Cardinality of empty set

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  • #1
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Is 0 I am told. Is this an axiom, or can it be proven?
 

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  • #2
pbuk
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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.
 
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  • #3
mathman
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By definition, the cardinality of any finite set is the number of elements.
 
  • #4
pbuk
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By definition, the cardinality of any finite set is the number of elements.
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?
 
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pbuk
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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 \} ##
 
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  • #6
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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?
 
  • #7
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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.
 
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economicsnerd said:
In our case, there is a bijection between [itex]\emptyset[/itex] and [itex]0=\emptyset[/itex], so we're good.
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.
 
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  • #9
pbuk
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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].
You have just defined the cardinality of the empty set to be 0, there is no need to start looking for a bijection.
 
  • #10
pwsnafu
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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?
Why don't you try to prove or disprove it? There is only one function (the empty function) to check.
 
  • #11
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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.
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?
 
  • #12
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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.
This equation ALWAYS holds if A or B is the empty set, no matter how caridinality is defined for ANY set
 
  • #13
pbuk
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This equation ALWAYS holds if A or B is the empty set, no matter how caridinality is defined for ANY set
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.
 
  • #14
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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?
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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