I was wondering if the empty set is at most countable. Specifically, if it has to be defined in some sense like finite or countable or if it can be proved from the definition of countability what it is. Then I thought if I could be able to show that there is at least a surjective function between the natural numbers and the empty set. So I started thinking.

Let $\alpha$ be a function: $\alpha: \mathbb N \rightarrow \emptyset$.

Then we have that $\alpha(n) \notin Im(\alpha)$ for all $\ n \in \mathbb N$, where $\alpha(n)$ is the value of the function $\alpha$ at $n$, i. e., $\alpha(1)$, $\alpha(2)$ ... , $\alpha(n),$ ...

In this manner, the image of $\alpha$ would be empty. But so it is the empty set, an there are no distinct empty sets. Therefore $\alpha$ is surjective (because the codomain equals the image of $\alpha$). As the empty set has no elements, it is at most countable, or better, it is finite.

What you guys think of this proof?

Thank you everybody, now I believe I know where I went wrong, in thinking that there are such functions!

  • $\begingroup$ What is your definition of "at most countable"? $\endgroup$ – Eric Wofsey Mar 11 '18 at 23:44
  • $\begingroup$ Finite or countable. $\endgroup$ – Guilherme Ottoni Mar 11 '18 at 23:45
  • $\begingroup$ Well then, what is your definition of finite? $\endgroup$ – Eric Wofsey Mar 11 '18 at 23:46
  • $\begingroup$ There is no function $f:\mathbb{N}\to\emptyset$. In fact, there is no function $f:A\to\emptyset$ for any nonempty set $A$. $\endgroup$ – John Griffin Mar 11 '18 at 23:49
  • $\begingroup$ This is a good point. How to define finitude for the empty set. Sincerely, I do not know. The standart one would be put the set in one-to-one correspondence between the set {a : a between 0 and n - 1}, if there is such n then the former is said to be finite. But one can also think of the set that is countable but not infinite, therefore needing the previous definition. $\endgroup$ – Guilherme Ottoni Mar 11 '18 at 23:54

Your argument shows that if $\alpha:\mathbb{N}\to\emptyset$ is a function, then $\alpha$ is surjective. However, this does not prove that there is a surjection from $\mathbb{N}$ to $\emptyset$: what if there are no such functions at all? In fact, there aren't: what would $\alpha(1)$ be, for instance?

In any case, if your goal is to prove the empty set is at most countable "by the definition", then you need to use the definition. The definition of "at most countable" is not that there exists a surjection from $\mathbb{N}$ (though it is a theorem that if such a surjection exists, then your set is at most countable). You said your definition is just that a set is at most countable if it is either countable or finite. So you need to prove either that $\emptyset$ is countable or that it is finite. It's not countable (assuming that by "countable" you mean countably infinite), so you'll have to prove that it's finite.

Again, if you want to prove this, you'll have to use the definition of "finite". There are several different possible definitions, but the most common is that a set $S$ is finite if there is a bijection between $S$ and the set $\{m\in\mathbb{N}:m<n\}$ for some $n\in\mathbb{N}$. In the case $S=\emptyset$, it is equal to the set $\{m\in\mathbb{N}:m<0\}$, so you can take $n=0$ and use the identity map as your bijection.

  • $\begingroup$ Well, have you tried writing one down? I'll let you start simple: just tell me what you want to define $\alpha(1)$ to be. $\endgroup$ – Eric Wofsey Mar 12 '18 at 0:01
  • $\begingroup$ The ideia was really use that theorem you have mentioned, I misused the term... I am sorry... I am trying to write $\alpha(1)$ as a contradictory assertion. $\endgroup$ – Guilherme Ottoni Mar 12 '18 at 0:04
  • $\begingroup$ On the other hand the theorem is equivalent to the definition, so it can be used as a definition. I hope! $\endgroup$ – Guilherme Ottoni Mar 12 '18 at 0:09
  • $\begingroup$ A thing like this: $\alpha(1)= \{1 : 1=2\} $. Something like that. $\endgroup$ – Guilherme Ottoni Mar 12 '18 at 0:14
  • $\begingroup$ $\alpha(1)$ would be the empty set, so it is defined on the empty set. The same for the others $\alpha(n)$... Am I going to far away? $\endgroup$ – Guilherme Ottoni Mar 12 '18 at 0:21

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