Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The empty set

  1. Sep 21, 2008 #1
    Hey. um i got a few questions about the empty set... i know for a fact that the empty set is a subset of all any set..
    a) but is it an element as well??
    b) can it be an element of itself ?
    c) or even a subset of itself??

    [tex]\oslash[/tex] [tex]\in\oslash[/tex] is this true??
  2. jcsd
  3. Sep 21, 2008 #2
    since it is a subset of all set ,it is a subset of itself .

    And an element of any power set.

    Let Φ={ x: x=/=x} and let ΦeΦ then we have that Φ=/=Φ Which it is not true therefor ~ΦεΦ So it does not belong to itself.

    So a and c are correct and b is false
  4. Sep 22, 2008 #3


    User Avatar
    Homework Helper

    The statement

    \emptyset = \{x | x \ne x \}

    is not correct - it is one way to generate an empty set, but not the only way. Here is another.

    \emptyset = \{x | x \text{ is an odd positive integer perfectly divisible by } 2 \}

    A safer way to think about [tex] \emptyset [/tex] is to say that it is the set that does not contain any elements (note I said think about, not define).
    This directly answers one question: since it contains no elements (had cardinality zero), it cannot be an element of itself.
    This also lends a way to argue it is an element of every other set. If [tex] A [/tex] is another set, suppose [tex] \emptyset \not\subset [/tex]. The, by definition of subset, there is some element [tex] x \in \emptyset, x \not\in A [/tex]. Since [tex] \emptyset [/tex] does not contain any elements, this is a contradiction and, therefor, [tex] \emptyset \subseteq A [/tex] is true.
  5. Sep 24, 2008 #4
    what about this??

    [tex]\oslash[/tex] [tex]\subseteq[/tex] {[tex]\oslash[/tex]}

    would this be true?
    because from what i know.... this is an empty set.... and it doesn't contain
    an element... but it a subset of all sets....... so shouldn't this be true??
    or would this even be a proper subset???
  6. Sep 24, 2008 #5
    For example, the first four natural numbers are defined as follows (see wiki for natural numbers and peano postulates)

    0 := ∅, the empty set

    0 = ∅
    1 = {∅} = {∅}
    2 = {0,1} = {∅, {∅}}
    3 = {0,1,2} = {∅, {∅}, {∅, {∅}}}

    0⊆ 1⊆ 2⊆ ....

    As shown above, ∅ ⊆ {∅}.
  7. Sep 25, 2008 #6


    User Avatar
    Science Advisor
    Homework Helper

    {} is a subset of {{}}.
    {{}} is a subset of {{}}.
    {} is a proper subset of {{}}.
    {{}} is not a proper subset of {{}}.
    {} is an element of {{}}.
    {{}} is not an element of {{}}.
  8. Sep 25, 2008 #7
    Some additional notes from set and category theory (see wiki for details of "Regularity Axiom").

    "No set is a member of itself."
    Thus, 0 ∉0, 1 ∉1,,, (0 and 1 is defined above ).

    It follows from the "Regularity Axiom", which says "Every nonempty set A has a member m with m ∩ A = ∅".

    Let X be such a set that X is an element of itself and define Y = {X}.
    We apply "Regularity Axiom" to Y. Then Y has only one element, which is X and it should be disjoint from Y by the "Regularity Axiom". However, X is both an element of itself and an element of Y, which shows X and Y is not disjoint, contradicting the "Regularity Axiom". Thus X should not exist.

    In category theory, the universe U of the category 'Set' of all small sets is not a small set (A small set is defined as a set u∈U). Otherwise U∈U, which contradicts the "Regularity Axiom". Thus 'Set' is not a small category (Both objects and arrows of a small category should be small sets).
  9. Sep 26, 2008 #8

    How would you prove that {} is a proper subset of { {} }
  10. Sep 26, 2008 #9


    User Avatar

    well, {} contains no elements. So
    1) every element of {} is an element of {{}} (a vacuous truth). Hence {} is a subset of {{}}.
    2) {{}} contains the element {}, so {} != {{}}

    so {} is a proper subset of {{}}.
  11. Sep 26, 2008 #10
    thats a proof by inspection
  12. Sep 26, 2008 #11
    By the axiom of specification we have that , B={ xεA: S(x) } ,where A and B are sets .
    and S(x) is made up according to specific rules .

    If we now put S(x) : x=/=x then, B={ xεA: x=/=x }.

    We define B to be the empty set and denote it by Φ.

