I have:

  • Axiom of existence: there exists a set with no elements
  • Axiom of extensionality (equality basically?): if two sets have the same elements they are identical
  • Lemma showing the empty set is unique
  • Axiom of schema of comprehension: $\{x\in A|P(x)\}$ exists where P is a property.
  • Lemma showing the set given in the comprehension axiom is unique (justifying my notation)
  • Axiom of a pair: for any A, B there is a set C such that $x\in C\iff[x\in A\text{ or }x\in B]$

The book then goes on to say "We introduce the notation {A,B} for the unordered pair of A and B, in particular if A=B we write {A} instead of {A,A}"

Is this suggesting that 1 element sets exist? Can I justify that with comprehension?

Or is it simply just notation "We're lazy and don't want to write {A,A}" (I'm hoping for this)

I ask because I'm trying to be rigorous, so far I am happy that: sets containing nothing exist, I can have pairs (but I must not think sets containing two things, because {A} is just one thing) right?

I do not like that pair means "set containing two things"


3 Answers 3


This is how I like to think about it. The first remark uses the axiom of extensionality.

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[Sorry for the picture; I prefer to typeset in my LateX editor with \newcommands + autocompletion]

  • $\begingroup$ I like this, so the answer is "I should proceed, and not let the fact I have the numbers 0 and 2 get to me" right? OOOHhh! My axioms say nothing about unique things in a set - that's my prior knowledge polluting me (I also can't count elements yet) - thanks! $\endgroup$
    – Alec Teal
    Commented Mar 1, 2015 at 3:17
  • $\begingroup$ You are right that numbers don't come until much later (at least until you introduce the axiom of infinity). I won't say anything more about that. I don't want to spoil the logical journey too much. I enjoyed it. $\endgroup$ Commented Mar 1, 2015 at 3:27
  • $\begingroup$ But just to confirm. I should totally ignore the fact "pair" means 2 and "singleton" means zero and write {A} knowing I am writing {A,A} and that there is no such thing as {A,A,A} or {A,B,C} YET right? $\endgroup$
    – Alec Teal
    Commented Mar 1, 2015 at 3:33
  • 1
    $\begingroup$ To be pedantic, "unordered pair" and "singleton" are English phrases that talk about already proven unique sets. We could have used the English phrase "unordered two-tuple" and "unordered one-tuple." Also, "$\{ A,B \}$" is short-hand notation for 'English phrases' (This of course depends if you have a formal language in place or not; but I find that logic comes after Set Theory. What I mean is that Set Theory can associate specific sets to symbols such as "$\neg$" and "$\wedge$." This ends up being a Model within Set Theory) $\endgroup$ Commented Mar 1, 2015 at 3:38
  • $\begingroup$ To answer your question, you are writing "$\{A,A\}$" and there is no such thing as "$\{ A,A,A \}$" or "$\{ A,B,C \}$" UNTIL you assign to each of them an 'English phrase' which then specifies a unique set that you're referring to. (you could skip the 'English phrase,' but then you're walking into Russell's Principia Mathematica purely symbolic approach which Godel proved is of no use by his Incompleteness Theorems; Anyway, mathematicians allow themselves 'meta-languages' such as English whereas Logicians do not) $\endgroup$ Commented Mar 1, 2015 at 3:50

Your axiom of a pair is slightly incorrect. The correct statement is (colloquially) "for any sets $A$ and $B$, $\{A,B\}$ is a set."

Using this, $\{\emptyset\}$ is a non-empty set.

  • $\begingroup$ To say that, you also need from logic the statement that $(P \text{ or } P) \Leftrightarrow P$ $\endgroup$ Commented Mar 1, 2015 at 3:11
  • $\begingroup$ It does say that, it's the same thing. Yes from what I have $\{\emptyset,\emptyset\}$ exists. But does $\{\emptyset\}$ exist? Without saying "rather than writing it twice, lets write it once" - I do not like how now I have the numbers 0 and 2. $\endgroup$
    – Alec Teal
    Commented Mar 1, 2015 at 3:15
  • $\begingroup$ Well, by the axiom of extensionality, $\{\emptyset, \emptyset\} = \{\emptyset\}$, since they have the same elements. $\endgroup$ Commented Mar 1, 2015 at 3:21
  • $\begingroup$ @WilliamStagner that second bit is what I'm not happy with, {A} is just a shorthand for {A,A} - I have no concept of a "1 element set" - YET $\endgroup$
    – Alec Teal
    Commented Mar 1, 2015 at 3:28

Since $A$ in the axiom of schema of comprehension, and $A$ and $B$ in the axiom of pair, are sets, nothing in the 5 axioms and lemmas you provided says that there exists a set other than the empty set (that is, pair axiom as you have stated it does not give you a set containing two elements).

So it is no surprise that you can't see it logically implying existence of a set with one element.

  • $\begingroup$ Thanks, but from this axiom (of a pair) $\{\emptyset,\emptyset\}$ exists, which we would write $\{\emptyset\}$ - even though it's a pair $\endgroup$
    – Alec Teal
    Commented Mar 1, 2015 at 3:13
  • $\begingroup$ I'm not the downvoter by the way, I'm not sure yet (I've upvoted one answer). See my comment to it. I am happy that $\{\emptyset,\emptyset\}$ exists. However "members of a set must be distinct" is nowhere in my axioms. That's prior knowledge seeping in. $\endgroup$
    – Alec Teal
    Commented Mar 1, 2015 at 3:19

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