Transfinite induction is similar to induction but the well-ordered 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 transfinite-induction argument.
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 well-ordered set.
For every set there exists a well-ordering of the elements of . (This is called the well-ordering theorem.)
Given any well-ordered set , there is exactly one ordinal such that is order-isomorphic 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 well-ordered 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 well-ordered) there is a least such ordinal . Therefore, every set can be well-ordered 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 one-to-one 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.)
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 well-ordering theorem, there is a well-ordering of the elements of . Therefore, there is an ordinal such that the elements of can be put into one-to-one 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 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 .
Example 1 might suggest that transfinite induction is just another way of carrying out Zorn's-lemma 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 well-ordering that one places on the set that one is examining. The next example illustrates this.
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, well-order the set of all lines in , which has cardinality , in such a way that each element has strictly fewer than predecessors in the well-ordering. 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 do not belong to any of those lines. (We want to avoid those intersections since we do not want to have three points of in the same 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 .
The big difference between the second argument and the first is that we insisted that each element of the well-ordering 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's-lemma 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.
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 well-ordered subsets of , no two of which are order-isomorphic.
We shall prove this by constructing for each countable ordinal a well-ordered subset of that is order-isomorphic 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 order-isomorphic 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 order-isomorphic copy of that lives in the interval (obtained as a linear image of , say). Next, remove from an initial segment that is order-isomorphic to and call this set . (To put this loosely, remove the first elements of .) Then is order-isomorphic to and every element of is larger than every element of . The union of the sets lives in and is order-isomorphic to the union of the sets , which equals the union of the sets , which equals . Thus, the inductive step is proved.
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 well-ordered subset of is the unique ordinal that is order-isomorphic to it. Another well-known 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.