### Quick description

Suppose you are asked to prove that a mathematical object of a certain type can be found that satisfies some property . It quite often happens that there is a partial order on the given kinds of objects such that if is less than and has property , then so does . If that is the case, then one can look for objects that are maximal in this partial order. And of course the same is true if we replace "less than" and "maximal" by "greater than" and "minimal".

This article is very far from finished: eventually there needs to be a rather large set of examples taken from all over mathematics (and not just algebra).

### Example 1

Let be a group with two generators and . This means that every element of can be written as a product of powers of and . (In other words, all elements of are things like .) Define the *length* of a product such as to be the sum of the absolute values of the powers of and (so the product has length 13, for example), and define the *ball of radius * about the identity to be the set of all elements that can be expressed as a product of length at most .

Now some products can be equal to others. For example, the group is generated by the two elements and . Using multiplicative notation, we then find that . In this group, the ball of radius about the identity consists of all points such that . From this it follows that the size of the ball of radius is roughly when is large.

Is it possible for the size of the ball of radius to grow much more rapidly than this as tends to infinity? This is not a very difficult question once one knows the right way of thinking about it. And the right way springs from the observation that if we want to define a group with two generators and in such a way that the size of the ball of increases as rapidly as possible, then we want as few as possible of the products involving and to be equal.

We can make this idea more precise as follows. Suppose that is a group generated by two elements and , and is a group generated by two elements and . Suppose also that there is a homomorphism from G to H (meaning a map from to such that for every pair of elements ) such that and . Then the ball of radius in is at least as big as the ball of radius in . Why? Because if two products are equal in then the corresponding products are equal in . For example, if we know that , then , and since is a homomorphism that tells us that , which in turn tells us that .

The structures we are considering here are of the form , where is a group and and are generators of . Let us write if there is a homomorphism such that and . It is easy to check that if and then and are isomorphic. It is also easy to check that if and then . Therefore, we have a (which basically means an ordering like "is a subset of" where not every pair of objects can be compared) on these structures. We have just shown that if then the ball of radius in is at least as big as the ball of radius in .

Therefore, If we are looking for an example where the ball of radius is big, then it makes sense to try to choose to be maximal in this partial order – if we can, that is. And it turns out that we can. The free group on two generators and is defined (loosely) as the group where the only products that are equal are those that are forced to be equal by the group axioms: for example, is equal to , since the group axioms allow us to cancel . It is intuitively clear, and not that hard to prove rigorously, that the free group on two generators and has the following *universal property*: given any group with two generators and there is a homomorphism from to such that maps to and maps to . The homomorphism takes a word such as and simply maps it to the corresponding word . One must check that this is well-defined. Why might it not be? Well, there would be a problem if two products involving and were equal but the corresponding products involving and were different. But this cannot happen, since the only equalities in are the ones forced by the group axioms, so if two products in are equal then the corresponding products in are equal.

It follows from the discussion above that the ball of radius in the free group is as big as it can possibly be. Since there are products that are made out of just and and not their inverses, and since no cancellation is possible in such a product, the ball of radius in the free group on two generators has size at least . That is, it is possible for the size of the ball of radius to grow exponentially in rather than polynomially as it did in .

The rate of growth of the ball of radius is a very important parameter of an infinite group and is and has been the subject of a great deal of research.

### Example 2

Is there an ordered field that does not satisfy the Archimedean property? That is, is there an ordered field that contains a positive element such that for every positive integer (where a positive integer is defined to be one of the field elements , , , etc.)?

To answer this question, let us try to build a minimal example. It is a nice exercise to prove that every ordered field contains a copy of the rationals, so let us start with the rationals and try to extend them in a minimal way.

Since we know that an ordered field that does not satisfy the Archimedean property must (by definition) contain a positive such that for every , and since no rational number has this property, we are forced to adjoin a new element with the property. Let us call it .

Since we are going for a minimal example, our next aim is to put into our field only what we are forced to put in by the field and order axioms. To be more accurate, we use closure properties of fields to determine what else we must add, and we use other field axioms and order axioms to determine how the field operations and order properties behave on our new elements.

