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Cardinality of empty set

  1. Sep 11, 2013 #1
    Is 0 I am told. Is this an axiom, or can it be proven?
     
  2. jcsd
  3. Sep 11, 2013 #2
    Neither; it is part of the definition of cardinality (the cardinality of the empty set is defined to be 0). If it was defined to be any other number, or left undefined, then (among other problems) the equality ## \mid A \cup B \mid = \mid A \mid + \mid B \mid - \mid A \cap B \mid ## would not hold if ## A ## or ## B ## is the empty set.
     
    Last edited: Sep 11, 2013
  4. Sep 11, 2013 #3

    mathman

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    By definition, the cardinality of any finite set is the number of elements.
     
  5. Sep 11, 2013 #4
    I thought that by definition the cardinality of any finite set that is not the empty set is the lowest ordinal number ## N ## such that there exists a bijection between the set and ## \{ 1, 2, ... N \} ##? This definition cannot be used for the empty set because no such bijection exists.

    If you define the cardinality of a set as the number of elements, how do you define the number of elements?
     
  6. Sep 11, 2013 #5
    Oops, I managed to conflate two alternative definitions there:
    1. the cardinality of any finite set that is not the empty set is the ordinal number ## N ## such that there exists a bijection between the set and ## \{ 1, 2, ... N \} ##
    2. the cardinality of any finite set that is not the empty set is the lowest ordinal number ## N ## such that there exists a injection of the set into ## \{ 1, 2, ... N \} ##
     
    Last edited: Sep 11, 2013
  7. Sep 12, 2013 #6
    Let me try to be more precise about what worries me about the cardinality of the empty set. A set X is countable if there exists an injection from X to N. So is the empty set countable? Well clearly if we define it to be 0, less than the cardinality of the empty set. But then in some exercise I did today, I used this property to conclude something about a set. When a definition is used in this way, what is then the difference between it being a definition and an axiom?
     
  8. Sep 12, 2013 #7
    There's a collection of sets which are called cardinal sets. I won't define a cardinal in general, but the finite ones are defined like:
    - [itex]0:=\emptyset[/itex]
    - [itex]1:= 0 \cup \{0\} = \{\emptyset\}[/itex]
    -[itex]2:= 1 \cup \{1\} = \{\emptyset, \{\emptyset\} \}[/itex]
    ...
    -[itex]k+1:=k\cup \{k\}[/itex]

    We name these things like numbers, but they're just sets like any other.

    By definition, a set [itex]A[/itex] has cardinality [itex]\kappa[/itex] if [itex]\kappa[/itex] is a cardinal set and there exists a bijection between [itex]A[/itex] and [itex]\kappa[/itex]. In our case, there is a bijection between [itex]\emptyset[/itex] and [itex]0=\emptyset[/itex], so we're good.
     
  9. Sep 12, 2013 #8

    Zafa Pi

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    Boy, have I heard people go round and round as to whether there is a bijection of [itex]\emptyset[/itex] to itself. You're correct, of course, but unless you are going to be very formal about your set theory, I think mathman has it right. Thus 0 is correct by definition.
     
    Last edited by a moderator: Sep 12, 2013
  10. Sep 12, 2013 #9
    You have just defined the cardinality of the empty set to be 0, there is no need to start looking for a bijection.
     
  11. Sep 12, 2013 #10

    pwsnafu

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    Why don't you try to prove or disprove it? There is only one function (the empty function) to check.
     
  12. Sep 12, 2013 #11

    Zafa Pi

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    I have two questions:
    1. Why doesn't $$1\cup \{1\} = \{\{\emptyset\}, \{\{\emptyset\}\} \}?$$
    2. Given MrAnchovy's response, are you happy with the confusion you've sown?
     
  13. Sep 12, 2013 #12
    This equation ALWAYS holds if A or B is the empty set, no matter how caridinality is defined for ANY set
     
  14. Sep 13, 2013 #13
    Oops - good point! How about...

    Neither; it is part of the definition of cardinality (the cardinality of the empty set is defined to be 0). If it was defined to be any other number, or left undefined, then (among other problems) the equality ## \mid A \cup B \mid = \mid A \mid + \mid B \mid ## for disjoint sets ## A, B ## would not hold if ## A ## or ## B ## is the empty set.
     
  15. Sep 13, 2013 #14
    1) If you'll let me be a bit pedantic... [itex]x\in 1\cup\{1\} \iff x \in 1 \text{ or } x \in \{1\} \iff x \in \{\emptyset\} \text{ or } x \in \{1\} \iff x=\emptyset \text{ or } x=1=\{\emptyset\} \iff x\in \{\emptyset, \{\emptyset\}\}[/itex]. Therefore, [itex]1\cup\{1\}=\{\emptyset, \{\emptyset\}\}[/itex]

    2) You're totally right. I thought I was clarifying things by explaining that "cardinality=0" and "cardinality=4" are in some sense more primitive definitions than "the cardinality of a set". It was my mistake.
     
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