5
$\begingroup$

The axiom of infinity is formulated as

$$\exists S ( \varnothing \in S \wedge (\forall x \in S) x \cup \{x\} \in S)$$

Can someone explain why the use of $\varnothing$ in the axiom of infinity makes sense, when the very existence of $\varnothing$ is predicated on it?

$\endgroup$
6
  • $\begingroup$ Existence of emptyset, in axiomatizations I have used, does not use the Axiom of Infinity. $\endgroup$ Nov 19, 2014 at 4:42
  • $\begingroup$ @AndréNicolas That's interesting. Are we talking about the usual axioms of ZFC? How would one prove the existence of the empty set without the axiom of infinity? $\endgroup$
    – user193756
    Nov 19, 2014 at 4:48
  • $\begingroup$ There are several axiomatiatins, all relatively mild variants of each other. One can prove (in any standard theory, set-theoretic or not) that there is an object. I think of it model-theoretically, the underlying set of any $L$-structure is non-empty. Then one can pick out the empty set using Separation by saying it consists of all objects not equal to themselves. $\endgroup$ Nov 19, 2014 at 4:56
  • $\begingroup$ When you say there are several axiomatizations, what do you mean? My book Set Theory by Jech and the Wikipedia page on ZFC says there are 9 axiom/axiom schemas which are extensionality, regularity, schema of specification, pairing, union, replacement, infinity, power set, and AC. Without the axiom of infinity, it does not follow that a single set exists. $\endgroup$
    – user193756
    Nov 19, 2014 at 5:01
  • $\begingroup$ Nvm. I found the answer. $\exists x (x = x)$ is apparently a theorem of first order logic, so it's not even needed as an axiom. $\endgroup$
    – user193756
    Nov 19, 2014 at 5:13

3 Answers 3

3
$\begingroup$

You don't need the axiom of the empty set for the axiom of infinity.

$\exists S(\exists x(x\in S\land\forall y(y\notin x)\land\forall z(z\in S\rightarrow\exists u(u\in S\land\forall w(w\in u\leftrightarrow w\in z\lor w=z))))$

The axiom states that there exists $S$ such that there is an element of $S$ which has no members, and $S$ is closed under successorship.

$\endgroup$
6
  • $\begingroup$ Why is it not $\forall w(w\in u\leftrightarrow (w\in z)\vee (w=z))$ ? I'm just trying to parse the last bit and I can't really see how $u\in z$ or $u=z$ implies that $w\in u$ for any set $w$. $\endgroup$
    – C Squared
    Jun 4, 2021 at 3:51
  • $\begingroup$ Yeah, that's definitely a typo. Thanks. $\endgroup$
    – Asaf Karagila
    Jun 4, 2021 at 9:16
  • $\begingroup$ You have made the same typo here as well. Was trying to derive some stuff about ordinals from ZFC axioms and came across these answers, which have helped a lot. $\endgroup$
    – C Squared
    Jun 4, 2021 at 9:22
  • $\begingroup$ I'm glad they helped, and thanks for the corrections. $\endgroup$
    – Asaf Karagila
    Jun 4, 2021 at 13:28
  • $\begingroup$ Just to confirm, no other axioms (for example replacement or separation) are required other than this version of axiom infinity to prove "there exists an empty set"? Then infinity + extensionality is enough to prove the existence of a unique empty set? $\endgroup$
    – Jagerber48
    Dec 19, 2022 at 3:55
0
$\begingroup$

The Axiom of Existence states that the empty set exists. If you don't accept the Axiom of Existence as axiomatic, the Axiom of Infinity implies the existence of $\varnothing$, though you need another axiom to "extract" it from $S$.

$\endgroup$
6
  • $\begingroup$ And which axiom is that? $\endgroup$
    – Asaf Karagila
    Nov 19, 2014 at 19:32
  • $\begingroup$ Axiom of Subsets works. $\endgroup$
    – GFauxPas
    Nov 19, 2014 at 20:00
  • $\begingroup$ Right. But since the axiom of infinity already tells us that there is a set such that one of its elements is empty, why do we need an additional axiom? If something is an element of another set, then it exists. $\endgroup$
    – Asaf Karagila
    Nov 19, 2014 at 20:04
  • $\begingroup$ "If something is an element of another set, then it exists." - I don't think you can state this without an axiom asserting it directly or indirectly. $\endgroup$
    – GFauxPas
    Nov 19, 2014 at 20:11
  • $\begingroup$ Let's go at it using quantifier instantiations. You agree that the axiom of infinity, as written in my answer on this page, asserts the existence of $S$ which is inductive and has an empty set as a member? $\endgroup$
    – Asaf Karagila
    Nov 19, 2014 at 20:14
0
$\begingroup$

The existence of the empty set can be proved from predicate calculus, the Rule of Generalization, and the Axiom of Separation; you don't need the Axiom of Infinity. You can find a formalized proof here:

http://us.metamath.org/mpegif/axnul.html

Of course in many treatments it is simply taken as an axiom.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .