先说规则。棋盘是平面上所有整数坐标的格子，没有边，也没有角。双方各有有限个棋子，走法和普通国际象棋一样，车、后、象可以一次走任意远。这里沿用研究这个游戏的 Hamkins 等人的约定：兵没有底线可去，所以不升变；也没有王车易位和吃过路兵；每方至多一个王，甚至可以没有王。白方的目标是将死黑王；逼和算和棋；如果对局无限地进行下去、始终没有将死，也算和棋。

这条规则让这个游戏有一个特别的形状：白方要赢，必须在某个有限的时刻完成将死；黑方要不输，只需要永远撑下去。用第五讲的话说，白方获胜的那些对局构成一个“开集”：一盘棋是不是白胜，有限步之内就见分晓。这种博弈叫开博弈，第五讲说过它总是决定的；这也是这一期能用序数来量的原因。

用集合论研究国际象棋的第一篇论文，作者正是公理集合论里的那位 Zermelo。1912 年在剑桥的国际数学家大会上，他做了一个报告，题目叫《论集合论在国际象棋理论中的一个应用》，次年印出来。

他关心的问题是：如果一个局面白方能强制取胜，是不是就能在确定的步数之内取胜？他的论证靠“局面不重复”，后来被 Dénes König（第四讲那位 Julius König 的儿子）指出有缺口；1927 年 König 用今天叫“König 引理”的办法补上，Zermelo 也写了修正。1928 年 Kalmár 又往前走了一步：如果每一步可以有无穷多种走法，强制取胜所需的步数可能是一个超限序数。这正是本期的主题。

今天教科书里说的“Zermelo 定理”，是后人整理出来的下面这条，证明用的倒推法也不是他原文的写法：

按国际棋联现行的规则，连续 75 步没有吃子也没有动兵、或者同一局面出现 5 次，对局自动判和，所以普通国际象棋是有限博弈，属于定理 1 的情形。至于到底是哪一种，没人知道：这棵树太大了。

不过小一点的残局能真的倒推出来。本文脚本把“王后加王对单王”在 $8\times 8$ 棋盘上的全部局面倒推了一遍：

注意这里每个局面都有一个具体的数：“$n$ 步将死”。在有边的棋盘上，这是理所当然的。

第一件怪事：在无穷棋盘上，王后加王连一个光杆黑王都将不死。

在普通棋盘上，王后加王的将死全都靠棋盘的边：边替白方挡住了黑王一半的退路。到了无穷棋盘中央，黑王周围永远有 8 个格子。本文脚本把黑王放在中央、把白王白后所有可能的位置都试了一遍，结论是：

所以无穷棋盘上的将死，需要更多棋子、更巧妙的布置。也正因为棋子能离得很远、能在一条线上走任意远，才有了下面的事。

先看一个比棋简单得多的玩具游戏。黑方先报一个正整数 $n$，随便多大；然后白方倒数：$n,n-1,\dots ,1$，数到头就算白方将死了黑方。

白方必胜，这毫无疑问。可是“白方几步能赢”？对任何一个具体的 $n$，黑方都能报一个更大的数，让白方走得比 $n$ 更久。所以没有一个有限的数能回答这个问题。可它也不是“无穷步”：每一盘都在有限步内结束。

第三讲的序数正好填补这个空档。给每个局面定一个“值”，表示白方离胜利还有多远：

倒数游戏里，“还剩 $n$ 下要数”的局面值为 $n$；黑方报数之前的那个局面，值是 $\sup \{1,2,3,\dots \}=\omega$。照这个思路，就能造出值为 $\omega +1$、$\omega \cdot 2$、${\omega }^2$ 的玩具游戏：

这个定义之所以好用，是因为下面这条定理：

这正是 Goodstein 序列那一套的另一种面目：白方手里的局面值，就是第三讲里那个“影子”。数在涨（对局可以拖得任意长），影子在降，而影子不能永远降下去。

那么，值什么时候会超过所有有限数？关键在于黑方每一步有多少种走法。

这正是 Dénes König 1927 年那个想法的现代版本。反过来说：要出现 $\omega$，黑方手里必须有一枚能“走任意远”的棋子，让他在一步之内从无穷多个选择里挑一个。

