The axiom of empty set.: Philosophy Forums

The axiom of empty set.

MathematicalPhysics Wizard
Quantum-Gravity Mechanic

Usergroup: Members
Joined: May 02, 2003

Total Topics: 105
Total Posts: 739

Posted 02/17/09 - 10:48 PM: Subject: The axiom of empty set.	#1
Now I don't want to shake the foundations but nontheless , the definition of the empty set is a contradiction, here is one way to define it: {x\|x>x}, now we know from logic that from a contradcition we can deduce everything, so doesn't it mean that set theory is inconsistent?
"What kind of a bomb is this? The exploding kind." Peter Sellers

unenlightened
everything is...

Usergroup: Administrators
Joined: Aug 10, 2007
Location: Wales

Total Topics: 36
Total Posts: 3286

Posted 02/17/09 - 11:00 PM:	#2
Do you mean a contradiction like 'the set of all bearded monkeys without beards"? The definition is a contradiction, but I think that's ok because the set is empty so there aren't any contradictory monkeys.
...most of our actions are the result of the past, or according to a future ideal. That's not action, that is just conformity. J Krishnamurti "Philosophy, to the Philistine, is an evolutionary process, watched over by some sort of brisk dynamic Providence, and culminating in the supreme insight of modern thought." John Cowper Powys

Incision
Professor

Usergroup: Administrators
Joined: Jan 04, 2008
Location: Utah

Total Topics: 23
Total Posts: 906

Posted 02/17/09 - 11:01 PM:	#3
"{x\|x > x}" is not contradictory, unlike "2 > 2." It sets an impossible standard for membership, sure, but that doesn't mean it implies a contradiction. Since there are no objects of which there are true contradictions, the set is empty.

MathematicalPhysics Wizard
Quantum-Gravity Mechanic

Usergroup: Members
Joined: May 02, 2003

Total Topics: 105
Total Posts: 739

Posted 02/18/09 - 01:52 AM:	#4
unenlightened wrote: Do you mean a contradiction like 'the set of all bearded monkeys without beards"? The definition is a contradiction, but I think that's ok because the set is empty so there aren't any contradictory monkeys. That's exactly what I meant, if the definition is contradictory how do we know that what we deduce from the axiom with other axioms won't be as well contradictory?
"What kind of a bomb is this? The exploding kind." Peter Sellers

MoeBlee
aka I. Kabruob

Usergroup: Members
Joined: May 19, 2005

Total Topics: 8
Total Posts: 1273

Posted 02/18/09 - 10:22 AM:	#5
The Bearded Monkey wrote: Now I don't want to shake the foundations but nontheless , the definition of the empty set is a contradiction, here is one way to define it: {x\|x>x}, now we know from logic that from a contradcition we can deduce everything, so doesn't it mean that set theory is inconsistent? (1) Z (and ZF, etc.) do not require an empty set axiom, since E!xAy ~yex is deducible from the axiom schema of separation and axiom of extensionality. (2) I don't know what '>' indicates in your remark. Presumably, you mean some 'greater than' relation, but what 'greater than' relation do you mean among sets in general? (3) An ordinary definition of the empty set is: 0 = x <-> Ay ~yex. That is not a contradiction. (4) For any contradictory formula P, we do have the theorem (whether as itself a definition or not) 0 = {x \| P}. However, that is not itself a contradiction.

MoeBlee
aka I. Kabruob

Usergroup: Members
Joined: May 19, 2005

Total Topics: 8
Total Posts: 1273

Posted 02/18/09 - 10:27 AM:	#6
The Bearded Monkey wrote: That's exactly what I meant, if the definition is contradictory how do we know that what we deduce from the axiom with other axioms won't be as well contradictory? Properly formed definitions are not contradictory. A contradiction may occur as a clause in some definition, but that doesn't make the definition contradictory. Merely that a contradictory clause occurs in a formula doesn't make the formula contradictory. For example (P & ~P) -> (P & ~P) has two occurreneces of the contradiction 'P & ~P', but the formula '(P & ~P) -> (P & ~P)' is not a contradiction. The definition of '0' in set theory is not contradictory.

muxol
yuletide

Usergroup: Members
Joined: Dec 07, 2003

Total Topics: 1
Total Posts: 1932

Posted 02/19/09 - 12:41 AM:	#7
The Bearded Monkey wrote: Now I don't want to shake the foundations but nontheless , the definition of the empty set is a contradiction, here is one way to define it: {x\|x>x}, now we know from logic that from a contradcition we can deduce everything, so doesn't it mean that set theory is inconsistent? {x:x>x} is not a formula of the language, it is a term, so it cannot be contradictory, as only formulas can be contradictory, not terms. Of course, proving the existence of some object denoted by a term might lead to contradictions (as in the Russell set), but this is not a problem here. THE formula (up to logical equivalence, and without being fussy) which defines the extension of the set in question is not contradictory either, being (y)(y e {x:x>x} <-> y>y). Likewise (x)(Rxy <-> P & ~P) is not contradictory.

MathematicalPhysics Wizard
Quantum-Gravity Mechanic

Usergroup: Members
Joined: May 02, 2003

Total Topics: 105
Total Posts: 739

Posted 02/25/09 - 12:08 AM:	#8
But one way to look at this set as definining it as the set which doesn't consist of items is: x in φ iff ~(x in φ ) and this is obviously a contradiction. As unelightened gave an example "the set of all bearded monkies which are shaved" is empty because it's a contradiction.
"What kind of a bomb is this? The exploding kind." Peter Sellers

MoeBlee
aka I. Kabruob

Usergroup: Members
Joined: May 19, 2005

Total Topics: 8
Total Posts: 1273

Posted 02/25/09 - 10:27 AM:	#9
No, that's wrong. That is NOT a DEFINITION of the empty set, and it's not a theorem either. Here is what we DO have: Definition: 0 = the x such that Ay ~yex. Theorem: xe0 -> ~xe0. But we do NOT have, as theorem or part of any definition: ~xe0 -> xe0.

MathematicalPhysics Wizard
Quantum-Gravity Mechanic

Usergroup: Members
Joined: May 02, 2003

Total Topics: 105
Total Posts: 739

Posted 02/25/09 - 10:02 PM:	#10
Yes, you are correct I realised it bymyself.
"What kind of a bomb is this? The exploding kind." Peter Sellers

Sorry, you don't have permission to post. Log in, or register if you haven't yet.