Self referance

  • #1
disregardthat
Science Advisor
1,854
33
Russels paradox says that the set of all sets not contained in themselves, i.e.

[tex]x=\{ y \ : \ y \not \in y \}[/tex]

neither is or is not contained in itself. The set can be created based on an axiom saying that
"The set of objects with a property Q exists".

Let`s assume Q : "Is not contained within itself" is a valid property. Then not Q also is a valid property. Assume z has the property not Q. Then

[tex]z \in z[/tex]

What we do when defining such a set is:

[tex]z=\{ y \ : \ y \ \text{has} \ P \} \cup \{ z \}[/tex] where P is some property.

Is this in any sense a valid definition? Before z is properly defined, we use it. Any definition must consist of properly defined terms, and in this case i argue it doesn`t. Hence not Q is not a valid property, and thus Q isn`t.

ZFC fixes this by saying that a set can be created as long as it isn`t equivalent with Russels paradox. Can`t we just say that the property must be well defined?
 

Answers and Replies

  • #2
811
6
Russels paradox says that the set of all sets not contained in themselves, i.e.
Russel's paradox says that the existence of a set [tex]
x=\{ y \ : \ y \not \in y \}
[/tex]
leads to a contradiction. It was for this reason why ZF was created, which has the axiom of regularity, preventing the question of self-inclusion all together.

If we're working in ZF, then your question about [tex]
z=\{ y \ : \ y \ \text{has} \ P \} \cup \{ z \}
[/tex] is uninteresting. The set z contains itself, and so within ZF, it does not exist.


There are other kinds of self-reference which do not create any sort of contradiction. It's possible in ZF to define functions, sequences, and other nice things recursively.
 

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