Quick description
Transfinite induction is similar to induction but the wellordered set is replaced by larger ordinals. This article tells you what you need to know about ordinals in order to be able to prove results by transfinite induction, gives examples of its use, and distinguishes between various different types of transfiniteinduction argument.
General discussion
The following facts about ordinals are all you need to know about them in order to carry out proofs by transfinite induction.

There is a total ordering, , defined on the class of all ordinals.

0 is an ordinal and there is no smaller ordinal.

Given any ordinal , the set is a wellordered set.

For every set there exists a wellordering of the elements of . (This is called the wellordering theorem.)

Given any wellordered set , there is exactly one ordinal such that is orderisomorphic to the set .

Every ordinal is either a successor ordinal or a limit ordinal. It is a successor if there is some such that is the smallest ordinal that is greater than . In this case we write . It is a limit if it is not a successor.

(Optional, but convenient.) The way ordinals are conventionally constructed, is in fact equal to (which implies in particular that is the empty set). Therefore, instead of talking about the cardinality of we can talk more simply about the cardinality of . We can also say that is a wellordered set. The successor of is then . If is a limit ordinal, then .

(Not needed in proofs, but you ought to know it just to be safe.) Each ordinal is a set, but there is no such thing as the set of all ordinals, because if there were then would be an ordinal as well. This is a contradiction since then we would have . However, it is all right to talk about the class of all ordinals.

Given any set , there is an ordinal with the same cardinality as , and hence (since is wellordered) there is a least such ordinal . Therefore, every set can be wellordered in such a way that the set of predecessors of any element of has strictly smaller cardinality than the cardinality of itself.

