- #1

- 1,170

- 3

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

- Thread starter aaaa202
- Start date

- #1

- 1,170

- 3

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

- #2

pbuk

Science Advisor

Gold Member

- 2,041

- 813

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:

- #3

mathman

Science Advisor

- 7,901

- 460

By definition, the cardinality of any finite set is the number of elements.

- #4

pbuk

Science Advisor

Gold Member

- 2,041

- 813

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.By definition, the cardinality of any finite set is the number of elements.

If you define the cardinality of a set as the number of elements, how do you define the number of elements?

- #5

pbuk

Science Advisor

Gold Member

- 2,041

- 813

Oops, I managed to conflate two alternative definitions there:

- 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 \} ##
- 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:

- #6

- 1,170

- 3

- #7

- 269

- 24

- [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.

- #8

- 631

- 132

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.economicsnerd said:In our case, there is a bijection between [itex]\emptyset[/itex] and [itex]0=\emptyset[/itex], so we're good.

Last edited by a moderator:

- #9

pbuk

Science Advisor

Gold Member

- 2,041

- 813

You have just defined the cardinality of the empty set to be 0, there is no need to start looking for a bijection.There's a collection ofsetswhich are calledcardinal sets. I won't define a cardinal in general, but the finite ones are defined like:

- [itex]0:=\emptyset[/itex].

- #10

pwsnafu

Science Advisor

- 1,080

- 85

Why don't you try to prove or disprove it? There is only one function (the empty function) to check.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?

- #11

- 631

- 132

I have two questions:setswhich are calledcardinal 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.

1. Why doesn't $$1\cup \{1\} = \{\{\emptyset\}, \{\{\emptyset\}\} \}?$$

2. Given MrAnchovy's response, are you happy with the confusion you've sown?

- #12

- 1,999

- 282

This equation ALWAYS holds if A or B is the empty set, no matter how caridinality is defined for ANY setNeither; 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.

- #13

pbuk

Science Advisor

Gold Member

- 2,041

- 813

Oops - good point! How about...This equation ALWAYS holds if A or B is the empty set, no matter how caridinality is defined for ANY set

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

- 269

- 24

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]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?

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.

- Last Post

- Replies
- 6

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 914

- Last Post

- Replies
- 9

- Views
- 3K

- Last Post

- Replies
- 5

- Views
- 4K

- Last Post

- Replies
- 15

- Views
- 8K

- Last Post

- Replies
- 20

- Views
- 5K

- Last Post

- Replies
- 7

- Views
- 505

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 12

- Views
- 3K