I heard this question and was intrigued. I think the answer is yes, since empty sets are not a problem for functions, but maybe I'm missing a border case somewhere.

Thanks in advance!

I heard this question and was intrigued. I think the answer is yes, since empty sets are not a problem for functions, but maybe I'm missing a border case somewhere.

Thanks in advance!

Submited on : | Fri, 9th of Nov 2018 - 05:34:05 AM |

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A legitimate function from X to Y must be able to deterministically assign one output in Y every time it is given an input from X. We can write this as ∀x∃y f(x) = y.

I believe X can be empty, since some sort of null function that doesn't do anything technically satisfies this criterion. A ∀ statement over an empty set is vacuously true. But if Y is empty, and X is not, it will fail since there are values of X that can't go anywhere. So there is no function from any nonempty set to the empty set.

That's it, thanks!

If a set B has k many members (and k can be an infinite cardinal) and A has l many members (again may be an infinite cardinal) then the number of functions from A to B depends only on k and l and that number is k^{l}. In fact, this is usually taken as the definition of cardinal exponentiation. So there exists a function from A to B if and only if k^{l} is not zero. From this we can see that there always exists a function from A to B unless A is not empty but B is (if they are both empty then the empty function - the function with nothing in its domain - is a function from A to B which is why 0^{0}=1 in cardinal arithmetic).

To see this more clearly note that if B is not empty then it must have some member, b, and so the constant function f defined by f(a)=b for all a in A is an example of a function from A to B.

Thanks a lot!

## ben1996123

there are no functions f : {x} -> {}

## PM-me-math-riddles

Oh that was what I was missing. The domain can be empty, but the codomain can't be if the domain is empty. Thanks!

## Nathanfenner

Note that there

isone function from {} to {}. It doesn't do much.## PM-me-math-riddles

Yeah, I thought about the empty function when I said that empty sets are not a problem for functions. I somehow didn't think about the case where

oneof the sets was empty.## [deleted]

But doesn't every set contain {}?

Every set contains the null set.

## ben1996123

no, every set has {} as a subset

## [deleted]

What's the difference?

Edit:

I found the answer:

https://math.stackexchange.com/questions/1103664/is-empty-set-element-of-every-set-if-it-is-subset-of-every-set

## Number154

The set of all even natural numbers is a subset of the set of all natural numbers, because every even natural number is a natural number. The set of all even natural numbers is not a member of the set of all natural numbers, because the set of all even natural numbers is not a natural number.

## PM-me-math-riddles

In simple terms, a set is a collection of objects. A subset of a set A is any set you can form using elements of A. You can form an empty subset from any set, since you can just not pick anything. But it doesn't mean that the empty set is an object of every set.

## [deleted]

Very good explanation.

## cluedit

A function is a relation that maps every point of the first set to exactly one point in the second set. Since {x}×∅=∅ it follows that the only relation we have from {x} to ∅ is the empty set itself, and that relation can't be a function since it does not map x to anything.