- #1

julypraise

- 110

- 0

Is the notion of injectivity undefined for the empty set function?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

In summary: How do you get from the fact that ∅×A is a subset of P(P(A)) to the absurd claim that ∅×A ⊆ P(P(A))?In summary, the notion of injectivity of functions is undefined for the empty set function. However, the statement that the empty set function is injective can be proven to be true.

- #1

julypraise

- 110

- 0

Is the notion of injectivity undefined for the empty set function?

Physics news on Phys.org

- #2

julypraise

- 110

- 0

http://en.wikipedia.org/wiki/Injective_function

...

- #3

A. Bahat

- 150

- 0

f(a)=f(b) implies a=b

is vacuously true when the domain is the empty set, because there are no a,b for this in the first place.

- #4

julypraise

- 110

- 0

Yes, I agree with you to the extenet that it is vacuously true. But the issue is that the f(a) in this case in undefined. Is it okay?

- #5

- 22,183

- 3,324

This can clearly not be satisfied since there cannot exist such a and b.

- #6

julypraise

- 110

- 0

Well, thanks.. I kinda get what you guys are trying to say; that the statement is true. But before talking about the truth of the statement, the statement itself seems a meaningless statement because in this case the notations f(a_1) and f(a_2) are meaningless. So in the logical point of view, the truth of this statement cannot be given. Isn't it?

- #7

- 22,183

- 3,324

julypraise said:

Well, thanks.. I kinda get what you guys are trying to say; that the statement is true. But before talking about the truth of the statement, the statement itself seems a meaningless statement because in this case the notations f(a_1) and f(a_2) are meaningless. So in the logical point of view, the truth of this statement cannot be given. Isn't it?

The notation is not meaningless. For every a in the empty set, the notation f(a) is defined. Indeed, there is no a in the empty set, so we don't have to define anything!

- #8

julypraise

- 110

- 0

Okay. I kinda get yout point, and it's kinda fresh. Thanks. Anyway, to be crystal clear, let me ask you about one more thing.

Definition. Suppose [itex]A,B[/itex] are set. Suppose [itex]\forall x \in A \exists ! y \in B ((a,b) \in f)[/itex]. Then it is true [itex]\forall x \in A \exists ! y \in B ((a,b) \in f \quad \mbox{and} \quad y=f(x))[/itex]

So is this way of defining things alright? If this is right, then as you said, f(x) is defined. Thus actually no problem on proving the injectivity of the empty function.

- #9

- 22,183

- 3,324

Yes, indeed. We write y=f(x) if exactly (x,y) is in f. S

- #10

mathwonk

Science Advisor

Homework Helper

- 11,774

- 2,041

This makes Euclid's proof of existence of infinitely many primes correct. I.e. he says if one has any finite set of primes, then to find another prime,take their product and add 1, and find a prime factor.

One cannot find a prime factor unless a number is greater than 1, and I used to think this was a gap in his proof, i.e. not producing at least one prime to begin, but even for the empty set of primes, the product plus one is 2!(The reason for this is not perhaps logically required, but it is a corollary of the property that products over index sets must be multiplicative for disjoint decompositions of the index set, hence for decomposing S, as S union empty, the empty product must be one.)

- #11

sponsoredwalk

- 533

- 5

Set-theoretically the Cartesian product of two sets is a subset of the power set of the power set of their union,

∅×A ⊆ P(P(∅∪A)) w/ {{x},{x,a}} = <x,a> ∊ ∅×A,

and a function is just a special kind of subset of ∅×A.

This is the way I would think about it:

An injective function maps to each element of the range a unique element of the domain.

A function is a special kind of relation.

A relation is a set of ordered pairs.

An ordered pair is an element of the power set of the power set of a set.

Thus for ∅×A:

An injective function maps each element of the domain to a unique element of the range.

A function is a special kind of relation.

A relation is a subset of ∅×A.

∅×A is a subset of P(P(∅∪A)), ∅×A ⊆ P(P(∅∪A)) w/ {{x},{x,a}} = <x,a> ∊ ∅×A.

Thus following the trail back far enough we find that for ∅×A to make sense we allow for the membership x ∊ ∅, which is he exact opposite of the definition of ∅. I don't know how you can even justify defining a relation, let alone a function, never mind injectivity.

Furthermore ∅∪A = A so ∅×A ⊆ P(P(A)) & the implications of this definition, if expanded upon via ordered pairs, leads to more of these problems.

So I don't see how such a concept is even defined as it stands.

A. Bahat said:

f(a)=f(b) implies a=b

is vacuously true when the domain is the empty set, because there are no a,b for this in the first place.

julypraise said:@A. Bahat

Yes, I agree with you to the extenet that it is vacuously true.

How can something be vacuously true for a function when you can't even define that function?

The empty set function is a mathematical function that takes in an empty set as its input and outputs an empty set. It is denoted as ∅ or {}.

A function is injective if each element in the domain of the function maps to a unique element in the range. This means that no two different inputs can result in the same output.

The empty set function does not have any elements in its domain, so there are no inputs to map to unique outputs. Therefore, the concept of injectivity does not apply to this function.

No, the empty set function cannot be injective because it does not have any elements in its domain. For a function to be injective, it must have at least one input that maps to a unique output.

The undefined injectivity of the empty set function means that it cannot be used in certain mathematical operations that require injectivity, such as finding the inverse of a function. This is because the empty set function does not have a well-defined inverse due to its lack of input-output relationships.

- Replies
- 2

- Views
- 2K

- Replies
- 19

- Views
- 3K

- Replies
- 4

- Views
- 2K

- Replies
- 18

- Views
- 2K

- Replies
- 15

- Views
- 2K

- Replies
- 7

- Views
- 1K

- Replies
- 5

- Views
- 1K

- Replies
- 9

- Views
- 2K

- Replies
- 1

- Views
- 2K

- Replies
- 27

- Views
- 3K

Share: