Explaining technicalities involving the empty set

[Uncanny]

TL;DR

I’m curious why and where it’s necessary to explicitly state whether or not a set must be excluded from potentially being the empty set.

For instance, I attached two problems in the the thumbnail below. I’m curious why A cannot be the empty set in 18b, but A is not excluded from being the empty set in 17a.
In 17a, if A is empty, then all the hypothesis can be satisfied (the composition will be empty too, obviously), but g need not be equal to h. Am I right? If so, why isn’t this addressed in some problems, but is in others?

 

[andrewkirk]

The answer is to do with logic, not set theory. In 18b, if A is empty, the 'for all ... if ... then ...' is what logicians call a vacuous truth, ie it says nothing useful, just as the vacuous truth 'If 1=2 then Lady Gaga is the pope' says nothing useful. Why is it vacuous? Because if A is empty there will be no functions g or h from A to B, so saying something is true for all such functions is saying nothing.

The same problem does not arise in 17a because there is no 'for all ...' statement.

Reading the linked article on vacuous truth should help make this clearer.

 

[Uncanny]

I understand what you're saying regarding vacuous truth, but isn't a function with domain being the empty set technically a function (the empty set)? If so, then if A is the empty set in problem 18a, wouldn't the corresponding statement similarly be "saying nothing?" Why isn't the restriction on A not included here?

 

[PeroK]

I understand what you're saying regarding vacuous truth, but isn't a function with domain being the empty set technically a function (the empty set)? If so, then if A is the empty set in problem 18a, wouldn't the corresponding statement similarly be "saying nothing?" Why isn't the restriction on A not included here?

If you allow $A$ to be the empty set in 18b, then you can find a counterexample to the stated result. You should try this.

For problem 17, if you allow $A$ to be the empty set, then you cannot construct a counterexample.

In other words, the results in 17 and 18a holds for empty $A$ but the result in 18b fails for empty $A$.

 

[Uncanny]

Yes, I think I noticed that for 18a. So is this a mistake in the problem statement (not restricting A to non-empty sets)?

 

[PeroK]

Yes, I think I noticed that for 18a. So is this a mistake in the problem statement (not restricting A to non-empty sets)?

Sorry, I got the numbers confused. I've edited now. The problems are right as they are.

 

[Uncanny]

Alright, gotcha. That makes sense now- so that’s the more practical reason the restriction is necessary: when, without it, the intended result(s) fails to hold?

 

[Uncanny]

17b) also requires A to be non-empty though, no? Otherwise, the definition of surjection can’t be satisfied.

 

[PeroK]

17b) also requires A to be non-empty though, no? Otherwise, the definition of surjection can’t be satisfied.

No. Let's analyse these scenarios with the assumtpion that $A$ is the empty set. This implies that $f$ is the empty function (if that's the right term).

17a) If $f: A \rightarrow B$ is onto, then $B$ must also be the empty set and $g, h$ must both be the empty function. The final condition and the result are trivial.

17b) In this case, $f$ and both compositions $f \circ g$ and $f \circ h$ are the empty function.

If $B$ is the empty set, then the condition is met trivially and the result holds. Since $g = h$ automatically and $f$ is onto.

If $B$ is not the empty set, then the premise cannot hold. I.e. $f \circ g = f \circ h$ for all functions $g, h$, even if $g \ne h$. There is no case to test for $f$ being onto. In other words, if $A$ is the empty set then you never have the case where:

$f \circ g = f \circ h \ \Rightarrow g = h$

So, you never have a case where $f$ must be onto.

 

[Uncanny]

Thanks! You hit the nail on the head with the case in 17b) in which B is non-empty, but A is. This is the case I was concerned with.

I agree with everything you’ve written, I suppose my “real” question, then, is: shouldn’t the problem statement avoid or warn against this outcome by including the additional premise that A must be non-empty (unless B is also empty)?

I understand this seems cumbersome and, perhaps, distracting to include, but the fact that A need be non-empty does arise (or is asserted) in the course of the proof, as long as the vacuous truth cases are omitted.

 

[PeroK]

Thanks! You hit the nail on the head with the case in 17b) in which B is non-empty, but A is. This is the case I was concerned with.

I agree with everything you’ve written, I suppose my “real” question, then, is: shouldn’t the problem statement avoid or warn against this outcome by including the additional premise that A must be non-empty (unless B is also empty)?

I understand this seems cumbersome and, perhaps, distracting to include, but the fact that A need be non-empty does arise (or is asserted) in the course of the proof, as long as the vacuous truth cases are omitted.

17b holds whether A is empty or not.

Let's take a different example. Suppose you had something like:

Let $a, b, c$ be numbers. Show that if $a = b^2$ then ...

You might be concerned about the case $a = 0$, where $b$ must also be zero.

What you're saying is that this needs to be changed to say:

Let $a, b, c$ be numbers where $a \ne 0$ unless $b \ne 0$. Show that if ...

This is an unnecessary condition. The relationship between $a$ and $b$ comes out of the other conditions.

The simple fact is, for 17b, unless you can construct a counterexample with $A$ as the empty set, then there is no reason to exclude it.

And, unless you have a counterexample, you really can't argue!

 

[Uncanny]

Ah, I think I understand now. If A is empty and B is non-empty, both directions of the biconditional result in vacuous truth. If we choose A empty, but not B, then f is not surjective, making the antecedent false. Analogous argument going in the other direction, per your reasoning in post #9. Have I “gotten it?”

 

[PeroK]

Ah, I think I understand now. If A is empty and B is non-empty, both directions of the biconditional result in vacuous truth. If we choose A empty, but not B, then f is not surjective, making the antecedent false. Similarly, going in the other direction, per your reasoning in post #9. Have I “gotten it?”

I think so, but I'm not that knowlegeable about the formal terminology for all this!