Suppose that we have put a set into onetoone correspondence with an ordinal . For each , write for the element of that corresponds to . Suppose that is some statement about elements of , that is true, and that is true whenever is true for every . Then is true for every . (This is the principle of transfinite induction.)
Example 1
Let us prove that every vector space has a basis. This result is also proved in the article on how to use Zorn's lemma (in a better way).
Let be a vector space. By the wellordering theorem, there is a wellordering of the elements of . Therefore, there is an ordinal such that the elements of can be put into onetoone correspondence with . For each , let us write for the element of that corresponds to . Now let be the set of all that do not belong to the linear span of any of their predecessors. It is clear that is linearly independent, since if , , and , then we have expressed as a linear combination of earlier s. It also spans, since every either belongs to or is a linear combination of earlier elements of , and every element of is for some .
General discussion
Example 1 might suggest that transfinite induction is just another way of carrying out Zorn'slemma arguments. And indeed, it can be used for this purpose, but Zorn's lemma is neater and does not require the machinery of ordinals. So Example 1 is not a very good advertisement for the transfinite induction. However, there are situations where transfinite induction really is needed. Typically, these involve a cardinality condition on the initial segments of the wellordering that one places on the set that one is examining. The next example illustrates this.
Example 2
A famous result of Sierpinski is the existence of a subset of that intersects every line exactly twice. Here is how the proof goes. First, wellorder the set of all lines in , which has cardinality , in such a way that each element has strictly fewer than predecessors in the wellordering. Let be the line associated with the ordinal . (If the continuum hypothesis is assumed, then must be countable, but we are not assuming the continuum hypothesis.)
We now build up our set as follows. Suppose that we have found for each a set such that for every we have and such that for every line we have . Suppose also that we have done this in such a way that if then . Finally, suppose that all the sets have cardinality strictly less than .
Let and note that the cardinality of is strictly less than (since for any infinite cardinal , a union of many sets of size has size ). Then for every line (since otherwise there would have to be some such that ), and for every . If , then let , and the induction has continued to stage . Otherwise, the number of lines that go through two points of is strictly less than , and each of these lines intersects at most once, so we can find infinitely many points in that does not belong to any of those lines. (We want to avoid those lines since we do not want three points in a line.) Add either one or two of these points to to create , and again the induction has continued to stage .
This proves that we can construct sets for every . The union of all these sets has the property that we want of .
General discussion
The big difference between the second argument and the first is that we insisted that each element of the wellordering had fewer predecessors than the cardinality of the set of all lines, which happened to be the same as the cardinality of each individual line. A cardinality condition such as this gives rise to an argument that cannot be straightforwardly converted into a Zorn'slemma argument.
Sometimes one needs a stronger cardinality assumption still: one wants the set of all predecessors of any given element to be countable. This requires the continuum hypothesis. The article on how to use the continuum hypothesis has some examples of arguments of this kind.
Example 3
Many interesting uses of transfinite induction use just the countable ordinals (without any assumption that the number of these is ). As a simple but representative example, let us prove that there is an uncountable collection of wellordered subsets of , no two of which are orderisomorphic.
We shall prove this by constructing for each countable ordinal a wellordered subset of that is orderisomorphic to . This we do using transfinite induction up to (but not including) the first uncountable ordinal. The induction starts with our setting . Now let be a countable ordinal. As our inductive hypothesis we will take the slightly stronger statement that for every we can choose to be a subset of . (But it isn't significantly stronger because is orderisomorphic to .)
It is convenient to split the proof into the two cases: either is a successor or is a limit ordinal. This splitting is a common but not universal feature of transfinite induction arguments over the countable ordinals.
If , then the construction of is easy: we just squash into and add an extra point. To be more definite, we can set to be .
We now need a small lemma, which is that for every countable limit ordinal there is an increasing sequence of ordinals such that . This is easy to prove. First, enumerate all the ordinals , of which there are countably many. Next, pick a subsequence that consists of every ordinal that is larger than all previous ordinals in the enumeration. This sequence cannot be bounded above by any , since there exists such that (as is a limit ordinal), so we would have a contradiction when we got to in the enumeration. Therefore, its union is indeed (because it contains every ).
Given such a sequence, construct as follows. For each , let be an orderisomorphic copy of that lives in the interval (obtained as a linear image of , say). Next, remove from an initial segment that is orderisomorphic to and call this set . (To put this loosely, remove the first elements of .) Then is orderisomorphic to and every element of is larger than every element of . The union of the sets lives in and is orderisomorphic to the union of the sets , which equals the union of the sets , which equals . Thus, the inductive step is proved.
General discussion
Transfinite induction over the countable ordinals is often used to give a measure of complexity. For example, we could say that the complexity of a wellordered subset of is the unique ordinal that is orderisomorphic to it. Another wellknown example is the complexity of Borel sets. We can define open sets and closed sets to have complexity 0, and then say that a set has complexity at most if it is a countable union or countable intersection of sets such that each has complexity at most for some . The complexity of is the least ordinal such that has complexity at most . This is a definition by transfinite induction. It is possible to prove, again by transfinite induction, that for every countable ordinal there is a Borel set that has complexity . Thus, there are uncountably many different kinds of Borel set.
Comments
Symbol for total order
Sat, 25/04/2009  00:21 — JoseBroxA total order is supposed to be total, in the sense that for every pair , we have either or . Choosing the symbol instead of for the total order can be deceptive because it usually means "lesser but not equal". It's clear that this order isn't total, because we don't have . So, either it is a mistake or it is a conscious choice of the symbol. I haven't changed the notation myself precisely because there could be a good explanation to do it like this. If I don't receive an answer in several days I will change it.
I always thought that a total
Sat, 25/04/2009  00:50 — gowersI always thought that a total order could be strict, and that the condition was that , or . Is that a very nonstandard convention?
The relation on the ordinals is more convenient to use than , since one wants many statements to be true for every , for some given . I suppose one could add the word "strict" in brackets before the first occurrence of "total order", but that feels a bit strange, given that you can convert orders of the type into orders of the type and back again. Indeed, because of this I describe them both as total orders and think of them as different ways of describing the same basic underlying object.
You are right
Sat, 25/04/2009  01:23 — JoseBroxThat's true, if you ask for the trichotomy law axiom, then both definitions are equivalent. I don't know exactly how nonstandard is this, but since Wikipedia and Planetmath both mention it (Mathworld doesn't!), it should be safe. Maybe in the transfinite context it is more used? (I just checked my copy of Kamke's "Theory of Sets". He uses your convention!).
Excuse me for my comment, I thought the notation was somewhat confusing (but it was more like I was the one confused!)