Question

Set theory operations from logical operations

Answer 1

\[A \cap B = C\]is equivalent to\[k\in C \iff (k\in A) \wedge (k \in B)\]I don't even know what the question is but...

Answer 2

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.

Answer 3

It's not really a tutorial, it's more of a discussion. You can ask questions but there is no actual question.

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

Hmmm

Answer 13

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

Answer 14

ok I think I understand you are saying.
For intersection
U X U -> U

for union the same

Answer 15

Yep

Answer 16

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

Answer 17

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

Answer 18

for a unary operation, you have U -> U

Answer 19

complement would be a unary operation

Answer 20

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

Answer 21

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

Answer 22

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?

Answer 23

I would rather have a system that is clear and intuitive , that may have a paradox, than a complicated system that I cannot use.

Answer 24

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

Answer 25

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

Answer 26

yeah

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

"In type theory, concepts like "and" and "or" can be encoded as types in the type theory itself."
This could correspond to memory addresses

Answer 35

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.

Answer 36

There is probably no way to avoid universe in some sense

Answer 37

Here is another idea: \[
f(\circ)A = y \equiv \forall x\in A: x \circ \bigg(f(\circ) (A\setminus x)\bigg) = y
\]

Answer 38

wait, that isn't quite right...

Answer 39

Here is another idea: \[
f(\circ)A = y \equiv \bigg( f(\circ)\emptyset = I_{\circ} \bigg) \land x\in A \implies x \circ \bigg(f(\circ) (A\setminus x)\bigg) = y
\]

Answer 40

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

Answer 41

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

Answer 42

\[
f(\circ) = \begin{cases}
\circ_{identity} & \emptyset \\
x\circ f(\circ) (A\setminus\{x\}) &\text{otherwise}
\end{cases}
\]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?

Answer 43

Also, for \(x\) to be an arbitrary element in \(A\), then I'm guessing this operation must be associative.

Answer 44

is your latter definition of \( f \circ \) different from your former definition above

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

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

an accumulator function

Answer 49

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

Answer 50

okay that looks valid, but your definition looks recursive

Answer 51

$$f(\circ) = \begin{cases}
\circ_{identity} & \emptyset \\
x\circ \color{red}{f(\circ)} (A\setminus\{x\}) &\text{otherwise}
\end{cases}
$$

Answer 52

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

Answer 53

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

Answer 54

\[
f(\circ) \{1,2,3\} \equiv 1 \circ(2\circ 3) = 1 \circ(3\circ 2) = 2\circ (1\circ 3) = 2\circ(3\circ 1) = \ldots
\]

Answer 55

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

Answer 56

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

Answer 57

Nope

Answer 58

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