nLab
Markov's principle

Markov's Principle

    In constructive mathematics, Markov's principle is the (classically trivial) statement that any infinite sequence of 00 and 11 that is not all 11s must have a 00 somewhere. Stated in a more logical form, if PP is a predicate on natural numbers, then

    (n,P(n)¬P(n))¬(n,P(n))n,¬P(n). (\forall n, P(n) \vee \neg{P(n)}) \Rightarrow \neg(\forall n, P(n)) \Rightarrow \exists n, \neg{P(n)} .

    Compare this to

    ¬(n,P(n))n,¬P(n), \neg(\exists n, P(n)) \Rightarrow \forall n, \neg{P(n)} ,

    which is a theorem of intuitionistic logic. More generally, a set SS may be called Markovian if this principle holds for all predicates on SS.

    In standard constructive mathematics (such as in the internal logic of a topos), it is possible that the only Markovian sets are the Kuratowski-finite sets. Thus, Markov's principle, stating that the set of natural numbers is Markovian, is nontrivial. (It is true, of course, in a Boolean topos; that is, Markov's principle follows from the principle of excluded middle.)

    А. А. Марков Jr? (the one who proved undecidability theorems, and son of the great stochastician) belonged to the Russian school of constructivism, which saw mathematics as about computability. From this perpsective, Markov's principle is justified as follows: We are justified in concluding n,¬P(n)\exists n, \neg{P(n)} if we can actually compute a value of nn such that P(n)P(n) can be proved; since PP is decidable, it's enough to compute nn such that P(n)P(n) is true. And to compute this, you just set a computer working, deciding P(0),P(1),P(2),P(0), P(1), P(2), \ldots, until it finds nn. Other constructivists find this argument unconvincing, since they're not convinced that the computer will ever stop, even though it's impossible that it continue forever.

    Equivalent forms:

    • If a Turing machine? does not run forever, then it halts.
    • If an extended natural number is not infinite, then it is finite.
    • If a Cauchy real number does not equal zero, then it is apart from zero in that it has a multiplicative inverse.

    Note that the contrapositives of these are all valid regardless of Markov's principle.

    Revised on October 15, 2012 03:35:37 by David Roberts (192.43.227.18)