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Does x={x}

  1. Apr 27, 2012 #1
    If x is an "element", then is x={x} a true statement?

    I mean, how can you define x as an element without reference to some set or another? It is at least a member of a set contain only itself, right? If not, then what's the difference between an element x and the set {x}? Thanks.
     
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  3. Apr 27, 2012 #2

    mathman

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    This is the way set theory is structured. x and {x} are just different.
     
  4. Apr 27, 2012 #3

    Hurkyl

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    "Most" x are not solutions to the equation x={x}. If you assume the axiom of foundation (and mathematicians usually do), the equation doesn't have any solutions at all.

    {x} is the singleton set containing x. x is just x
     
  5. Apr 27, 2012 #4
    Maybe I'm just not thinking hard enough, but I'm having difficulty thinking up an example of when I refer to x but do not mean {x}. If x were not an "element", then I can say it is outside the scope of set theory. But as soon as I say x is an element, then I am referring to x as being a member of some set. Since the least set that x can be a member of is {x}, then I think this is what is meant by referring to x as an "element".

    Can you think of a situation in which you consider x as an element apart from being an element of at least {x}? For if x is an element of a larger set, say {x,y,z}, this still means, {x} U {y,x}, so x being an "element" means at least {x}. Do you disagree?

    Come to think of it, can you name something that doesn't mean that you are not referring to that thing as belonging to a category with some specific properties that do form a set of similar objects where that set may only have one member, namely that thing? How can you refer to something that does not belong to any category or have any general propoerties that cannot be referred to as belonging to the set of those objects that have those properties?

    To labor the point a bit. I'm thinking it is impossible to describe an object, abstract or concrete, that does not have some sort of unique property, where the extention of that property (Predicate logic) is a set.
     
    Last edited: Apr 27, 2012
  6. Apr 27, 2012 #5
    In my left hand I have an orange.

    In my right hand I have a paper bag containing an orange.

    Do you see these as two different things? One is an orange. The other is a paper bag containing an orange.

    Just think of the curly braces as a paper bag.

    {} is an empty paper bag.

    {{}} is a paper bag containing an empty paper bag.

    x is an orange.

    {x} is a paper bag containing an orange.

    Does that help? Because that's the best intuitive way to think about it.
     
  7. Apr 27, 2012 #6

    Hurkyl

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    If you ever thought to say "x is a set with 3 elements", you can be very sure you didn't mean "{x} is a set with 3 elements", because the latter is clearly wrong. ({x} is a set with 1 element)

    Similarly, if you ever thought to say "x is an _____" where the notion "_____" can include things which aren't sets, you can again be pretty sure you didn't mean "{x} is an ______".
     
  8. Apr 29, 2012 #7
    x∈ { x }, but x != { x } because x isn't not a set and equality is defined for set comparison, but not a non-set to set comparison.
     
    Last edited: Apr 29, 2012
  9. Apr 29, 2012 #8
    What if x is a set? What if x is, say, the set of natural numbers? Or the set {3,5,43}? Does your reasoning still apply?
     
  10. Apr 29, 2012 #9

    mathman

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    Yes - a set contains elements (which may be sets), but a set cannot be one of its own elements. This is one of the conundrums of set theory - what is the set of all sets?
     
  11. Apr 29, 2012 #10
    Was it not clear to you that my post was a rhetorical question for the person who thought that x could not be a set?

    There are variants of set theory in which [itex]x \in x[/itex] is allowed. This has nothing to do with the set of all sets, which is disallowed for different reasons.

    If fact the reason we don't usually have [itex]x \in x[/itex] is simply because of the Axiom of Regularity.

    http://en.wikipedia.org/wiki/Axiom_of_regularity

    This axiom is in no way "obvious" or "necessarily true" or any such intuitive formulation. In fact it's simply a technical condition. The article notes that there are set theories in which we allow [itex]x \in x[/itex].

    You might find this article of interest. It describes set theories in which [itex]x \in x[/itex] is allowed.

    http://en.wikipedia.org/wiki/Non-well-founded_set_theory

    This is all beyond the scope of the OP's question.

    But still ... what was unclear about my question, "Does your reasoning still apply?" directed to a specific poster? What made you think I was asking a general question?

    By the way, mathman, I attempted to send you a private message but you have that functionality blocked. If there's some way for me to communicate with you privately for a moment, it would be very helpful today.
     
    Last edited: Apr 29, 2012
  12. Apr 29, 2012 #11
    I get what you're asking, thank you for the great question, I was wrong, x could be a set, but by including it in set notation, you're automatically wrapping it another set, so no matter what, x != {x}.
     
  13. Apr 29, 2012 #12
    Yes, exactly.
     
  14. May 1, 2012 #13

    mbs

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    I think there is a reason differentiate between a singleton set containing one individual and the individual itself. The axiom of extensionality, in effect, defines sets based on the members they contain. Individuals by definition are non-set-objects that have no members and are exempted from the axiom of extensionality.

    If you were to consider a singleton set containing an individual as being the same object as the individual itself, then for consistency you'd also have to consider a singleton set containing another set as the same object as the set it contains.

    I.e., you'd have to have...

    [itex]x = \left\{ x \right\} = \left\{ \left\{ x \right\}\right\}=\left\{\left\{\left\{ x \right\}\right\}\right\} = \cdots[/itex]

    by the extensionality axiom. This seems weird to me, but I suppose it could work. I'd have to think about it.
     
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