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I Empty set confusion..

  1. Oct 14, 2016 #1
    The empty set has a cardinality of 0, yet it is found in all sets and as we know any sets within a set are counted as a single element. But it never counted as an element in the cardinality of a set?

    For example: The set {1,2,3} contains 8 distinct subsets including the empty set, as taken by the formula and method of deriving a sets power set.

    We know the empty set to be a member of all sets, a proper subset of all except itself. Any set within a set is counted as its element.

    So why is the empty set not contained within the cardinality? I know that the empty set has no elements, but the definition of an element of a set means even the empty set is an element.

    Maybe I am confused again
     
  2. jcsd
  3. Oct 14, 2016 #2
    for example a= { {a,b,c}, d, 0, {}} contains 4 elements, right?
     
  4. Oct 14, 2016 #3

    fresh_42

    Staff: Mentor

    Yes. And these four elements are of different kinds: two sets, a character and a number. In addition one set has a character element "a" which also denotes the set as a whole. This is a potential source of misunderstandings.

    The same goes for your OP. The empty set as an element of another set, is fundamentally different from the empty set in its own right.

    Do you have a certain question, because I'm not sure that I understood you correctly. E.g. how do cardinals come into play?
     
  5. Oct 14, 2016 #4
    My question is, shouldnt the cardinality of the set {1,2,3} be 4? as the empty set is also a subset and thus element of this set?
     
  6. Oct 14, 2016 #5

    fresh_42

    Staff: Mentor

    No, because the empty set is not an element of ##\{1,2,3\}##.
    It is an element of the power set ##\mathbb{P}(\{1,2,3\})## in which it is counted.
    And it is also in element of your set a, which has cardinality four, i.e. it is also counted.
     
  7. Oct 14, 2016 #6
    ok..but the empty set is a proper subset of all sets except itself, but it is not an element of all sets unless explicitly mentioned?
     
  8. Oct 14, 2016 #7

    fresh_42

    Staff: Mentor

    Proper? Yes. But ##\{\}\subseteq\{\}## is also true. Not much worth, but true.
    Yes. It is only an element if either mentioned or as an element of any power set.

    Edit: I better withdraw my comment ##\{\}\subseteq\{\}## being not much worth. Far too dangerous here.
    At least it is responsible for ##|\mathbb{P}(\{\})|=2^0=1##.
     
  9. Oct 14, 2016 #8
    thank you.
     
  10. Oct 14, 2016 #9
    may i ask why the empty set is a subset of all sets but NOT an element unless mentioned explicitly?
     
  11. Oct 14, 2016 #10

    fresh_42

    Staff: Mentor

    You could ask as well why the entire set as a subset of itself, isn't an element. Or any other subset. Would make it difficult to distinguish between sets and power sets though.
    What you can say is: All elements of the empty set are as well elements of an arbitrary set, because all statements about the elements of the empty set are true. Or as I like to put it: The elements of the empty set have purple eyes.

    Normally one defines a set by starting with "{" and ending with "}". Anything between is considered as the elements of it. Either by a list or a closed expression, where the notation goes on with ##x \in < \text{ living space, I mean set, e.g. } \mathbb{Q} >\,\vert \, <\text{ expression }>##, e.g. ##1 \leq x \leq 2##. Of course any other symbol than ##x## will do. So there is simply no empty set anywhere near. To be an element, it's thus necessary, that at least one element of the set is a set itself. This is automatically true for power sets, or by a definition like yours for the set ##a##.

    In an environment, where all sets are sets of sets, it might be the case, that the empty set is part of it, in which case it counts for the cardinality.
    E.g. some definitions of natural numbers start with ##\{\}## as representation of zero.

    In most other areas, sets are usually not elements of other sets. They are subsets or supersets.

    For short: being a set and being an element are two different attributes. To combine them in any way, has therefore explicitly to be done - nevertheless still to be distinguished!
     
