1
$\begingroup$

I read this from Introduction of the 1st edition of Principia Mathematica by Whitehead and Russell:

Since the orders of functions are only defined step by step, there can be no process of "proceeding to the limit," and functions of infinite order cannot occur.

Here is a summary of orders of functions:

  1. An nth-order function is one that takes (n-1)th and lower order functions as arguments.

  2. A first order function is one that takes only "individuals" as arguments.

  3. A second order function is one that take first order function and "individuals" as arguments.

  4. By "individuals" we mean constituent objects that are neither a function or a proposition and will not disappear like a class or a set after analysis.

At this point, I think an nth-order function is perfectly eligible for mathematical induction. I can't see the connection between something defined step by step and its inability to "proceed to the limit." It seems that the authors assume the readers are familiar with some well-known rules that specify the necessary condition for the process of "proceeding to the limit." I wonder what these rules are.

$\endgroup$
  • $\begingroup$ Ohh, that is what you mean by order. It wasn't completely clear from your post. $\endgroup$ – Asaf Karagila Jan 27 '14 at 1:51
  • $\begingroup$ Sorry, I'll re-edit it. Every jargon in this post in within the scope of Whitehead and Russell's Principia Mathematica. $\endgroup$ – George Chen Jan 27 '14 at 1:54
  • $\begingroup$ Yes. I realize that now. You meant order as in type theory, or high-order logic. $\endgroup$ – Asaf Karagila Jan 27 '14 at 1:55
  • $\begingroup$ Yes, it's part of type theory. $\endgroup$ – George Chen Jan 27 '14 at 2:01
2
$\begingroup$

Since the usual set-theoretic suspects don't seem to be responding I would like to comment briefly that you seem to be confusing two notions of limit. The limit you have in calculus can by all means be taken, and is taken many times every second if you try to estimate the number of calculus students around the globe. The "conditions" for taking such limits can be found in every calculus textbook.

Meanwhile, PM is talking specifically about a situation where one cannot take a "limit" where of course limit is understood in a more nebulous way. Namely, the types exist only for finite orders, and restricting them to finite orders may be precisely what is needed so as to escape the paradoxes of naive set theory that Russell is famous for spotting.

$\endgroup$
  • $\begingroup$ Taking the hint from ❋9, I guess mathematical induction at this stage is not yet applicable. $\endgroup$ – George Chen Feb 11 '14 at 4:18
  • 1
    $\begingroup$ Induction on $n$ allows you to prove a statement for all finite values of $n$. It does not give you a way of forming an infinitary object (without additional axioms at any rate). $\endgroup$ – Mikhail Katz Feb 11 '14 at 13:04
  • $\begingroup$ I finally understand your last comment and I agree. Mathematical induction turns out to be a defining property of inductive numbers, as opposed to non-inductive or reflexive numbers. Infinity belongs to the latter category. $\endgroup$ – George Chen Jul 23 '14 at 9:10
0
$\begingroup$

Short answer:

Mathematical induction can never proceed to infinity. The necessary condition for mathematical induction is that the property in question is possessed by an inductive number and is hereditary with respect to the relation $+_c1$.

Long answer:

This book is like building a tower from ground up without scaffolding, crane or any outside support. Everything that touches the ground must be explicitly spelt out and every brick must be supported by the ones from beneath not above. In other words, primitive ideas and primitive propositions are all the foundations this book has.* And every proposition is either a primitive one or must be proved by numbers less than itself. The formulae of induction are not reached until ✳91·17·171·373. The hierarchy of types is introduced in ✳12 which precedes the formulae of induction.

✳91·17 $\vdash:. P \in Potid‘R:\phi S. \supset_S. \phi(S|R):\phi(I {\restriction C‘R}): \supset \phi P $

Which states that if the property $\phi$ is hereditary with respect to $|R$, then if $\phi$ belongs to $I \restriction C‘R$ it belongs to any member of $Potid‘R$

Based on summary of Part III Section C, mathematical induction is a definition rather than a principle. In other words, some cardinals are inductive, some other cardinals are non-inductive. The defining property of inductive cardinals is this: An inductive cardinal is one which obeys mathematical induction starting from $0$, i.e. it is one which possesses every property possessed by $0$ and by the numbers obtained by adding $1$ to numbers possessing the property. In other words:

$\vdash :: \alpha \in NC induct .≡:. \xi \in \mu .\supset_\xi. \xi +_c 1 \in \mu : 0 \in \mu \supset_\mu. \alpha \in \mu $

An inductive number by definition is one which can be reached from $0$ by successive additions of $1$. Such a process can never reach infinity because ✳123 shows that the smallest infinite number $\aleph_0$ is not an inductive number.

*Due to the speculative nature of philosophy, the authors admit that "there must always be some elements of doubt, since it is hard to be sure that one never uses some principle unconsciously." --see Part I. Section A, on first page of theory of deduction.

$\endgroup$
  • $\begingroup$ AS a footnote, I cannot even say the total number of types is $\aleph_0$, because in order for something to have a number, that something must be a collection. But it is impossible to put different types into one collection, therefore, the number of types is meaningless. $\endgroup$ – George Chen Aug 31 '14 at 4:17
  • $\begingroup$ It can't be the relation number of $Cl$ either, because $C‘Cl$ is meanlingless. $\endgroup$ – George Chen Aug 31 '14 at 5:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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