"There is no set containing everything"?

Question

I was reading this question regarding codomains, and I found something interesting in User134824's answer:

"On the other hand, owing to the set-theoretic fact that "there is no set containing everything," it's not possible to pick a single universal codomain for functions."

Why is it impossible to have a set containing everything? Why can't we define $U=\mathbb{R} \cup \mathbb{C} \cup ....$(all possible sets)?

P.S. This is a soft-question, so I am looking for intuitive, non-technical answers; I do not know any set theory

We can have a universal set. However, because of the Russell paradox and related paradoxes, we have to alter some useful notions, such as Comprehension. The cost is for most mathematicians too high. In the most common formal set theory (ZFC), one can prove there is no universal set. — André Nicolas, Apr 17 '16 at 02:02
Sets have cardinality, and if $A$ is a set, then the power set of $A$ has larger cardinality (and is a bigger set) -- Cantor proved this. Thus, there would be a "set" bigger than the "set of everything". The way out of this contradiction is to realize that not every "collection" is a set. — Christopher Carl Heckman, Apr 17 '16 at 02:04
I don't think that $U$ is a set. What counts as a set varies from application to application. (After all, the Axiom of Infinity is required in order to show that ${1,2,3,\ldots}$ is a set. Finitists, who do not work with this axiom, would say that ${1,2,3,\ldots}$ is not a set.) — Christopher Carl Heckman, Apr 17 '16 at 02:11
http://math.stackexchange.com/questions/162/why-is-the-set-of-all-sets-a-paradox — MathIsNice1729, Apr 17 '16 at 02:15
A set that contains everything would have to contain itself, which from an intuitive viewpoint seems impossible. Intuitively, you can place many things inside a bag, but one thing you can never place inside a bag is the bag itself. — littleO, Apr 17 '16 at 03:12
What do the three dots at the end of your expression mean? It would seem that it means "running through the collection of all possible sets". But that begs the question: You can't define "the set of everything" by invoking the idea of "the set of all possible sets", because those two ideas are identical. — mweiss, Apr 17 '16 at 03:28
Without the Axiom of Foundation (which has other names) it is consistent that a set can belong to itself. — DanielWainfleet, Apr 17 '16 at 04:06

score 4 · Answer 1 · answered Apr 17 '16 at 02:04

4

With your $U$, then $\mathcal{P}(U)\subset U$ but, $\# \mathcal{P}(U)=2^{\#U}$, absurd

answered Apr 17 '16 at 02:04

Martín Vacas Vignolo

4,710

2

Strictly speaking, this isn't true - in set theories like $NFU$ where there is a universal set, we have $\mathcal{P}(U)$ is strictly smaller than $U$. Roughly speaking, this is because the map $x\mapsto {x}$ doesn't exist! – Noah Schweber Apr 17 '16 at 03:22
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@NoahSchweber: Specifically, that happens in $\mathsf{NFU+Choice}$. $\mathsf{NF}$ just proves $\mathcal{P}(U)=U$, which is slightly less odd looking. – Malice Vidrine Apr 17 '16 at 03:25
@MaliceVidrine Yes, I meant NFU. Is Choice needed for this, though? Just the existence of urelements is enough, I think . . . – Noah Schweber Apr 17 '16 at 03:27
@Noah: I overstated the case; at least as of the last revision of Elementary Set Theory with a Universal Set, Holmes was not aware of any proof that did not use Choice, but comments that it is unknown whether you can get that result without Choice. I believe last time I spoke to him this hadn't changed, but I wouldn't put much stock in my memory.... :P – Malice Vidrine Apr 17 '16 at 03:35
@MaliceVidrine Hang on, I'm confused. Clearly as soon as there is a single urelement, $U$ properly contains $\mathcal{P}(U)$ (since every urelement is in $U$ but not $\mathcal{P}(U)$), so I don't see how choice even enters the picture. Or was Holmes using it to prove a stronger statement? – Noah Schweber Apr 17 '16 at 05:05
@NoahSchweber: $U$ properly containing $\mathcal{P}(U)$ is a different statement than $|\mathcal{P}(U)| < |U|$. The former is definitely true with urelements even without choice, but the latter doesn't obviously follow without Choice. – Malice Vidrine Apr 17 '16 at 05:23
@MaliceVidrine Doy! I'm tired. – Noah Schweber Apr 17 '16 at 05:25

score 2 · Answer 2 · edited Apr 17 '16 at 04:13

The problem via Cantor's paradox has already been noted. It is also the case that the most common set theories prove the existence of "the set of all $x\in A$ such that $x\notin x$". If $A$ is the universe, then there is a set $R$ containing every set that is not a member of itself; but $R\in R \iff R\notin R$, which is a paradox (Russell's, specifically).

More trivially, common set theories accept the Axiom of Foundation, which implies that no set can be a member of itself. But a set containing every set must have itself as a member.

There are, as someone mentions in the post linked to in the comments, consistent set theories with universal sets, but these theories must reject each of Foundation, the existence of $\{x: x\in A \wedge x\notin x\}$ for all $A$, and Cantor's theorem that $A < \mathcal{P}(A)$. The consequences of axiom systems that disprove these in favor of the existence of a universe can be counterintuitive or cumbersome; Currying a binary function might not work in $\mathsf{NFU}$, or complementation might not work in $\mathsf{GPK}$.

"There is no set containing everything"?

2 Answers2