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• 8.7k
Yet I am sure that a set that contains itself can be defined - I'm just not clever enough to think of it.

Not given the axiom of regularity and the axiom of pairing.
• 5.1k
set that contains itself seems like the Ouroboros making the last bite. How is that managed?

A set that has only the moon as a member is a distinct entity from the moon itself. A set is an abstract object.

A set is criteria, kind of like a club.

So the set of all non-penguins, NP, intuitively has itself as a member because NP is not a penguin.
• 218
I'm not seeing how you can "without X" and still have any X left - in terms of the notation.

It is not 'without X' it is 'without {X}' as a set. {X} is not the same as X, my bad notation in the beginning notwithstanding. X\{X} is every set in X but not the set {X} itself.

X\{X} = {{a}, {b}, {c},...} but not {X}, regardless of whether {X} can be a member of X.

Excluding {X} is not the same as excluding X.

The paradox asks if {X} is a member of X but I am disposing of the paradox by defining X as X\{X} so there is no contradiction.
• 1.2k
No, I am saying there are infinite collections of things that are not a set.

That's an interesting discussion there. Most of us here are non-mathematicians, and among mathematicians only a small fraction are working in or at least interested in foundations.

Let's say Set X = {{a}, {b}, {c},....}

If {X} is a member of X then

Set X = {{a}, {b}, {c},....{X}}

Your notation is confusing. If you want to say that a is a member of X (a ∈ X), you would write that as

X = {a, ...}

which is not the same as

X = {{a}, ...}

{a} is a singleton set with a as the sole member.
• 559
Yet I am sure that a set that contains itself can be defined - I'm just not clever enough to think of it

X={X}

:roll:

"...and among mathematicians only a small fraction are working in or at least interested in foundations." How true!
• 218
Your notation is confusing. If you want to say that a is a member of X (a ∈ X), you would write that as

X = {a, ...}

which is not the same as

X = {{a}, ...}

{a} is a singleton set with a as the sole member.

Yes, but X is a set of sets so X = {{a}, {b}, {c},...} but {a, b, c, ...} might be correct too as long as the logic of what I'm saying holds up. Link: https://truebeautyofmath.com/lesson-4-sets-of-sets/
• 4.4k
So the set of all non-penguins, NP, intuitively has itself as a member because NP is not a penguin.

Bingo! - that's what I was trying to think of!

Would you say this set is NP-complete?
• 5.1k
What does that mean?
• 218
If we define things as follows it might make it clearer-

a = {x}
b = {y}
c = {z}

Set X is the set of sets a, b, c so

Set X = {{x}, {y}, {z}}

If X is included

X = {{x}, {y}, {z}, {{x}, {y}, {z}}}

If X is not included

X = {{x}, {y}, {z}}

So X\X is {{x}, {y}, {z}} which is what I originally meant by X\X or X\{X}
• 1.2k
Yes, but X is a set of sets so X = {{a}, {b}, {c},...} but {a, b, c, ...} might be correct too as long as the logic of what I'm saying holds up.

I don't see what logic could imply that {{a}, {b}, {c},...} is the same as {a, b, c, ...}

You keep making the same mistake over and over again:

No!

X ≠ {X}

{X} is a set with one member: X

Set X = {{x}, {y}, {z}}

If X is included

X = {{x}, {y}, {z}, {{x}, {y}, {z}}}

No, that's not how it works.

X = {{x}, {y}, {z}}

X' = {{x}, {y}, {z}, {{x}, {y}, {z}}}

X ≠ X'

X ∈ X'

X' ∉ X'
• 4.4k
• 218
No!

Let 'All sets that do not contain themselves as members' be

a = {x}
b = {y}
c = {z}
d = ... and these sets go on for as long as is necessary, e, f, g, h,...

Set X = {{x}, {y}, {z},...}

Suppose for some set h, h = {X}

I am saying X = {{x}, {y}, {z},...}\h

That is, X = {{x}, {y}, {z},...}\{X}

There may be h such that h = {X} or there may not.

I am saying X\h regardless and this is the definition of X.

In simple language X = "All sets that do not include themselves as members, except {X}"

You seem to be assuming that {X} is included in X but by definition it is not.

Or suppose Set V = {{x}, {y}, {z},...{X}}

Set X = V\{X} and there you have it.
• 5.1k
Ah, tim wood does a math joke.
• 218
No, that's not how it works.

X = {{x}, {y}, {z}}

X' = {{x}, {y}, {z}, {{x}, {y}, {z}}}

X ≠ X'

X ∈ X'

X' ∉ X'

No, I am saying IF X is included in X then

X = {{x}, {y}, {z},... {{x}, {y}, {z}}}

But IF X is not included

X = {{x}, {y}, {z},...}

I am only saying this to clarify things. But by definition X is NOT included so

X = {{x}, {y}, {z},...}

Precisely X = {{x}, {y}, {z},...}\{{x}, {y}, {z}}

{x}, {y}, {z} and {{x}, {y}, {z}} are different sets so excluding the set {{x}, {y}, {z}} does not exclude {x}, {y}, or {z}
• 559
Would you say this set is NP-complete?

Create an algorithm to list all the penguins. Then take the complement in the Universal Set.
• 1.2k
Let 'All sets that do not contain themselves as members' be

a = {x}
b = {y}
c = {z}
d = ... and these sets go on for as long as is necessary, e, f, g, h,...

Why are they all singletons?
• 218
Ok, I'll put it this way. List the sets that are not members of themselves as

a1, a2, a3, ...

X is going to be {a1, a2, a3, ...}

But for some i,

ai = {a1, a2, a3, ...} = X

So the sets in question are-

a1, a2, a3, ...ai...

= a1, a2, a3, ...{a1, a2, a3, ...}...

So X is to be defined as {a1, a2, a3, ...}\ai

= X\{a1, a2, a3, ...} = X\X

That is, X is defined as not being a member of itself.

Don't worry about the notation. X is defined as not being a member of itself, that is all.
• 218
X ∈ X'

X' ∉ X'

By the way, is how do I type these set symbols? Latex? Is there a guide?

Edit: Found it. $\in$ Logic and Philosophy of Mathematics sub forum
• 1.2k
Those symbols are just Unicode characters that you can copy/paste from anywhere (e.g. the first Google hit for "set symbols"). But this site also supports Latex.
• 218
Those symbols are just Unicode characters that you can copy/paste from anywhere

Thanks. The first thread in the Logic and Philosophy of Mathematics forum (this forum) explains how to use the math tag and how to create symbols.
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