated points, countably many (sketch)
Figure 4·2Peel off the isolated points; one peeling may expose new isolated points. Keep peeling (this is exactly the process through which Cantor invented the ordinals, as told in Lecture 3), and what remains is the perfect kernel.

The history of this theorem is a little tangled. In 1883 Cantor first stated a version without proof; the same year Bendixson pointed out, in a letter to him, that part of it was wrong, corrected it and gave a proof. That a perfect set has exactly points was proved by Cantor in 1884. In 1916 Alexandrov and Hausdorff independently extended the result to more complicated sets, the Borel sets, which satisfy CH as well. The trouble is that the subsets of the real line go far beyond these “visible” sets.

2
WHAT ZFC CAN SAY

What little the axioms can say

The usual axiom system, ZFC, can prove pitifully little about how big really is. The most important result is due to Julius König, and there is a story behind it. At the 1904 International Congress of Mathematicians he announced that the real numbers cannot be well-ordered at all, so that CH would be meaningless. A lemma in his argument was soon found to have a gap, and the next year, in a paper, he explicitly withdrew that conclusion, keeping only the part that holds up, stated below.

Theorem 2König (1905)

The set of real numbers is not a countable union of sets “smaller than it”: if the sets each satisfy , then their union cannot be all of .

ProofLecture 2 proved , so we may replace by the set of all sequences of real numbers, . Suppose , where . Projecting onto the th coordinate gives , whose size is at most , so there is a real number . Let . For each , the th coordinate of is not in , so . Hence is not in the union, a contradiction.∎

Coord.
The th coordinate avoids , so lies in no
Figure 4·3Another diagonal: the th coordinate exists only to defeat the th set.

A corollary follows at once: . For is exactly the union of the countably many smaller cardinals . The same reasoning rules out , and every that “can be approached in countably many steps.” König himself proved only this countable case; the general form (an infinite sum of cardinals is smaller than the corresponding product) was written down by Zermelo in 1908.

Beyond this, ZFC can say almost nothing about . As we will see, this is not because we are not clever enough.

3
GÖDEL

Gödel: it cannot be refuted

“Proving that a statement cannot be proved” sounds like a tongue-twister. The method is actually quite plain: find a “world” (a model) that satisfies all the axioms but in which the statement is false. If the statement could be derived from the axioms, it would have to hold in every world that satisfies them.

In 1938 Gödel built a world in which CH holds, called the “constructible universe” .

: all sets
Every real in appears before
Each level takes only the “definable” subsets of the one below
Figure 4·4The universe of all sets, , is a cone that widens as it rises; is a tall, thin sub-universe inside it (schematic).

is built level by level: is the empty set; each level takes only those subsets of that “can be defined by a first-order formula”; at limit ordinals, take the union of all the levels before. Compared with the universe of all sets, is very “stingy”: it lets in only the sets that can be put into words.

Theorem 3Gödel (1938)

satisfies all the axioms of ZFC, and CH holds in (so does the generalized continuum hypothesis). Hence, if ZF is consistent, so is ZFC plus CH: ZFC cannot refute CH.

Why CH holds in The key step is the “condensation lemma”: every real in (that is, every subset of ) already appears at some countable level (). The reason is that describing such a real takes only countably many “ingredients,” and gathering up the ingredients it uses collapses down to a single countable level. Each countable level has only countably many elements, and there are levels, so has at most reals. (Only the idea is given here, not a proof.)

4
COHEN

Cohen: it cannot be proved either

The other half is harder: building a world in which CH fails. makes the universe “thinner”; Cohen had to make it “fatter,” stuffing new real numbers into the universe. But stuff in what? The universe already contains “all” the real numbers, so where would a new one come from?

Cohen's method of 1963 is called forcing. An analogy first: to add to the rational numbers, you don't need to know what “really is”; it is enough to know which equation it satisfies, and you can work out the extended world completely. Forcing follows the same idea: describe a new real using only “finite pieces of information,” then prove that these pieces fit together into a real different from every real already there.

