# Is emptyset a function?

• Apr 4th 2011, 08:08 AM
MSUMathStdnt
Is emptyset a function?
My discrete book is defining a function, f, as a special type of relationship in which if both $(a,b) \in f$ and $(a,c) \in f$, then $b=c$ (and a relation is defined as a set of ordered pairs).

So, is the empty set not a function because it doesn't have any ordered pairs, or is it a function because it does not violate the definition of a function?

(1) Is it a function? If not, explain why.
(2) If yes, what are it's domain and range?
(3) Is the function one-to-one? If not, explain why.
(4) If yes, what is the inverse function?

f. $f=\emptyset$
(1) Yes (trivially), because there are no ordered pairs in $f$, it does not violate the definition of function.
(2) dom $f$ = im $f$ = $\emptyset$
(3) Yes (trivially), since the definition of one-to-one is not violated
(4) $f^{-1}=\emptyset$

Above is how I wrote up my homework (but it's not due until Thu.), but I'm not confident it's the right answer.
• Apr 4th 2011, 09:54 AM
MoeBlee
Quote:

Originally Posted by MSUMathStdnt
My discrete book is defining a function, f, as a special type of relationship in which if both $(a,b) \in f$ and $(a,c) \in f$, then $b=c$ (and a relation is defined as a set of ordered pairs).

So, is the empty set not a function because it doesn't have any ordered pairs, or is it a function because it does not violate the definition of a function?

I wouldn't put it as "not violating" but rather as "does fulfill."

When we say "S is a set of ordered pairs" we mean "For all x, if x in S, then x is an ordered pair." Well, the emptyset fulfills that requirement. For all x, if x is in the empty set then x is an ordered pair. So, in that very specific sense (which is the only sense that matters toward this question), yes, the empty set is a set of ordered pairs.

Quote:

Originally Posted by MSUMathStdnt
Above is how I wrote up my homework (but it's not due until Thu.), but I'm not confident it's the right answer.

All correct, except I would modify (1) as I mentioned above.
• Apr 4th 2011, 04:22 PM
MSUMathStdnt
Quote:

Originally Posted by MoeBlee
I wouldn't put it as "not violating" but rather as "does fulfill."

When we say "S is a set of ordered pairs" we mean "For all x, if x in S, then x is an ordered pair." Well, the emptyset fulfills that requirement. For all x, if x is in the empty set then x is an ordered pair. So, in that very specific sense (which is the only sense that matters toward this question), yes, the empty set is a set of ordered pairs.

All correct, except I would modify (1) as I mentioned above.

I understand what you're saying. But I still don't see how to word it (although I'll probably get full credit as long as I've got the idea right). How does this sound:

(1) Yes (trivially). There are no ordered pairs in $f$, therefore; for all ordered pairs in $f$, there are none that have the same first value and a different second value.
• Apr 4th 2011, 04:55 PM
TheChaz
This reminds me of a thread I saw recently.
http://mymathforum.com/viewtopic.php?f=22&t=18683
• Apr 5th 2011, 07:05 AM
MoeBlee
But you missed mentioning that every member of the empty set is an ordered pair.

I'll do it in English [where '0' stands for the empty set]:

For all x, if x is in 0 then, x is an ordered pair. So 0 is a relation. And for all x, y, z, if <x y> and <x z> are in 0 then y=z. So 0 is a relation that is moreover a function.

In symbols:

Ax(x in 0 -> x is an ordered pair).
So 0 is a relation.
Axyz((<x y> in 0 & <x z> in 0>) -> y=z).
So 0 is a function.

/

If you want to get more detailed, you can mention that

Ax(x in 0 -> x is an ordered pair)

Axyz((<x y> in 0 & <x z> in 0>) -> y=z)

are true because the antecedent in each is false.