真正的棋盘上也有这样的局面。2014 年，Evans 和 Hamkins 发表了一个只有七个棋子、值恰好是 $\omega$ 的局面（白方甚至没有王）：

黑方被围在一个小笼子里，最好的办法是把中间那只车往上开，开多高随他：$1\dots$车 e6–e$n$。之后白方开始“把黑王往车上推”：车在黑王所在的那一行横着将军，后从左下方斜着将军，两者轮流。黑王每次只能沿 e 列往上挪一格，一直挪到撞上自己的车。论文给的例子是车开到 e10：

本文脚本自己写了无穷棋盘上的走子规则，把这条线逐步核对了一遍：每一步都合法，最后确实是将死。再往一般情形推：车开到 e$n$（$n=7,8,\dots ,40$ 都试了），按同样的套路，白方再走 $n-5$ 步就将死，而且白方每次将军之后，黑方都只剩唯一一种合法着法。黑方如果不开车、去动别的子，论文列出的几种走法都会在两三步内被将死，脚本也一一核对过。

所以：黑方先报一个数（车开多高），白方再花差不多那么多步将死。对任何固定的 $n$，黑方都能让白方走得更久，可每一盘都在有限步内结束。这个局面的值就是 $\omega$，和第 3 节的倒数游戏一模一样。

再往上，就得放开“有限个棋子”的限制，在棋盘上铺无穷多个子（但要能用程序描述）。Evans 和 Hamkins 在同一篇论文里造出了值为 ${\omega }^2$、${\omega }^2\cdot k$ 和 ${\omega }^3$ 的局面；2017 年，他们和 Perlmutter 又造出了 ${\omega }^4$。

这些局面的设计很有画面感，论文里的部件名字也很好玩：一道无穷长的兵墙挡住白方的“后的大军”，黑方先决定把一只车升多高，白方每推进一格兵墙，黑方都能先发动一轮任意长的“骚扰将军”。这正是玩具游戏里的“黑先报要报几次数，再一次次报数”，值是 ${\omega }^2$。${\omega }^4$ 那篇里还有“王座间”和“象炮台”，后者让黑方每次拖延的花样从 $\omega$ 变成 ${\omega }^2$，于是 ${\omega }^2\cdot {\omega }^2={\omega }^4$。

有限个棋子能走多高，至今是开放问题。已发表的论文只到 $\omega$；Elkies 在 MathOverflow 上给过 $\omega \cdot n$ 一类的构造，2023 年 Tsevas 在同一网站上贴出过一个有限棋子、值为 ${\omega }^2$ 的局面，但都还没有正式发表。

自然要问：值最大能到多少？把所有（有限个棋子的）局面的值取上确界，记作 ${\omega }_1^{\text{Ch}}$。

这个名字是有意起的：它故意模仿 ${\omega }_1$。可它比 ${\omega }_1$ 小得多，究竟是哪一个可数序数，到今天还不知道。已知的上界更具体：它不超过 Church–Kleene 序数 ${\omega }_1^{\text{CK}}$，也就是“能被计算机程序描述出来的良序”所能达到的长度的上确界（Blass 指出，Welch 也独立发现）。Evans 和 Hamkins 猜想两者恰好相等。如果允许三维棋盘和无穷多个棋子，他们证明了每一个可数序数都是某个局面的值；2026 年 2 月的一篇预印本声称，二维棋盘、无穷多个棋子也能做到这一点，目前尚未经过评审。

还有一个看起来朴素的问题。2010 年 6 月，组合数学家 Richard Stanley 在 MathOverflow 上问：无穷棋盘上“白方能不能赢”，能不能用计算机判定？这个问题至今没有答案。退一步，“白方能不能在 $n$ 步内将死”呢？棋盘无穷大，搜索空间无穷，直觉上也不好说。

可判定的是“$n$ 步将死”。至于 Stanley 原来那个问题，“白方到底能不能赢”，因为值可以是超限序数，就不能靠对 $n$ 一个个试来回答；三位作者猜想它不可判定，而且难得远超过一般的不可判定问题。

这一期把第三讲和第五讲连在了一张棋盘上，还捎带着第一讲的可数性。第三讲说：序数是良序的长度，它不能无穷下降。第五讲说：开集上的博弈总是决定的，总有一方有必胜策略。这一期说：在无穷棋盘上，把这两件事放在一起，“还剩几步”就自然地变成了一个序数。$n$ 步将死是有限序数；黑方能先报一个数，就是 $\omega$；能报好几次、能先决定报几次，就往 $\omega \cdot 2$、${\omega }^2$ 上爬。

