# The notion of injectivity is undefined on the empty set function?

1. Mar 10, 2012

### julypraise

When defining the notion of injectivity of functions it uses notation, for example for a function $f$, $f(a_{1})=f(a_{2})$ where $a_{1},a_{2}$ are in the domain of the function. Since the empty set function, i.e., $\emptyset \subseteq \emptyset \times A$ for some set $A$, has the empty domain, the notation $\emptyset (a)$ is undefined, therefore the notion of injectivity is undefined. The reason why I'm concerned with this is that most textbooks when defining the notion of injectivity of functions do not specify the domains of the functions. Let me summarize my question:

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

2. Mar 10, 2012

### julypraise

3. Mar 10, 2012

### A. Bahat

The condition for injectivity:
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. Mar 10, 2012

### julypraise

@A. Bahat

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. Mar 10, 2012

### micromass

Look at it this way: a function is NOT injective iff there exists $a\neq b$, such that $f(a)\neq f(b)$.
This can clearly not be satisfied since there cannot exist such a and b.

6. Mar 10, 2012

### julypraise

@micromass

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. Mar 10, 2012

### micromass

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. Mar 10, 2012

### julypraise

@micromass

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

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. Mar 10, 2012

### micromass

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

10. Mar 10, 2012

### mathwonk

here;s one i find less obvious, the product of all numbers in the empty index set is equal to 1.

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. Mar 11, 2012

What does ∅×A even mean?
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.

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