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Is 0 I am told. Is this an axiom, or can it be proven?
By definition, the cardinality of any finite set is the number of elements.
economicsnerd said:In our case, there is a bijection between [itex]\emptyset[/itex] and [itex]0=\emptyset[/itex], so we're good.
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].
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?
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.
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
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?