黑方的处境挺有意思：他知道自己一定会输，甚至能自己决定输得多慢；可不管他报多大的数，都只是把白方的胜利推迟到某一个有限的时刻。这大概是“数到无穷之后”最直观的样子。

你一定会输，
只是我说不出第几步：
答案不是数，是 $\omega$。

First, the rules. The board is made of the squares at all integer coordinates in the plane, with no edges and no corners. Each side has finitely many pieces, which move as in ordinary chess; rooks, queens and bishops can travel any distance in one move. We follow the conventions of Hamkins and others who have studied this game: pawns have no last rank to reach, so they never promote; there is no castling and no en passant; each side has at most one king, and may have none at all. White's goal is to checkmate the black king; stalemate is a draw; and if the game goes on forever without checkmate, that is a draw too.

This last rule gives the game a special shape: to win, White must deliver mate at some finite moment; to avoid losing, Black only has to hold out forever. In the language of Lecture 5, the games White wins form an “open set”: whether a game is a win for White is settled within finitely many moves. Such games are called open games, and Lecture 5 noted that they are always determined; that is also why this issue can measure them with ordinals.

The first paper to study chess with set theory was written by none other than the Zermelo of axiomatic set theory. At the 1912 International Congress of Mathematicians in Cambridge he gave a talk titled “On an Application of Set Theory to the Theory of the Game of Chess,” printed the following year.

The question he cared about was this: if White can force a win from a position, can White force it within a fixed number of moves? His argument relied on “positions not repeating,” and Dénes König (son of the Julius König of Lecture 4) later pointed out a gap in it; in 1927 König closed the gap with the method now called “König's lemma,” and Zermelo also wrote a correction. In 1928 Kalmár went a step further: if a move can have infinitely many options, the number of moves needed to force a win can be a transfinite ordinal. That is exactly the subject of this issue.

What textbooks today call “Zermelo's theorem” is the following statement, distilled by later writers, and the backward induction used to prove it is not how he wrote it either:

Under the current FIDE rules, a game is automatically drawn after 75 moves with no capture and no pawn move, or when the same position occurs 5 times, so ordinary chess is a finite game and falls under Theorem 1. Which of the three cases it is, nobody knows: the tree is far too large.

Smaller endgames, though, can really be worked out backwards. This article's script worked backwards through every position of “king and queen against a lone king” on the $8\times 8$ board:

Note that every position here comes with a definite number: “mate in $n$.” On a board with edges, that goes without saying.

The first oddity: on an infinite board, king and queen cannot even mate a lone black king.

On an ordinary board, every king-and-queen mate relies on the edge of the board: the edge takes away half of the black king's escape squares for White. In the middle of an infinite board, the black king always has 8 squares around him. This article's script put the black king in the middle and tried every possible placement of the white king and queen. The conclusion:

So mate on an infinite board needs more pieces and cleverer set-ups. And it is precisely because pieces can be far apart and can slide any distance along a line that what follows becomes possible.

First, a toy game far simpler than chess. Black first names a positive integer $n$, as large as he likes; then White counts down: $n,n-1,\dots ,1$, and when the count runs out, White is deemed to have checkmated Black.

White is sure to win, no doubt about it. But “in how many moves does White win”? For any particular $n$, Black can name a larger number and make White take longer than $n$. So no finite number answers the question. Yet it is not “infinitely many moves” either: every game ends after finitely many moves.

The ordinals of Lecture 3 fill exactly this gap. Give each position a “value” that says how far White is from victory:

In the countdown game, a position with “$n$ numbers still to count” has value $n$; the position before Black names his number has value $\sup \{1,2,3,\dots \}=\omega$. Along the same lines one can build toy games of value $\omega +1$, $\omega \cdot 2$ and ${\omega }^2$:

What makes this definition useful is the following theorem:

This is the Goodstein-sequence argument in another guise: the value of White's position is the “shadow” from Lecture 3. The number goes up (the game can be dragged out as long as you like), the shadow goes down, and the shadow cannot go down forever.

So when does the value go beyond every finite number? The key is how many moves Black has to choose from on each turn.

This is the modern form of Dénes König's 1927 idea. Put the other way round: for $\omega$ to appear, Black must have a piece that can “travel any distance,” letting him choose among infinitely many options in a single move.