For instance, since fields are closed under multiplication, must also be an element of our field, and the order axioms imply that it will be positive and smaller than , and therefore not equal to or any rational. Similarly, will be smaller than any real number that is greater than . If one explores consequences of this kind, one finds that the elements that are forced to belong to the field are all elements of the form where and are polynomials and is not the zero polynomial. And of course if then and are the same element. As for the ordering, we can work it out as follows. We say that is positive if the leading coefficient of is positive, and otherwise it is negative. And is positive if and have the same sign. And finally, is less than if is positive. It is not too hard to check that this is a well-defined ordering that gives us an ordered field, and that the entire ordered field was indeed determined by minimality and the one fact that is smaller than any constant polynomial. So now we have our ordered field in which the Archimedean property is false. Let us call it .

Here is a more formal "universal property" that this construction satisfies: if is any ordered field that fails the Archimedean property and is any positive element of that is smaller than every , then there is a unique field embedding from to that takes to . This gives us a precise sense in which our example is "smallest".

A similar method can be used to construct a smallest ordered field that contains and does not satisfy the Archimedean property. This field is discussed for a different reason in an article on demonstrating that certain approaches to certain problems cannot work, where it appears as Example 3.

### Example 3

Every normed space is contained in a Banach space. That is, to be slightly pedantic about it, for every normed space there exists a Banach space that has a subspace isometric to .

How do we prove this? Well, if is a Banach space that contains and is a Banach space that contains , then is a Banach space that contains . So the philosophy behind universal constructions tells us that it makes sense to aim for a *minimal* example of a Banach space that contains . The result turns out to be unique, and it is called the *completion* of (the smallest space that "makes complete").

Let's assume then that we have a Banach space that contains , and see what is forced to contain. Of course, we will eventually have to prove that it is possible for a Banach space to contain , or our argument will be circular, but for now our aim is to describe what a minimal example would be like *if* there is an example.

A completely obvious condition that will have to satisfy is this: if is a Cauchy sequence, then it must converge to a limit in . If it didn't then wouldn't be complete. So a natural idea would be to adjoin to an extra element for each Cauchy sequence that does not converge in . That way, we would "force" the sequence to converge in the bigger space.

However, we are being a little hasty here. If we adjoin an extra element in order to make one sequence converge, there may be some good news and some bad news. The good news is that we may make other sequences converge at the same time, and the bad news is that we may create new Cauchy sequences that do not converge. Let us think about these two possibilities in turn.

Suppose that is a Cauchy sequence in , and that is another Cauchy sequence. When would we expect them to converge to the same limit? A neat answer is this: form the sequence . If that sequence is also Cauchy, then we want it to converge to a limit, and that limit will have to be the limit of both the and the . Conversely, if the interwoven sequence is *not* Cauchy, then we know that the two original sequences *cannot* tend to the same limit in any space, and in particular in our minimal space .

Let us define two Cauchy sequences that do not converge in to be equivalent if the sequence formed by interweaving them is Cauchy. The remarks of the previous paragraph imply that we would be in trouble if this relation was not an equivalence relation, but it is not hard to check that it is one. So now let us adjoin to a whole lot of new elements, one for each equivalence class of nonconvergent Cauchy sequences.

There is more that we have to do, however. We can't just adjoin elements unless we say how to add them, multiply them by scalars, and define their norms. In each case it is obvious what to do: for example, if is a Cauchy sequence that doesn't converge, and is the corresponding adjoined element, then we define to be the limit of as . And in each case, we must check that what we do is well-defined: here we need to know that if we had taken an equivalent sequence, we would have ended up with the same definition of . And in each case, the verification that the definition is well-defined is a very straightforward exercise.

What this proves, if we do all the routine verifications, is that can be embedded into a larger normed space in such a way that every Cauchy sequence in has a limit in . But does every Cauchy sequence in have a limit in ? If it didn't, then we would have to repeat the process of adjoining new elements, but it turns out that it does. Indeed, suppose that is a Cauchy sequence in . Then each is close to some in . If tends to then the sequence is a Cauchy sequence in , so it has a limit in , and this is the limit of the as well.

The universal property that has is that if is isometric to a subspace of a complete normed space , then there is a closed subspace of such that and is isomorphic to .

A very similar argument proves that every metric space can be embedded in a minimal way into a complete metric space. Again the resulting space is called the completion.

7 weeks 1 dayago8 weeks 1 dayago37 weeks 3 daysago1 year 6 weeksago2 years 28 weeksago