Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It is common in mathematics to see definitions of the following form:

  1. we begin with a certain object $A$.
  2. we perform some construction depending on a choice of some parameter $\lambda\in\Lambda$ for some set $\Lambda$ which yields a new object $A_\lambda$.
  3. we show that this new object does not depend on the choice of particular $\lambda$, that is: for each pair $\lambda_1,\lambda_2\in\Lambda$ we have $A_{\lambda_1}=A_{\lambda_2}$.
  4. at last we define $A'$ to be $A_\lambda$ for some $\lambda\in\Lambda$.

So, to each object $A$ we have associated some object $A'$.

For example, if $r>0$ and $s\in\mathbb{R}$ are real numbers, $r^s$ is commonly seen to be defined by $r^s := \lim_{n\to\infty}r_n^{s_n}$ where $\lbrace(r_n,s_n)\rbrace_n$ is a sequence in $\mathbb{Q}\times\mathbb{Q}$ converging to $(r,s)$.

How are such definitions justified logically?

The step that bothers me, is step 4. In order for this to really define something, shouldn't we first pick a specific element $\lambda_A\in\Lambda$ for each object $A$ and then define $A':=A_{\lambda_A}$? Since otherwise I have a constant feeling of having to make such specific choices of $\lambda$ again and again when I should really be focusing on other things. In other words with such definitions I feel I never really know what exactly I have defined ...

Is there indeed some logical issue here, or am I perceiving things that aren't there?

Wouldn't it be better to instead do something like: define $A_\Lambda:=\lbrace A_\lambda|\lambda\in\Lambda\rbrace$, then prove $\mathrm{card} A_\Lambda = 1$ and then define $A':=\mathrm{UE}(A_\lambda)$, where $\mathrm{UE}(C)$ is defined to be the unique element of $C$ if $C$ indeed has exactly one element (and $\mathrm{UE}(C)=\emptyset$ otherwise), thus eliminating the need to choose?

Thanks in advance.

share|improve this question
1  
We have made some choice of $\lambda$. It's definitely fixed. We simply don't bother saying what our choice is because that detail is irrelevant. At times we may be sneakily invoking the axiom of choice here, although this is rarely actually necessary (but hey, we have a hammer, might as well use it). –  Alex Becker Mar 11 '12 at 14:59
    
@Alex: We don't always have that hammer! :-) –  Asaf Karagila Mar 11 '12 at 15:11
    
@AsafKaragila You set theorists can pry Zorn's lemma from my cold, dead hands... :) –  Alex Becker Mar 11 '12 at 15:16
2  
@Alex: I don't have pry that. I can just force you to give it up! Muhaha! :-) –  Asaf Karagila Mar 11 '12 at 15:16
    
@AsafKaragila Okay I admit that was clever. –  Alex Becker Mar 11 '12 at 15:18
add comment

3 Answers 3

up vote 5 down vote accepted

What happens is that even if for each $\lambda$ we might get a slightly different result, the differences are not important for the purpose of the definition (e.g. if we only care about the limit of the sequence we don't really care about the first element).

So the point is that you don't need to choose. You can work with the equivalent classes as your objects and whenever you calculate just pick representatives.

Since calculations are finite you only choose finitely many at each time, and since the choice of representative does not matter - you don't have to worry about which representative you chose.

This is why when you define, for example, the real numbers as classes of Cauchy sequences of rational numbers you don't choose a sequence from each class. You define the real numbers as the equivalence classes themselves.

It is a bit "cheaty" because we are sort of saying "Oh, yes let us do the calculation in another place and pull the result back here.", but because the result is independent of choice of representatives it is valid to do that.

share|improve this answer
    
So, if I understand correctly, in the case of my example above, I'd do something like this: define $\omega$ to be the smallest inductive set and $\mathbf{N} = \omega\setminus\lbrace0\rbrace$. Then define $\mathbf{Z}$ as the usual quotient of $\mathbf{N}\times\mathbf{N}$ and $\mathbf{Q}$ as the usual quotient of $\mathbf{Z}\times\mathbf{Z}$. Then define $\mathbb{R}$ as equivalence classes of Cauchy sequences in $\mathbf{Q}$ and $\mathbb{Q}$ as he corresponding subset of $\mathbb{R}$ and then a new set of real numbers $\mathcal{R}$ as the equivalence classes of sequences in $\mathbb{Q}$. –  Dejan Govc Mar 11 '12 at 15:44
    
Then I use the obvious one to one correspondence, work in $\mathcal{R}$ where there is no need of choosing, and then I pull the definitions in $\mathcal{R}$ back to $\mathbb{R}$. Seems quite convoluted. I think I like it. =) –  Dejan Govc Mar 11 '12 at 15:44
    
@Dejan: That is the plan, more or less. You often want to define canonical embeddings (like $\mathbb Z\to\mathbb Q$ by $p\mapsto [(p,1)]$ etc.) to allow yourself stay within one framework. –  Asaf Karagila Mar 11 '12 at 15:47
    
Thinking about this some more this actually seems to be the same construction as the one with $\mathrm{UE}$ in my question, viewed from a different angle. Thanks. –  Dejan Govc Mar 11 '12 at 17:29
add comment

The issue you're facing, I think, is a lack of abstraction.

When we define an $A'$, we are producing an object we intend to think about as an $A'$, not as a $A_\lambda$. For example, we define $r^s$ because we want to think about exponentiation of real numbers, not because we want to think about limits of exponentials of rational numbers.

