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The empty set is a function. Let A $\neq$ $\emptyset$. We know that the following is true:

0) $f$: $\emptyset$ $\rightarrow$ A is a function & $g$: A $\rightarrow$ $\emptyset$ is not a function.

We also know the following are true:

1) $g$ $\subseteq$ A $\times$ $\emptyset$ = $\emptyset$. So $g$ = $\emptyset$. Hence: A $\times$ $\emptyset$ = $g$.

2) $f$ $\subseteq$ $\emptyset$ $\times$ A = $\emptyset$. So $f$ = $\emptyset$. Hence: $\emptyset$ $\times$ A = $f$.

Some obvious consequences of this data: $f$ is a function, $g$ is not a function, $f$=$\emptyset$ & $f$=$g$ (by extensionality). This gives us:

3) $\emptyset$ is a function & $\emptyset$ is not a function (since $f$ and $g$ are the same object).

This is a contradiction.

Question: It's clear that $f$ vacuously satisfies the conditions required for a set to be a function and $g$ fails to satisfy those conditions. But since $f$ and $g$ just are the same set (by 1) & 2) they are the empty set), I'm having difficulty articulating why this doesn't lead to a contradiction. Where have I gone wrong?

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    $\begingroup$ The string of symbols "$g\!:A\to\emptyset$" makes no sense. You cannot talk of $g$ being empty because there is no such $g$. $\endgroup$ – Andrés E. Caicedo Aug 2 '18 at 19:58
  • $\begingroup$ I'm confused by this answer. I can't tell if you mean: 1) ''We can prove that no function has a nonempty domain and an empty range. $g$ was defined as denoting a function with this property but no such function exists. So the string of symbols '$g$: A $\rightarrow$ $\emptyset$' does not denote any set at all. So it cannot denote a set that is identical a set that can be known to exist (e.g. the empty set). A sentence asserting such an identity in the language of ZFC is one that expresses a falsehood.'' $\endgroup$ – CTH Aug 2 '18 at 22:11
  • $\begingroup$ or 2) ''The string of symbols '$g$: A $\rightarrow$ $\emptyset$' is like writing '} $\emptyset$'. There are a set of rules for determining whether an expression is well formed in the language of ZFC. This string of symbols violates those rules. So it is meaningless because it is not an expression that permissible in the language of ZFC. Consequently, it makes no sense to say that $g$ = $\emptyset$'' Did you mean 1), 2) or something else? If you meant 2), can you point me to where I find these rules? Thanks! $\endgroup$ – CTH Aug 2 '18 at 22:12
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You are mixing two "incompatible" conventions to talk about functions.

  • The set-theoretic construction of functions as certain sets of ordered pairs.
  • A function $f:A\to B$ as triple $(f,A,B)$ with domain $A$, codomain $B$ and mapping rule $f$.

If you decide to use the set-theoretic construction of functions, then the set $f$ of ordered pairs does not carry all the information that the notation $f:A\to B$ does. Given the set $f$, we might be able to extract the domain

$$\mathrm{dom}(f)=\{x\mid \exists y:(x,y)\in f\},$$

and we might be able to extract the range

$$\mathrm{range}(f)=\{y\mid \exists x:(x,y)\in f\},$$

but we cannot extract the codomain $B$ that you use in the notation $f:A\to B$. So, for example, the functions $\sin:\Bbb R\to[-1,1]$ and $\sin:\Bbb R\to \Bbb R$ are indistinguishable in the set-theoretic construction, as they are described by the same set, while they are usually considered different: one is surjective, the other one is not.

In the same way, only given the set, you might need more context to reconstruct the function. $f=\varnothing$ can stand for any function $f:\varnothing\to A$ for an arbitrary set $A$.

Conclusion: $\varnothing$ is only considered a function if the context says so, and then you need more information to pin it down uniquely.