    Next step is to Prove that, [tex]\forall x[/tex]( [tex] x\not\in[/tex]Φ) i.e Φ has no elemens

    Let xεΦ then x=/=x,but for all x ,x=x ( an axiom in equality) and thus we have a contradiction and so ~xεΦ and [tex]\forall x[/tex](~xεΦ)..

    Next step is to prove the very important theorem ,[tex]\forall x[/tex](~xεA) <===> A=Φ.

    From the above one can see who is right and who is wrong.

    What is the S(x) in your set according to what rules is made up??
  13. Sep 27, 2008 #12
    Do we make any broad distinctions between sets which contain [tex]\Phi[/tex] as an element, compared with those sets that only have [tex]\Phi[/tex] as a subset? All power sets have it as an element. Are there any special properties such sets share?
  14. Sep 27, 2008 #13


    User Avatar
    Homework Helper

    poutsos.A: I'm not sure what, exactly, you think your work accomplishes - I'm not even sure it's valid, but that is partly because I can't follow your hybrid notation (my shortcoming). If you read my post carefully, you will find that I stated that there are several ways to define an empty set; the one that you proposed is only one such way.
    My reference to
    "set with no elements" was not proposed as a definition, it was provided as a way to think about, visualize, the empty set.
    Regarding your comment that seems to imply that "a set must be made up of specific rules" - technically, you haven't defined 'rule'. But, in what seems to be the vein of your comment, the statement

    [tex] x [/tex] is an odd positive integer perfectly divisible by [tex] 2 [/tex]

    is a type of rule, is it not?

    Finally, I think using [tex] x \ne x [/tex] as the only "rule" for defining the empty set is too narrow. Consider

    \{f | f \in \text{ functions differentiable on } [0,1] \text{ and } f \text{ not continuous} \}

    qualifies as an empty set, since if a function is differentiable it must be continuous, so there is no such beast. Specifying a property that is self-contradictory is not the same as specifying something that is not equal to itself. (Perhaps a condition that is self-contradictory is what you actually meant?)

    Again, I did not say you were wrong in your example, only in your implication that your construction is the only one possible: it is not.
  15. Sep 27, 2008 #14


    User Avatar
    Science Advisor
    Homework Helper

    I'm not sure what problem you have with that definition. I think that
    [tex]\emptyset:=\{x:x\neq x\}[/tex]
    is quite sensible, and that
    [tex]\emptyset=\{f | f \in \text{ functions differentiable on } [0,1] \text{ and } f \text{ not continuous} \}[/tex]
    would follow as a consequence of the above definition.
  16. Sep 27, 2008 #15


    User Avatar
    Homework Helper

    The difference is this.

    x \ne x

    puts the onus of definition on a specific object: my point is that you do not need to be that specific - it narrows the applicability of the concept (and so narrows the usefulness in applications) of an empty set.

    Again: I didn't say it was wrong - I said i wasn't the only approach. That point seems to be missed.
  17. Sep 27, 2008 #16

    Axiomatic definition of empty set is as follows:

    Empty Set Axiom (ZF):

    ∃B∀x ~(x∈B)

    which is interpreted as "There is a set having no members."

    It is the only set with no members because of "Axiom of Extensionality".

    The "Axiom of Extensionality" says
    "If A and B are sets such that for every object t, t∈A iff t∈b,then A = B."

    For sets with no element, the above statement is vacuously true and it is unique as any two such set must coincide by the "Axiom of Extensionality".
    Last edited: Sep 27, 2008
  18. Sep 28, 2008 #17


    User Avatar
    Science Advisor
    Homework Helper

    Yes, and you can weaken the Axiom of Extensionality to allow nontrivial ur-elements in your theory, which are essentially nonequal empty sets.
  19. Oct 2, 2008 #18
    is not that contradictory??
  20. Oct 3, 2008 #19


    User Avatar
    Homework Helper

    try to read things, and comment on them, in their complete context.

    My "is not correct" comment referred to your assertion that the statement in question is the only way to define an empty set. If you cannot realize that fine, but you will be wrong in your belief.

    "I did not say you were wrong in your example" refers to the fact that your statement does provide one way to generate an empty set.

    Read things carefully next time.
  21. Oct 3, 2008 #20
    The empty set ø is a subset of any set A because [tex]\forall x \in \emptyset , x \in A[/tex]

    So you see nothing is in the empty set which is exactly why the above statement is true no matter what set you are talking about.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook