I'm new to set theory and came across this question in our notes. I don't know how to continue. The question is:

Build a model of a universe of sets where for each set $a$, the collection $\{a\}$ forms a set, while the axiom of pairing fails.

The axiom of pairing says:

The axiom of pairing states that for any two sets $a$ and $b$, there exists a set having $a$ and $b$ as its only elements. We write \begin{equation} \{a,b\} = \{x | [x=a]\vee [x=b]\} \end{equation} When $a=b$ we also write: \begin{equation} \{a\} = \{a,a\} \end{equation}

How can we show that there does not exist a set containing two other sets as its only elements? As soon as I try to place restrictions on either of these two sets, it gets invalidated by the $\vee$ because it's just a collection of all the elements of either sets. So even if we add a constraint on one set that, for example, says "this set cannot contain any elements from $b$", it doesn't matter because the $\vee$ basically ignores that. I don't know if that made sense...

Can anyone give me some tips on how to construct universes that break axioms in general?

Here's a list of all the axioms we know of (in class):

  1. Axiom of equality of sets (Axiom of extensionality)
  2. The empty set axiom
  3. The axiom of restricted comprehension
  4. The axiom of pairing
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    $\begingroup$ And should all other axioms of ZF(C) hold in that model? $\endgroup$ – Berci Jul 30 at 23:03
  • $\begingroup$ I don't think so. We haven't done many of the axioms yet. I'll add a list of all the axioms we know of in the question. $\endgroup$ – Imak Jul 30 at 23:06

I would guess by the wording of the problem that nothing needs to be satisfied other than what is written. One could start here by finding a simple, explicit universe where for any $a,$ $\{a\}$ always exists (so instead of trying to violate pairing, try to make the other condition succeed). Here is the simplest one that comes to mind: Start with an empty set $\varnothing$. Then add a set $\{\varnothing\}$ that just contains the empty set, then the set $\{\{\varnothing\}\}$ that just contains that, and so on. (In other words, just take the closure of $\varnothing$ under the operation $a\mapsto \{a\}.$) If this universe violates pairing, you're done, and it should be pretty clear that it does.

A more abstract example would be a universe $\{a,b\}$ where $a\ne b$ and $a=\{a\}$ and $b=\{b\}.$ (Alternatively, you could take $a=\{b\}$ and $b=\{a\}$.) These kinds of sets violate our intuition since if you try to 'unpack their contents' you wind up in an infinite loop, but they're legal unless you assume the axiom of foundation.

  • $\begingroup$ Thank you for your answer. I am, however, still a bit confused. You say that your first example violates pairing. In what way? Maybe I am confused by what the axiom does... So, from what I see, you are defining a universe to contain only the empty set and the sets (of sets...) thereof. So because of that, when we try and pair, say, $\{ \emptyset \}$ with $\emptyset$, we expect to find the set $\{\{ \emptyset \}, \emptyset \}$, but from our definition, this set does not exist in our universe? $\endgroup$ – Imak Jul 30 at 23:59
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    $\begingroup$ Yes, exactly. Our universe does not contain any set matching the description $\{\varnothing,\{\varnothing\}\},$ so it violates pairing. $\endgroup$ – spaceisdarkgreen Jul 31 at 0:01
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    $\begingroup$ @Imak regarding 2, it contains exactly two sets, $a$ and $b.$ But $a=\{a\}$ and $b=\{b\},$ so it contains $\{a\}$ and $\{b\}.$ $\endgroup$ – spaceisdarkgreen Jul 31 at 0:03
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    $\begingroup$ @Imak: a model can be anything you say it is. It is just a collection of sets. The hard part comes when it has to satisfy complicated axioms, like ZFC. Here you only have one axiom to satisfy, then one to fail to satisfy. $\endgroup$ – Ross Millikan Jul 31 at 0:13
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    $\begingroup$ @Imak No problem, the second example is pretty abstract and more could be said about how it "works" (the key is that membership is an abstract relation: we just consider two distinct objects each related to itself and only itself by the membership relation). The first example $\{\varnothing, \{\varnothing\}, \{\{\varnothing\}\},\ldots\}$ is a better example of the more intuitive universes we typically work with. $\endgroup$ – spaceisdarkgreen Jul 31 at 0:15

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