# Math Help - The Set of all Sets

1. ## The Set of all Sets

Hi, today in class my professor mentioned something about the set of all sets being a paradox. I did a little research and I'm not entirely sure why this is (or how they are proving it). Could someone please help me understand why it is a paradox. Thanks.

2. ## Re: The Set of all Sets

Originally Posted by Assassin0071
Hi, today in class my professor mentioned something about the set of all sets being a paradox. I did a little research and I'm not entirely sure why this is (or how they are proving it). Could someone please help me understand why it is a paradox. Thanks.
This is known as Russell's Paradox.
Russell famously wrote to Frege with one question "Is a set of teaspoons a teaspoon?"
Frege it is said instantly realized that two volumes of his set theory was founded on a paradox.

3. ## Re: The Set of all Sets

I can kind of understand it for the examples they give, where they have the set of all sets that are not members of themselves and the set of all sets that are normal. What I don't get is that how could you show that the set of all sets (in general, with no condition like the examples) is a paradox?

4. ## Re: The Set of all Sets

it's not quite Russell's paradox, but it quickly leads to it.

the problem is this: if the set of all sets is actually a set, then since it contains ALL sets, one of its elements is itself. this leads one to consider two kinds of sets:

a) those that contain themselves as a member
b) those that do not

which category does the set of all sets that do not contain themselves as a member belong to?

in fact, in "modern" set theory, the set of all sets is prohibited by the axiom of foundation (or axiom of regularity) which says (i am paraphrasing, here):

every non-empty set A contains an element B disjoint from A.

since the set {A} has only the single element A, it must be that A and {A} are disjoint, so A is not an element of A (or else A and {A} would have the common element A).

the axiom of regularity prevents "infinitely nested descending sets"..."the buck stops somewhere".

however, the collection of all "well-defined sets" (whatever THAT may mean) seems to be somewhat "set-like": it has elements and we can perform set operations on its elements. so to keep set theory "in the family" so to speak, a man named Alexander Grothendieck came up with a compromise: a special sort of set called a "universe" in which the sets we use are deemed to live. intuitiviely, a grothendieck universe is a set "big enough to have the sets we need in it". elements of a grothendieck universe are called "small sets", for usually only when considering the consistency of set theory as a whole do we actually need to consider EVERY set.

another approach is to consider collections of sets that are "too big" to be sets as another, similar kind of object: a proper class. there is even a kind of entity that is too big to be considered a class (if my memory serves me correctly, these are called "ensembles"). once can continue indefinitely in this manner, creating a hierarchy of objects that subsume the lesser kinds. one obtains a bewildering array of objects of dizzying size (not just infinite, but several kinds of uncountably infinite).

it really is a can of worms, and the hope is: that as long as we stay within "objects that satisfy the axioms of set theory" (bona-fide sets), we're "safe", but even this is not known for sure.

5. ## Re: The Set of all Sets

Originally Posted by Assassin0071
I can kind of understand it for the examples they give, where they have the set of all sets that are not members of themselves and the set of all sets that are normal. What I don't get is that how could you show that the set of all sets (in general, with no condition like the examples) is a paradox?
Paul Halmos in his little book Naive Set Theory puts it this way:
Nothing contains everything.
Suppose $G$ is the set of all sets that are not members of themselves.
$X\in G\text{ if and only if }X\notin X$.
Question does $G\in G~?$
If $G\in G$ then by definition $G\notin G$.
If $G\notin G$ then by definition $G\in G$.
So $G$ cannot exits.

6. ## Re: The Set of all Sets

Originally Posted by Plato
Paul Halmos in his little book Naive Set Theory puts it this way:
Nothing contains everything.
Suppose $G$ is the set of all sets that are not members of themselves.
$X\in G\text{ if and only if }X\notin X$.
Question does $G\in G~?$
If $G\in G$ then by definition $G\notin G$.
If $G\notin G$ then by definition $G\in G$.
So $G$ cannot exits.
o.o

so the null set is that russell set i've been looking for (i've noticed that all of the members of the null set have every conceivable property, and yet they don't, as well. for example it has all those integers that are both even and odd. how strange!).

the idea that "nothing contains everything" sounds suspiciously similar to the zen idea of the void: formless, empty, yet brimming with the potential to be "anything". of course, as soon as it becomes "something", its somewhat less than it was: actual things are only "partially" everything.

or, i might add with a sly grin, it seems that it may well be impossible to define mathematics itself, because every time we describe its contents, we have to append the description with a description OF the description. how inconvenient.

7. ## Re: The Set of all Sets

Thanks, I think this makes sense now. That part about the axioms definitely helped.

8. ## Re: The Set of all Sets

Plato, the book you mentioned, Naive Set Theory by Paul Halmos,...I took a look at it today and found the part you were talking about. It was very well phrased and it made the problem I was having make sense. Thanks for mentioning it.