In the end, it usually doesn't matter how an object is defined -- just that it is defined, and that we know enough properties of the object to work with it. (and whiel nice, you don't always need the former....)

Anyways, you can certainly translate constructions of the form you describe to use unique elements of singleton sets; it really is effectively same thing.

Do note that even if you go with the translation, the definition still depends upon the ability to make a choice -- e.g. to expand on Alex Becker's comment, a collection of objects defined in this fashion is only well-defined if we have a choice function on their underlying sets. If not, then the parameter space of your $\lambda$'s is collectively empty, so the image of the construction is also empty.

(edit: it may be possible to work around this final comment -- I haven't fully thought it through. Thanks Asaf for making me think a second time, at least)

share|improve this answer
    
The use of the axiom of choice is absurd at times. In the Solovay model where all sets of reals are measurable, we cannot construct a Vitali set. Just because we cannot choose a set of representatives for $\mathbb R/\mathbb Q$ does not mean that the quotient group and its operations are not definable, it is just not a "set of representatives from $\mathbb R$" but rather a set of equivalence classes of real numbers. –  Asaf Karagila Mar 11 '12 at 15:19
    
@Asaf: If we have a family of objects so constructed, then naively $\prod_i \Lambda_i$ is involved and would cause problems if empty. However, I believe I've convinced myself that we can use the axiom of replacement to make things work out. –  Hurkyl Mar 11 '12 at 15:34
    
The definition $a\mapsto [a]$ (some equivalence class) indeed assures us that the set of equivalence classes is nonempty (and of course, it assures it is a set!) and so by going back and forth from our space to the equivalence classes for the calculations we can easily avoid choosing anything. –  Asaf Karagila Mar 11 '12 at 15:38
add comment

A prototypical example is when the set $\{A_\lambda\}$ is some equivalence class. For example, suppose it is the class of all polynomial expressions (composed of $0,1,+,*,x$) that are provably equal by ring axioms, e.g. $\: x\:(x + 1) = x*x + x\:$ (or, semantically, equal if they are ring identities, i.e. true when evaluated in any ring). Using the ring axioms, we can prove that every polynomial is equivalent to one in normal form $c_0 + c_1\:x+\cdots+c_n\:x^n\:$. Then we can define the polynomial ring $\mathbb Z[x]$ to be the set of these normal forms (vs. set of equivalence classes), and the induced ring operations are well-defined on the normal forms, the ring axioms are satisfied, etc.

This is rigorous because the normal forms really denote their associated equivalence classes, and everything works fine at the equivalence class level. One is simply transporting the ring structure from the set of equivalence classes to a complete system of normal form representatives.

There are many reasons for working with normal forms vs. equivalence classes. First, conceptually they are more concrete. Thus modular arithmetic in $\mathbb Z/m$ is more concrete via normal form reps $i\in \{0,1,\cdots,m-1\}$ vs. cosets $\:i + m\:\mathbb Z.\:$ Further, normal forms facilitate effective algorithms, e.g. for deciding equality ("word problem"), e.g. reduce polynomials to normal form and compare their coefficients. Finally, normal forms help one forget internal (constructional) structure that we wish to abstract away from when viewing the set as an abstract algebraic structure (here a ring).

From an algebraic perspective it's crucial to forget about any internal structure possessed by the elements of the structure. Such internal structure is an artefact of the particular construction employed. Such representational information is not an essential algebraic property. It matters not whether the elements are represented by sets or not, or by sequences, matrices, functions, differential or difference operators, etc. Instead, what matters are how the elements are related to one another under the operations of the structure. Thus the isomorphism type of a ring depends only upon its additive and multiplicative structure. Rings with the same addition and multiplication tables are isomorphic, independent of whatever 'names', representations or other internal structure the elements might possess. Andy Magid emphasizes this nicely in his Monthly review of Jacobson's classic textbook Basic Algebra I. Here is an excerpt:

"This reviewer will not attempt a definition of this essence of algebraic thought. But perhaps the reader will excuse a little soft speculation. Algebra seems to be about the holistic properties of collections of things which, while they have no independent status, derive their significance from the relations and operations that exist on the collection as a whole. It makes little difference that the elements of our ring, say, are matrices, or differential operators, or formal linear combinations of group elements; in fact it can even be a positive hindrance to think of them that way. For example, consider the groups of Galois theory: it is vital, in the end, to think of these groups as groups of substitutions in the roots of the equation being examined, but the technical problems which would attend an attempt to prove the criterion for solvability by radicals while working exclusively within this particular representation would be enormous. Far better to ignore the nature of the elements and to think of the group as a thing in itself. To take another example: the notion that the cosets of a normal subgroup of a group, while they have intrinsic meaning as subsets of the original group, are best thought of as unities, as elements of a new group, the quotient group, is often the pons asinorum of the Basic Algebra course. Those who cross it successfully usually do learn to think algebraically. It is probably unfair to claim this thought mode - ignoring the essence of elements of a structure and focusing on their relations - as exclusive to Algebra. This is the basis of much modern abstraction, and not only in mathematics; see for example [Piaget: Structuralism]. But Algebra does seem to appear whenever structures dominate a piece of mathematical thought."
-- Andy Magid, Amer. Math. Monthly, Oct. 1986, p.665
-- excerpt from a review of Nathan Jacobson: Basic Algebra I

share|improve this answer
    
Thanks, that's a nice example. –  Dejan Govc Mar 11 '12 at 16:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.