  12. Oct 14, 2016 #11
    does this have to do with Bertrand Russell and set theory ?
     
  13. Oct 14, 2016 #12

    fresh_42

    Staff: Mentor

    Well as soon as we talk about the concept of sets, it has automatically to do with set theory (by definition) and Bertrand Russell (by history and logic).
     
  14. Oct 14, 2016 #13
    ok. I feel like sometimes I should give up on mathematics, as I am never able to grasp the concepts quickly or at the speed I would like. tODAY i only covered 5 pages and could manage 25 questions.
     
  15. Oct 14, 2016 #14

    fresh_42

    Staff: Mentor

    Nobody is at the speed he would like to be. Maybe you should take some things for granted while you are reading stuff, by which I mainly mean the examples at the far end of a definition. These are things one could perfectly think about on a walk through the park or similar.
    E.g. There is a good reason why units (invertible elements, e.g. ##1##) are not allowed to be called prime. But while learning them and the theorems that come along, it is not really necessary to investigate these reasons. One might as well take it simply as part of the definition.
     
  16. Oct 15, 2016 #15

    jbriggs444

    User Avatar
    Science Advisor

    "Subset of" and "element of" are different things. "Element of" is more basic.

    An item (which may or may not be a set) is an element of a set if it is listed as a member of that set.

    A set is a subset of another set if every one of its elements are also elements of the other set. If there are no elements then the "every one" is vacuously satisfied. That is how the empty set can be a subset of every set.

    {1,2,3} is not an element of {1,2,3,4} because it is not listed as a member.
    {1,2,3} is a subset of {1,2,3,4} because each of its members is on the list.

    {1,2,3} is an element of { {1,2,3}, Frankenstein's monster } because it is listed as a member.
    {1,2,3} is not a subset of { {1,2,3}, Frankenstein's monster } because its first member (1) is not equal to {1,2,3} and is also not equal to Frankenstein's monster.
     
    Last edited: Oct 15, 2016
  17. Oct 15, 2016 #16

    fresh_42

    Staff: Mentor

    Just to complete confusion:
    ... but {{1,2,3}} is a subset of { {1,2,3}, Frankenstein's monster }.
     
  18. Oct 15, 2016 #17
    Agreed. I just read the BOOK OF LOGIC..It is very good and has cleared out these dounts. We have to be very careful with our langauge.

    Yes {1,2,3} is not an element of {1,2,3,4}. But {1,2,3} is a subset of {1,2,3,4}.

    1 is not a subset OR an element of {N} but it is an element of N! Sets must always be enclosed by brackets, if defined in any other way than CAPITAL letters. If anoyne is confused pleased read the book of logic. :)

    Sets withing a set like {1 {2,3}} are counted and treated as single elements! {2,3} IS NOT a subset of {1, {2.3}} it is just an element.

    The empty set is never an element of another set unless mentioned explicitly, but by definition it is a subset of all sets.

    Some other examples to think about for others (I hope confused like me :P)
    {Z} is not a subset or an element of Z.
    Z is an element of {Z} but not a subset.
    {{3,4}} is not an element of {1,2,{3,4}}

    evXIr98.jpg
     
    Last edited: Oct 15, 2016
  19. Oct 15, 2016 #18
    To say that {{1}} is an element of {1, {1}} is wrong as it equivalent to saying that the ENTIRE
    box
    that contains a smaller box with the number one is also contained in a box that contains 1 and a smaller box of one. An element relation is never specified with brackets unless the element is a set and for some reason you cannot jsut use a large capital letter to denote it, and then it has to be the right amount of brackets otherwise it means something else completely.

    {1} is an element of {1. {1}}

    {{1}} is an element of {1,{{1}}}
     
  20. Oct 15, 2016 #19

    fresh_42

    Staff: Mentor

    I suggest the following way out: simply use "is in" :wink:
     
  21. Oct 15, 2016 #20
    some books and websites are just not good enough..I think many assume too much about the intelligence and deductive capabilities of students! The book of logic and physicsforums

    are very good sources though. Very thorough.
     
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