To be more concrete: start from a countable model of ZFC. A “condition” is a finite string of 0s and 1s, standing for the first few digits of the new real; a longer string is a stronger condition. A set in is called “dense” if every condition can be extended to one in .

Theorem 4Generic reals (a special case of the Rasiowa–Sikorski lemma)

Let be countable. Then there is an infinite 0/1 string that has an initial segment in every dense set belonging to . Such a is different from every real in .

Proof is countable, so it has only countably many dense sets; list them as . Start from the empty string, and at step extend the current string into (density guarantees this can be done). The strings get longer and longer, and together they make , whose initial segments have fallen, one after another, into every .

For any real in , let be the set of all finite strings that “differ from at some digit.” Any finite string can be extended by one more digit, chosen opposite to the digit of in that place, so is dense, and it belongs to . Hence some initial segment of is in , differs from at that digit, and .∎

Gold: the first few digits of the generic real
Blue circles: some reals in ; parts from each at some digit
Figure 4·5 works its way down the binary tree, stepping into each dense set it meets, and so avoids every real in . Look closely: this is the diagonal from Lecture 1 again, only now the opponent is a countable model.

Add to , together with every set that can be built from it, and you get a larger model . The central theorem of forcing says that still satisfies ZFC, and that what holds in can be worked out inside using the “conditions.”

Theorem 5Cohen (1963)

If ZF is consistent, then so is ZFC plus . Hence ZFC cannot prove CH.

Proof sketchInstead of adding one new real, add of them at once (a condition now specifies finitely many digits of the new reals). The remaining difficulty: once they are added, might the original and collapse to countable, so that “ of them” no longer lives up to its name? Cohen proved that they do not. These conditions have a combinatorial property called the “countable chain condition”: every family of pairwise incompatible conditions is countable, and this guarantees that cardinals do not collapse. (Only the idea is given here, not a proof.)

Cohen sent his proof to Gödel to check; Gödel read it, wrote back in praise, and personally recommended it to the Proceedings of the National Academy of Sciences for publication. Gödel plus Cohen: CH is independent of ZFC, and can be neither proved nor refuted. For this Cohen received the Fields Medal in 1966. In 2019 Han and van Doorn formalized the whole argument in the proof assistant Lean.

5
ALMOST ANYTHING GOES

Almost anything goes

Cohen's method can do more than make . By varying the number of new reals added, you can make it equal to all sorts of . Which ones, exactly?

Theorem 6Solovay, Cohen (1963–65); Easton (1970)

For any that cannot be written as the “limit” of countably many smaller cardinals (that is, whose cofinality is greater than ), the statement “” is consistent with ZFC. Easton went further and showed that for all regular cardinals , the values of can be prescribed almost at will, subject only to monotonicity and König's restriction.

Yes
Yes
Yes
No
Yes
Rule: can be any not a limit of countably many smaller ones
Figure 4·6What König's theorem rules out is exactly everything that can be ruled out; for every other position there is a world that satisfies ZFC.

In other words: apart from “it is larger than ,” everything ZFC knows about is that one fact in Theorem 2.

6
IS IT TRUE?

So is it true or not?

“Independent” answers the question “can it be proved?” and not the question “is it true?” If you believe that there is one definite universe of sets, then CH is either true or false in it; the axioms of ZFC are simply too few to tell which. That is what Gödel thought. In 1947 he wrote that CH was most likely independent, and that he expected new axioms to refute it; in a manuscript from late in his life he even conjectured that the number of real numbers is .

There is a lovely equivalent form that gives CH a kind of “geometric intuition.”

Theorem 7Sierpiński (1919)

CH holds if and only if the plane can be split into two parts and : every horizontal line meets in only countably many points, and every vertical line meets in only countably many points.

Proof(⇒) Under CH, well-order in order type as , so that each real has only countably many reals before it. Let and . Fix : the points of on that horizontal line are , countably many. Fix : the points of on that vertical line are , countably many.

