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I just read about inductively defining a set as follows :

  • Take a set of objects U.
  • Take a set of starting objects B ⊆ U.
  • Let C be the smallest subset of U that contains B and is closed under all operations on some class F.

    Any set S that contains B and is closed under all operations in F is called inductive.

For example , consider the set of natural numbers as :

  • U = R , B = {0} and F = {S} where S(x) = x+1

My Question is can we define positive Real Numbers as follows :

U = R , B = [0 , 1) and F = {S} where S(x) = x+1

If so , can we prove property P(x) of positive real numbers inductively as follows :

  • P is true in interval [0,1).
  • If P(x) is true ,P(x+1) is also true.
Conifold
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Aditya Singh Rathore
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  • Well, kinda? It's definitely not a single step induction, but you can transfinitely induct on the reals. So no, not exactly the properties you want, but similar. – Rushabh Mehta Jan 20 '20 at 05:31
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    Unfortunately, when you transfinitely induct, you lose any sense of the normal order on the reals. That may not be desirable. – Rushabh Mehta Jan 20 '20 at 05:32
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    When you "define" reals this way most of the work is swept under the rug in the initial set $[0 , 1)$. A proof of $P(x)$ for $x\in[0 , 1)$ will likely go through for all real $x$ without any induction. There is a more meaningful "induction" on reals, where you prove 1) $P(x_0)$; 2) If $P(x)$ then $P(x+h)$ for all small $h$; 3) If $P(x_n)$ and $x_n\to x$ then $P(x)$. Then $P(x)$ follows for all $x\geq x_0$, see real induction. – Conifold Jan 20 '20 at 06:19
  • @Conifold Is the definition correct ? – Aditya Singh Rathore Jan 20 '20 at 06:23
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    It isn't a definition since fractional reals are taken as a given. A meaningful definition has to start from something independently constructible, like integers or rationals. – Conifold Jan 20 '20 at 06:25
  • There is a truly inductive definition that generates reals (and beyond) due to Conway, see surreal numbers. But the induction will not correlate with the usual order on the reals. – Conifold Jan 20 '20 at 06:30
  • https://math.stackexchange.com/questions/4202/induction-on-real-numbers – Alessandro Codenotti Jan 20 '20 at 07:12

1 Answers1

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There is something analogue to the induction principle, called transfinite induction. It is not quite as intuitive as simple induction on $\omega$, but you make do with what you have. It works like this: let $P$ be a property, which you want to proof is true for all ordinal (or cardinal) numbers. You want to show that 1)-P(0) is true 2)-For all ordinals $a < b$, if $P(a)$ is true, $P(b)$ is true

Usually, the latter is divided in two stages, the successor ordinal and the limit ordinal: for the successor, you apply induction sort of in the classical way, whereas for the limit ordinal you have to rely on the fact that it is defined as $sup\{b|b > a\}$, and use 2)

Nennee
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