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\[
A f(\circ) B = C\equiv \forall x:(x \in A) \circ (x\in B) \iff x \in C
\]Hmmmmm

So \(f(\land) = \cap\).

so would set theory disallow this? Since it requires that there exist a set of all sets?

How did you go from operating on a set to a set of all sets.

Well, what would you say is the domain and codomain of \(\cap\)?

https://en.wikipedia.org/wiki/Intersection_%28set_theory%29

Hmmm

power set still has 'all sets' as a domain, codomain

ok I think I understand you are saying.
For intersection
U X U -> U
for union the same

Yep

but you are using the domain as U, not UXU ?

taking the cross product of such a large set can present its own problems

for a unary operation, you have U -> U

complement would be a unary operation

Why did it bother the early 20th century mathematicians so much, allowing a universal set.

Yeah, this is something we can do in programming, yet we can't do it in math. How bizarre!

you can allow self reference in programming, as you said with memory earlier?

yeah

what you described above with logic and set theory is whats called an isomorphism

You can have an array of arrays, where one of the elements is itself

with some qualifications , the logical *and* corresponds to intersection

can you find a logical operation that is equivalent to A subset B

Well \(\implies \) corresponds to \(\supseteq\)

with some tweaking it could work.
note that you need for all x for the subset condition

Here is something I found that is interesting: https://en.wikipedia.org/wiki/Type_theory

There is probably no way to avoid universe in some sense

wait, that isn't quite right...

The idea is \(f(+)\) corresponds to \(\sum\).

Yep, they are not the same \(f\).

Right now I'm doing something along the lines of: \[
g(+)A =\sum_{x \in A}x
\]

Like \[
g(\cap) A =y \equiv \left(\bigcup_{x\in A}x \right)= y
\]

an accumulator function

While \(f(\land) = \cap\) was more of a setwise function.

okay that looks valid, but your definition looks recursive

Eventually you will run out of elements to use and hit the base case of the empty set.

can you show me how it operates on \( \{1,2,3 \} \)
$$\Large f(\circ) \{ 1,2,3\}$$

Well, technically: \[
f(\circ)\{1,2,3\} \equiv 1\circ (2\circ (3\circ \text{idenity} ))
\]

okay so this is defining an associative operation
using a binary operation

Nope

It's like this: \[
f(+)\{1,2,3\} \equiv \sum_{k\in\{1,2,3\}}k = 1+ 2+ 3
\]