(⇐) Suppose CH fails and such a decomposition exists. Take a subset of of size , and call it . On each vertical line (), only countably many points belong to ; taken together, their y-coordinates number at most , fewer than , so some is not among them. On this horizontal line, none of the points () is in , so they all lie in , and the line meets in points, uncountably many, a contradiction.∎

(well-ordered in type )
on verticals, on horizontals: countable initial segments
Figure 4·7Order both axes by a well-ordering of type , and the diagonal cuts the plane in two. (Schematic: the actual well-ordering has nothing to do with the usual order of the reals by size.)

In 1986 Freiling turned this equivalent form into a thought experiment. Throw two darts at random at the real line, getting and . If CH holds, then by the decomposition above, either lands among “the countably many bad points on the vertical line through ,” or lands among “the countably many bad points on the horizontal line through .” But a random point lands in a countable set fixed in advance with probability 0. By “dart-throwing intuition,” neither should happen, so CH should be false.

The flaw in this argument is that those “countable sets” come from a non-measurable decomposition, where probability theory does not apply. So it is an intuition, not a proof; believe it or not, as you like.

Set theorists today are divided on this question. Some are looking for new axioms that everyone can accept and that decide CH: a family of strong “forcing axioms” implies , and in 2021 Asperó and Schindler connected two of the main candidate axioms. Another line of work hopes to find an “ultimate ,” which would make CH true. Still others think there is no single universe at all: CH is true in some universes and false in others, much like the parallel postulate across different geometries. That is the topic of the next lecture.

7
CODA

In closing

The main character of this lecture is a question with no answer. Cantor conjectured ; all closed sets satisfy it, which makes the conjecture look very believable; König could rule out only positions like ; Gödel built a thin universe in which it holds, and Cohen a fat universe in which it fails. And the key to proving “it cannot be proved” turned out to be, once again, the diagonal from Lecture 1.

If the axioms are not enough, should we add new ones? Which ones? Can “bigger infinities” help? The next lecture, the last in this series, climbs to the very top of infinity.

What the axioms cannot decide
need not lack an answer;
the question itself may have several.
✎ A puzzle for the comments

Which of these could be (in some world that satisfies ZFC): , , or ? Hint: check whether it can be approached in countably many steps. The answer comes in the next lecture.

References
  1. D. Hilbert, “Mathematische Probleme” (1900), Problem 1
  2. G. Cantor, Math. Ann. 21 (1883); Acta Math. 4 (1884); I. Bendixson, Acta Math. 2 (1883); P. Alexandroff, C. R. Paris 162 (1916); F. Hausdorff, Math. Ann. 77 (1916)
  3. J. König, “Zum Kontinuum-Problem,” Math. Ann. 60 (1905); E. Zermelo, Math. Ann. 65 (1908)
  4. K. Gödel, PNAS 24 (1938); The Consistency of the Continuum Hypothesis (1940)
  5. P. J. Cohen, “The independence of the continuum hypothesis” I and II, PNAS 50 (1963), 51 (1964)
  6. H. Rasiowa, R. Sikorski, Fund. Math. 37 (1950)
  7. R. M. Solovay, “ can be anything it ought to be” (1965); W. B. Easton, “Powers of regular cardinals,” Ann. Math. Logic 1 (1970)
  8. W. Sierpiński, Bull. Int. Acad. Sci. Cracovie (session of February 24, 1919); Hypothèse du continu (1934)
  9. C. Freiling, “Axioms of symmetry: throwing darts at the real number line”, J. Symbolic Logic 51 (1986)
  10. K. Gödel, “What is Cantor’s continuum problem?”, Amer. Math. Monthly 54 (1947)
  11. D. Asperó, R. Schindler, “Martin’s Maximum⁺⁺ implies Woodin’s axiom (*)”, Ann. of Math. 193 (2021)
  12. J. M. Han, F. van Doorn, “A formal proof of the independence of the continuum hypothesis,” CPP (2020)
  13. T. Jech, Set Theory (3rd edition, 2003)
Note: All figures are schematic illustrations made for this article.
犹剑在心
北极甜虾(剑心犹在!)
— End of Lecture 4 —

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