Such positions exist on a real chessboard too. In 2014 Evans and Hamkins published a position with only seven pieces whose value is exactly $\omega$ (White doesn't even have a king):

Black is shut in a small cage, and his best plan is to send the middle rook up, as high as he likes: $1\dots$Re6–e$n$. Then White starts “pushing the black king onto the rook”: the rook checks sideways along the black king's rank and the queen checks diagonally from the lower left, taking turns. Each time, the black king can only step one square up the e-file, until he bumps into his own rook. The example in the paper has the rook go to e10:

This article's script implements its own move rules for the infinite board and checked this line move by move: every move is legal, and the final position really is checkmate. Going on to the general case, with the rook on e$n$ (all of $n=7,8,\dots ,40$ were tried), White follows the same pattern and mates in $n-5$ more moves, and after each White check Black has exactly one legal move. If Black does not send the rook up but moves something else, each of the alternatives listed in the paper is mated within two or three moves, and the script checked these one by one as well.

So Black first names a number (how high the rook goes), and White then needs roughly that many moves to mate. For any fixed $n$, Black can make White take longer, yet every game ends after finitely many moves. The value of this position is $\omega$, exactly like the countdown game of section 3.

To go higher, we must drop the “finitely many pieces” restriction and spread infinitely many pieces over the board (though they must be describable by a program). In the same paper Evans and Hamkins built positions of value ${\omega }^2$, ${\omega }^2\cdot k$ and ${\omega }^3$; in 2017, together with Perlmutter, they built one of value ${\omega }^4$.

These positions are wonderfully visual in design, and the papers give their components fun names: an infinitely long wall of pawns holds back White's “army of queens”; Black first decides how high to raise a rook, and each time White advances the pawn wall by one square, Black may first launch an arbitrarily long round of “harassing checks.” This is exactly the toy game in which “Black first names how many numbers to name, then names them one by one,” with value ${\omega }^2$. The ${\omega }^4$ paper also has a “throne room” and a “bishop cannon battery”; the latter raises Black's options for stalling each time from $\omega$ to ${\omega }^2$, and so ${\omega }^2\cdot {\omega }^2={\omega }^4$.

How high finitely many pieces can go is still an open problem. The published papers reach only $\omega$; Elkies has given constructions of the $\omega \cdot n$ kind on MathOverflow, and in 2023 Tsevas posted on the same site a position with finitely many pieces and value ${\omega }^2$, but none of these has been formally published yet.

The natural question is how large the value can be. Take the supremum of the values of all positions (with finitely many pieces) and call it ${\omega }_1^{\text{Ch}}$.

The name is deliberate: it imitates ${\omega }_1$. But it is far smaller than ${\omega }_1$, and exactly which countable ordinal it is remains unknown to this day. The known upper bound is more concrete: it is at most the Church–Kleene ordinal ${\omega }_1^{\text{CK}}$, the supremum of the lengths of the well-orderings “that a computer program can describe” (pointed out by Blass, and found independently by Welch). Evans and Hamkins conjecture that the two are equal. If three-dimensional boards and infinitely many pieces are allowed, they proved that every countable ordinal is the value of some position; a preprint from February 2026 claims that this can also be done on a two-dimensional board with infinitely many pieces, but it has not yet been peer-reviewed.

There is also a question that looks innocent. In June 2010 the combinatorialist Richard Stanley asked on MathOverflow whether a computer can decide “can White win?” on an infinite board. That question is still open. Taking a step back, what about “can White mate in $n$ moves?” The board is infinite and so is the search space, so intuition doesn't give a clear answer either.

What is decidable is “mate in $n$.” As for Stanley's original question, “can White win at all?”: since values can be transfinite ordinals, it cannot be answered by trying each $n$ in turn; the three authors conjecture that it is undecidable, and far harder than ordinary undecidable problems.

This issue has joined Lectures 3 and 5 on a single chessboard, with the countability of Lecture 1 thrown in. Lecture 3 said: an ordinal is the length of a well-ordering, and ordinals cannot decrease forever. Lecture 5 said: games on open sets are always determined; one side or the other has a winning strategy. This issue says: on an infinite chessboard, put the two together, and “how many moves are left” naturally becomes an ordinal. Mate in $n$ is a finite ordinal; if Black may first name a number, it is $\omega$; if he may name several, or first decide how many to name, it climbs to $\omega \cdot 2$ and ${\omega }^2$.

Black's situation is rather curious: he knows he is sure to lose, and can even decide for himself how slowly; yet however large a number he names, he only postpones White's victory to some finite moment. This is perhaps the most tangible picture of “counting past infinity.”

You are sure to lose;
I just can't say on which move:
the answer is not a number but $\omega$.