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  • $\begingroup$ Ah. I didn't know this. This is an extremely helpful answer. Thanks! $\endgroup$ – CTH Aug 2 '18 at 22:31
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What you are missing is that "is a function" does not have any meaning in isolation (well, it can have a meaning, but not in the way you are using it here). What does have a meaning is "is a function $B\to C$" for two specific sets $B$ and $C$. So, the empty set is a function $\emptyset\to A$. It is not a function $A\to\emptyset$. This is no contradiction, because a function $\emptyset\to A$ is not the same thing as a function $A\to\emptyset$.


I cannot resist making an additional remark which is tangential to your question but is really important. It's not correct to write something like

$f$: $\emptyset$ $\rightarrow$ A is a function & $g$: A $\rightarrow$ $\emptyset$ is not a function.

This sentence has no meaning, because you have not introduced the variables $f$ and $g$ yet. This is a statement about two free variables $f$ and $g$, which may or may not be true for any particular values of $f$ and $g$. You furthermore cannot then go on to reason about $f$ and $g$ as you have done, since you still have not ever defined what these variables refer to!

What you want to say instead is something like:

There exists a function $\emptyset\to A$, namely the empty set. However there does not exist a function $A\to\emptyset$, and in particular the empty set is not a function $A\to\emptyset$, since there does not exist any function $A\to\emptyset$.

I recommend reading this nice blog post by Tim Gowers for more information about the proper use of variables.

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  • $\begingroup$ This is helpful and the post by Gowers is very useful. Have an upvote! $\endgroup$ – CTH Aug 2 '18 at 22:28
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When you say "$g:A \to \emptyset$ is not a function", you should pay close attention to what you mean. Importantly, you do not mean "no subset of $A \times \emptyset$ is a function". You mean "there does not exist a set $g \subseteq A \times \emptyset$ so that $g$ is a function with domain $A$". This is an important distinction, because the second version is true and the first version is not. Recall that a function is just a set of ordered pairs no two of which have identical first elements; $\emptyset$ trivially satisfies this definition.

Phrasing the rest of your argument equally carefully, here's what you're saying:

0) Let $f \subseteq \emptyset \times A$ be a function, and let $g \subseteq A \times \emptyset$ be not a function.

1) $f \subseteq \emptyset \times A = \emptyset$, so $f = \emptyset$.

2) $g \subseteq A \times \emptyset = \emptyset$, so $g = \emptyset$.

3) So $f = \emptyset = g$. $f$ is a function, but $g$ is not, so $\emptyset$ is both a function and not a function.

Now, there's one missing step: when you reach a contradiction at the end of a proof, you must conclude that one of your premises is wrong. So the proper conclusion here is

4) Either there is no function $f \subseteq \emptyset \times A$, or every $g \subseteq A \times \emptyset$ is a function.

Indeed, the latter is true: if $g \subseteq A \times \emptyset$, then $g = \emptyset$ (as you observed in 2) and therefore $g$ is trivially a function. It just doesn't have domain $A$, because you're correct that there are no functions $A \to \emptyset$.

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  • $\begingroup$ This is an excellent answer. Thank you. $\endgroup$ – CTH Aug 2 '18 at 22:33
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Let me summarize what others have written before me.

The statement $f\colon X\to Y$ is a statement about the set $f$. It says that $f$ is a function, its domain is $X$, and its range is a subset of $Y$.

Whereas $\varnothing$ is in fact a function, it is not a function whose domain is non-empty. Therefore $\varnothing\colon A\to\varnothing$ is a false statement, it makes no sense. On the other hand, $\varnothing\colon\varnothing\to A$ is just fine.

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  • $\begingroup$ This is a lovely summary of some of the previous posts. What a slick proof. Thanks! $\endgroup$ – CTH Aug 2 '18 at 22:40
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There is no $g: A\rightarrow \emptyset$ not even $\emptyset$ since there is no way to map every element in $A$ to nothoing.


Edit: May be I didn't explain it well. I think that the misunderstanding in the question is not related to functions. It is about logic.

When we say that there is no number in [1,1) we can write $\forall x\in [1,1), x \text{ is not a number}$ but we can't use it lately unless we prove that such $x$ does exist (i.e. $\exists x\in [1,1)$) which is not true.

The same thing here, we cannot declare $g$ as being element of an empty set then use it to prove a contradiction.

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