The set of all things. A thing itself?

If the universe is the set of all things. Does it contain itself? In other words is it a thing itself?

I know its a stupid question, but it really grinds my gears.


Edit 8.12

Okey, someone here said that it cant exist. So what if it would be a proper class does, that change anything? And when I say “thing” I mean to be distinguishable.

Hope I didnt confuse someone even more!

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Typically the way you define this is you let $\Omega$ be the set of all elementary things, and then you pick from the set of all subsets of $\Omega$ (a.k.a. the power set of $\Omega$, denoted either as $2^\Omega$ or $\mathcal{P}(\Omega)$.

Then, $\Omega \in 2^\Omega$ and $2^\Omega \not \in 2^\Omega$, so this setup causes no casuistic problems.

However in your case, defining $\Omega$ as the set of all things you must first define is a set of things is a thing itself or not.

Most set theories I’m aware of have no problems with a set of urelements of arbitrary size (provided the theory permits urelements, of course). So taking urelements to represent the physical objects of the universe, we could easily have a set of them. Whether the physical universe is among them depends on, say, the merological structure you suppose the universe to have.

If you want to include sets and non-sets together, then there’s only two theories that seem to me to be candidates. Here I’m assuming that we want our set theoretic universe to still be quite mathematically expressive. The first, and mathematically most satisfactory, is NFU. I recommend reading Randall Holmes’s Elementary Set Theory with a Universal Set (available on his home page) to see if this sounds like a plausible mathermatical universe. The odd snag is that NFU+Infinity+Choice has more urelements than sets.

You can avoid that outcome of NFU by taking NF+”there’s a set of Quine atoms of cardinality $\kappa$” and take Quine atoms as representatives of the non-set universe. The caveat here is that such a set has to be quite small; any set of urelements of a cardinality that’s some finite number of power set applications away from the cardinality of the universe will allow one to prove Cantor’s paradox. Though you could have a countably infinite set of Quine atoms with only desirable side effects, you would still have to deal with NF being inconsistent with Choice.

The moral here is that if you do have a set of everything, then you have to put odd conditions on what the universe of sets looks like. Whether those results are ontologically acceptable is a different, non-mathematical issue.

If you are talking about the physical universe (and why wouldn’t you be, come to think of it), you might make the case that it is not an element of itself. This would work if a property of every set was that it is not a “physical object” however you may define it, and if the physical universe was the set of all physical objects. Likewise, it could be shown that the set of all non-physical objects is an element of itself.

Suppose that all sets are not physical objects.

$\forall x: [Set(x)\implies \neg P(x)]$

where $Set$ is the “is a set” predicate, and $P$ is the “is a physical object” predicate.

Suppose further that the physical universe is the set $U$ of all physical objects.

$Set(U)$ and $\forall x: [x\in U\iff P(x)]$

If such a set actually exists, we would have $\neg P(U)$ and, therefore, $U\notin U$.

Now, suppose that $U’$ is the set of all non-physical objects.

$Set(U’)$ and $\forall x: [x\in U’\iff \neg P(x)]$

If such a set actually exists, we would have $\neg P(U’)$ and, therefore, $U’\in U’$.


The trouble is, $U\cup U’$ is just the usual universal set of all things that, as I have shown in my previous answer, cannot exist in my system (or in ZFC). So, either $U$ or $U’$ or both cannot exist. Another paradox?

The set of all things (physical or abstract) cannot exist. Put another way, every set must exclude something.

Suppose the set $U$ of all things did exist.

$\forall a: a\in U$

Then, there would have to exist a subset $r$ of $U$ that is the set of all things that are not elements of themselves (the so-called Russell set).

$\forall a:[a\in r \iff a\in U \land a\notin a]$

In the manner of Russell’s Paradox, since $r\in U$, we could then obtain the the contradiction

$r\in r \iff r\notin r$

Therefore, $U$ as defined here cannot exist.