Set theory operations from logical operations

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Set theory operations from logical operations

Mathematics
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\[A \cap B = C\]is equivalent to\[k\in C \iff (k\in A) \wedge (k \in B)\]I don't even know what the question is but...
The set operations seem to mimic logical operations. \[ A\cap B = C \equiv \forall x:x\in A\land x\in B \iff x \in C \]In this sense \(\cap\) corresponds to \(\land\), \(\cup\) corresponds to \(\lor\), \(\subseteq\) corresponds to \(\implies \), \(\setminus \) is more of a combination of two operations, or as \(a \land \lnot b\), which is \(\lnot (a \implies b)\). I'm just sort of thinking aloud here.
It's not really a tutorial, it's more of a discussion. You can ask questions but there is no actual question.

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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?
Let \(B = \{ true, false\}\). We can say \(\circ\) is an operation on \(B\), but we can't say \(\cap\) is an operation/mapping because it acts on all sets, and you can't have 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 \(\cap\)?
I don't know if we can think of intersection as a function or an operation in the traditional sense. the power set i think is an operation
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
Yep Well maybe what we need is a new math concept, say "collection" , like a set, but less strict. Less strict to the sense that we can say the "collection" of all sets
Why did it bother the early 20th century mathematicians so much, allowing a universal set.
And a collection would nee the ability to let you know if something is in it or not, to iterate through it, have no duplicates?
I would rather have a system that is clear and intuitive , that may have a paradox, than a complicated system that I cannot use.
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
"In type theory, concepts like "and" and "or" can be encoded as types in the type theory itself." This could correspond to memory addresses
Here is some of the history behind type theory Types were created to prevent paradoxes, such as Russell's paradox. However, the motives that lead to those paradoxes – being able to say things about all types – still exist. So many type theories have a "universe type", which contains all other types.
There is probably no way to avoid universe in some sense
Here is another idea: \[ f(\circ)A = y \equiv \forall x\in A: x \circ \bigg(f(\circ) (A\setminus x)\bigg) = y \]
wait, that isn't quite right...
Here is another idea: \[ f(\circ)A = y \equiv \bigg( f(\circ)\emptyset = I_{\circ} \bigg) \land x\in A \implies x \circ \bigg(f(\circ) (A\setminus x)\bigg) = y \]
Hmm, what I'm trying to do is describe a map. It takes a binary operation with an identity element. For the empty set, it returns that element. For a non empty set, it removes any element x from the set, finds the result for the set minus x, then performs the operation on x and this result
The idea is \(f(+)\) corresponds to \(\sum\).
\[ f(\circ) = \begin{cases} \circ_{identity} & \emptyset \\ x\circ f(\circ) (A\setminus\{x\}) &\text{otherwise} \end{cases} \]I wonder if this is problematic in ZF set theory. Can you pick out an arbitrary element, and keep doing this until you have an empty set?
Also, for \(x\) to be an arbitrary element in \(A\), then I'm guessing this operation must be associative.
is your latter definition of \( f \circ \) different from your former definition above \( A f(\circ) B = C\equiv \forall x:(x \in A) \circ (x\in B) \iff x \in C\)
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
$$f(\circ) = \begin{cases} \circ_{identity} & \emptyset \\ x\circ \color{red}{f(\circ)} (A\setminus\{x\}) &\text{otherwise} \end{cases} $$
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 \( \{1,2,3 \} \) $$\Large f(\circ) \{ 1,2,3\}$$
\[ f(\circ) \{1,2,3\} \equiv 1 \circ(2\circ 3) = 1 \circ(3\circ 2) = 2\circ (1\circ 3) = 2\circ(3\circ 1) = \ldots \]
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: \[ f(+)\{1,2,3\} \equiv \sum_{k\in\{1,2,3\}}k = 1+ 2+ 3 \]

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