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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.

Thanks!

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!

Questboy
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  • Hi, Questboy, welcome to Math.SE – gt6989b Dec 05 '13 at 18:06
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    Related: http://math.stackexchange.com/questions/162/why-is-the-set-of-all-sets-a-paradox?rq=1 – Spenser Dec 05 '13 at 18:07
  • Do you mean this universe? http://en.wikipedia.org/wiki/Universe_%28mathematics%29 Otherwise this has very little to do with mathematics – Najib Idrissi Dec 05 '13 at 18:07
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    This isn't really a question about mathematics. It might be a better fit at http://philosophy.stackexchange.com. –  Dec 05 '13 at 18:09
  • Yeah should have paid more attention, that this would fit somewhere else than mathematics. But thank you all. Real nice people here. – Questboy Dec 05 '13 at 19:25

4 Answers4

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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.

gt6989b
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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.

Malice Vidrine
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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, $U'\in U'$.

EDIT:

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?

Dan Christensen
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  • It's trivially true that not both can exist in ZFC because Russel's paradox would disprove the existence of $U'$, since it would be a set containing all sets... ZFC+urelements has very to say about $U$. – Malice Vidrine Dec 09 '13 at 21:56
  • I thought it was clear that I wasn't assuming, as ZFC does, that all things are sets. – Dan Christensen Dec 10 '13 at 04:47
  • Yes, that's clear. $U'$ doesn't need to contain only sets to yield Russel's paradox, it just needs to contain all of them. I'm unsure what you're trying to get at in your edit that's "paradoxical" for reasons besides the usual Russel's paradox. – Malice Vidrine Dec 10 '13 at 04:54
  • Good point. The existence of a set of all sets (not assuming all things are sets) also leads to a contradiction in my system, and therefore, that set cannot exist. I am not immediately able to derive a contradiction from the existence of $U$, however. Any thoughts? – Dan Christensen Dec 10 '13 at 14:27
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    I suspect the set $U$ can be consistently assumed so long as there's no formula of the language that can establish a bijection between physical objects and sets (i.e. if there's no way for Replacement to get us in trouble). Not entirely sure how I'd check that, but I think it's a safe bet that it's okay. The conclusion that it wouldn't be a member of itself is solid, though. – Malice Vidrine Dec 10 '13 at 16:44
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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.

Dan Christensen
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    Actually there are several set theories that have universal sets. The point is that you can't have arbitrary separation and a universal set. Further, the argument above doesn't work as an argument that there can be no set of all physical things, say a set of all urelements; the subset of all urelements that aren't members of themselves is the set of all urelements, which wouldn't be a member of that set anyhow. – Malice Vidrine Dec 07 '13 at 18:23
  • @MaliceVidrine I would think that arbitrary separation (constructing subsets) would be far more useful to mathematicians than any supposed universal set, which, to my knowledge, never comes up in algebra, analysis or geometry. Also, I have introduced no notion of a set of all urelements. And not all urelements are physical things. So, I don't think your argument works. – Dan Christensen Dec 08 '13 at 03:52
  • You don't need all urelements to be physical things. The point is the Russel class of a set of urelements doesn't lead to paradox (any more than the Russel class of any set in ZF does), so using Russel's paradox as evidence that you can't have a set of all physical things is a non-starter.

    And the lack of utility of a universal set has no bearing on whether there are consistent theories that can assume it.

     – Malice Vidrine Dec 08 '13 at 11:25
  • @MaliceVidrine I use Russell's paradox to prove that there cannot exist a set of all things, not just physical objects. And how do you prove the existence of any function, say the identity function f(x)=x on a set, without constructing arbitrary subsets? Any set theory that doesn't allow this possibility would seem to be a bit of a non-starter to me. – Dan Christensen Dec 09 '13 at 03:28
  • You don't need arbitrary subsets to get identity functions (or lots of other things); you need the right (sub)sets. In NF you can define the identity function directly (and globally) as ${\langle x, x\rangle: x\in V}$. And it occurs to me I've been making a silly misreading of the beginning of your answer... – Malice Vidrine Dec 09 '13 at 10:09
  • It seems to me that arbitrary separation is somehow lurking in the background of your construction. Doesn't NF get around RP by simply banning self-membership? What if anything is gained by having a universal set? To my knowledge, it never comes up in ordinary mathematics (number theory, real analysis, etc.). – Dan Christensen Dec 10 '13 at 14:48
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    If arbitrary separation were available we would have Russel's paradox, and if self-membership were banned we wouldn't have a universal set. NF uses the schema of naive comprehension restricted to stratified formulae. So you can get many instances of separation, just not all of them. Where having a universal set is nice is making many constructions, especially of inductive structures, much more natural. Ordinary mathematics is besides the point; you don't need every corner of the ZFC universe for ordinary math either. – Malice Vidrine Dec 10 '13 at 15:19
  • Thanks. I like the part about constructing inductive structures. I have found them to be a real pain and anything but intuitive using arbitrary subsets in my system. But, apart from foundational work, it seldom comes up ordinary math. If I understand correctly, the notion of stratification would require the user to maintain a table of type-values (numbers?) for every variable at all times -- very unlike ordinary mathematics that students will find in textbooks. In practice, the type-value could even change if you reused a variable name in a long proof – Dan Christensen Dec 10 '13 at 18:03
  • NF is an untyped system, so while you need some temporary notion of type to tell you if you can assume an instance of comprehension, once you have the set you can ignore type. For instance, $x\cup{x}$ exists for all $x$ because the defining conditions for ${x}$ and for $x\cup y$ are stratified; but the axioms do not give you ${y:\exists x(y=x\cup{x})}$. On the other hand, if all the variables in a term abstract are bound, you can assign it any type in another abstract. I recommend Rosser's "Logic For Mathematicians" (which is pretty affordable) for more info. – Malice Vidrine Dec 10 '13 at 20:04