Here we consider simple variations of the famous Barber Paradox (3 versions) and demonstrate the need for set theory in a very simple and direct way. We will prove using set theory that the fabled village barber can actually shave those and only those men in that village who do not shave themselves *if and only if* that barber is *not* a man in the village. This would seem to be impossible to prove using only first-order predicate logic.

Last updated Feb. 10, 2017

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Last updated: Nov. 18, 2015

Here we turn Hilbert’s Hotel on its head. Instead of frantically

shuffling infinitely many guests for one room to the next in

Hilbert’s mythical infinite hotel, we start with a leisurely walk

through an ordinary and quite finite village.

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Last updated: Oct. 12, 2015

*The Drinker’s Theorem: Consider the set of all drinkers in the world, and the set of all people in a given pub. Then there exists a person who, if he or she is drinking, then everyone in that pub is drinking. *

It doesn’t matter how many people are there. Or how many are drinking. Or how few. No one needs to be taking their cues from some “lead drinker,” but in every pub, in every town and village, it just happens! How is this possible?

There are several possible approaches to this problem. Here, we will turn to British philosopher and mathematician, Bertrand Russell (1872 – 1970). His famous Paradox is the key.

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Last updated: Nov. 1, 2016

The Definition of the set of natural numbers

is given by nothing more or less than

Peano’s Axioms.

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Last updated: Jan. 12, 2015

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Now featuring a formal construction of the

partial function for exponentiation on *N*x*N.*

The Cantor-Bernstein-Schroeder Theorem (CBST), is one of the most important and widely applied results in set theory:

If set

Xcan be mapped one-to-one into setY(an injection), and setYcan be mapped one-to-one into set X (an injection), then X can be mapped one-to-oneonto Y(a bijection).

Though seemingly self-evident, some proofs of CBST can make your head spin! Every line of my machine-verified, formal proof (updated 2014-09-11) is justified by one of a very limited list of simple axioms and rules of inference (indicated in a grey font at the end of each line).

Such complete rigour does come at a price, however. While it is completely free of any of the sort of the hand-waving that plagues many informal versions of this proof, like most formal proofs of any complexity, it is *very* long. My commentary (indicated by a blue font) is inserted throughout using *DC Proof.* I believe this makes my proof considerably more readable than machine-generated formal proofs in other systems.

This proof was written to demonstrate he capabilities of the *DC Proof* system. Though designed for ease of use by the complete beginner, *DC Proof* is quite capable of some mathematical heavy lifting.

Last updated: Oct. 21, 2016

According to the Pigeonhole Principle, if you have more pigeons than pigeonholes (as in photo), then at least two pigeons will be in the same hole. Here we present a *non-numeric* version.

Usually, *more* pigeons than pigeonholes is taken to mean that the *number* of pigeons is *greater than* the *number* of pigeonholes. Here, we take more pigeons than pigeonholes to mean that the set of pigeonholes cannot be mapped *onto* the set the pigeons.

In this sense then, we can prove that, if we put a non-empty set of pigeons into a set of pigeonholes and there are more pigeons than holes, then we will have put at least two pigeons in the same hole. Note that there is no requirement here that there be finitely many pigeons or pigeonholes.

See formal proof (70 lines) at The Pigeonhole Principle: A non-numeric version.