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anonymous
 one year ago
Set theory operations from logical operations
anonymous
 one year ago
Set theory operations from logical operations

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0\[A \cap B = C\]is equivalent to\[k\in C \iff (k\in A) \wedge (k \in B)\]I don't even know what the question is but...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's not really a tutorial, it's more of a discussion. You can ask questions but there is no actual question.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ A f(\circ) B = C\equiv \forall x:(x \in A) \circ (x\in B) \iff x \in C \]Hmmmmm

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So \(f(\land) = \cap\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so would set theory disallow this? Since it requires that there exist a set of all sets?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let \(B = \{ true, false\}\). 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.

perl
 one year ago
Best ResponseYou've already chosen the best response.0How did you go from operating on a set to a set of all sets.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, what would you say is the domain and codomain of \(\cap\)?

perl
 one year ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0power set still has 'all sets' as a domain, codomain

perl
 one year ago
Best ResponseYou've already chosen the best response.0ok I think I understand you are saying. For intersection U X U > U for union the same

perl
 one year ago
Best ResponseYou've already chosen the best response.0but you are using the domain as U, not UXU ?

perl
 one year ago
Best ResponseYou've already chosen the best response.0taking the cross product of such a large set can present its own problems

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for a unary operation, you have U > U

perl
 one year ago
Best ResponseYou've already chosen the best response.0complement would be a unary operation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep 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

perl
 one year ago
Best ResponseYou've already chosen the best response.0Why did it bother the early 20th century mathematicians so much, allowing a universal set.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And a collection would nee the ability to let you know if something is in it or not, to iterate through it, have no duplicates?

perl
 one year ago
Best ResponseYou've already chosen the best response.0I would rather have a system that is clear and intuitive , that may have a paradox, than a complicated system that I cannot use.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, this is something we can do in programming, yet we can't do it in math. How bizarre!

perl
 one year ago
Best ResponseYou've already chosen the best response.0you can allow self reference in programming, as you said with memory earlier?

perl
 one year ago
Best ResponseYou've already chosen the best response.0what you described above with logic and set theory is whats called an isomorphism

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can have an array of arrays, where one of the elements is itself

perl
 one year ago
Best ResponseYou've already chosen the best response.0with some qualifications , the logical *and* corresponds to intersection

perl
 one year ago
Best ResponseYou've already chosen the best response.0can you find a logical operation that is equivalent to A subset B

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well \(\implies \) corresponds to \(\supseteq\)

perl
 one year ago
Best ResponseYou've already chosen the best response.0with some tweaking it could work. note that you need for all x for the subset condition

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Here is something I found that is interesting: https://en.wikipedia.org/wiki/Type_theory

perl
 one year ago
Best ResponseYou've already chosen the best response.0"In type theory, concepts like "and" and "or" can be encoded as types in the type theory itself." This could correspond to memory addresses

perl
 one year ago
Best ResponseYou've already chosen the best response.0Here 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.

perl
 one year ago
Best ResponseYou've already chosen the best response.0There is probably no way to avoid universe in some sense

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Here is another idea: \[ f(\circ)A = y \equiv \forall x\in A: x \circ \bigg(f(\circ) (A\setminus x)\bigg) = y \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait, that isn't quite right...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Here is another idea: \[ f(\circ)A = y \equiv \bigg( f(\circ)\emptyset = I_{\circ} \bigg) \land x\in A \implies x \circ \bigg(f(\circ) (A\setminus x)\bigg) = y \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The idea is \(f(+)\) corresponds to \(\sum\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ f(\circ) = \begin{cases} \circ_{identity} & \emptyset \\ x\circ f(\circ) (A\setminus\{x\}) &\text{otherwise} \end{cases} \]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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Also, for \(x\) to be an arbitrary element in \(A\), then I'm guessing this operation must be associative.

perl
 one year ago
Best ResponseYou've already chosen the best response.0is your latter definition of \( f \circ \) different from your former definition above \( A f(\circ) B = C\equiv \forall x:(x \in A) \circ (x\in B) \iff x \in C\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep, they are not the same \(f\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right now I'm doing something along the lines of: \[ g(+)A =\sum_{x \in A}x \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Like \[ g(\cap) A =y \equiv \left(\bigcup_{x\in A}x \right)= y \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0an accumulator function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0While \(f(\land) = \cap\) was more of a setwise function.

perl
 one year ago
Best ResponseYou've already chosen the best response.0okay that looks valid, but your definition looks recursive

perl
 one year ago
Best ResponseYou've already chosen the best response.0$$f(\circ) = \begin{cases} \circ_{identity} & \emptyset \\ x\circ \color{red}{f(\circ)} (A\setminus\{x\}) &\text{otherwise} \end{cases} $$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Eventually you will run out of elements to use and hit the base case of the empty set.

perl
 one year ago
Best ResponseYou've already chosen the best response.0can you show me how it operates on \( \{1,2,3 \} \) $$\Large f(\circ) \{ 1,2,3\}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ f(\circ) \{1,2,3\} \equiv 1 \circ(2\circ 3) = 1 \circ(3\circ 2) = 2\circ (1\circ 3) = 2\circ(3\circ 1) = \ldots \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, technically: \[ f(\circ)\{1,2,3\} \equiv 1\circ (2\circ (3\circ \text{idenity} )) \]

perl
 one year ago
Best ResponseYou've already chosen the best response.0okay so this is defining an associative operation using a binary operation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's like this: \[ f(+)\{1,2,3\} \equiv \sum_{k\in\{1,2,3\}}k = 1+ 2+ 3 \]
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