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anonymous

  • one year ago

Set theory operations from logical operations

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  1. ParthKohli
    • one year ago
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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...

  2. anonymous
    • one year ago
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    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.

  3. anonymous
    • one year ago
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    It's not really a tutorial, it's more of a discussion. You can ask questions but there is no actual question.

  4. anonymous
    • one year ago
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    \[ A f(\circ) B = C\equiv \forall x:(x \in A) \circ (x\in B) \iff x \in C \]Hmmmmm

  5. anonymous
    • one year ago
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    So \(f(\land) = \cap\).

  6. anonymous
    • one year ago
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    so would set theory disallow this? Since it requires that there exist a set of all sets?

  7. anonymous
    • one year ago
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    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.

  8. perl
    • one year ago
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    How did you go from operating on a set to a set of all sets.

  9. anonymous
    • one year ago
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    Well, what would you say is the domain and codomain of \(\cap\)?

  10. perl
    • one year ago
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    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

  11. perl
    • one year ago
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    https://en.wikipedia.org/wiki/Intersection_%28set_theory%29

  12. anonymous
    • one year ago
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    Hmmm

  13. anonymous
    • one year ago
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    power set still has 'all sets' as a domain, codomain

  14. perl
    • one year ago
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    ok I think I understand you are saying. For intersection U X U -> U for union the same

  15. anonymous
    • one year ago
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    Yep

  16. perl
    • one year ago
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    but you are using the domain as U, not UXU ?

  17. perl
    • one year ago
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    taking the cross product of such a large set can present its own problems

  18. anonymous
    • one year ago
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    for a unary operation, you have U -> U

  19. perl
    • one year ago
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    complement would be a unary operation

  20. anonymous
    • one year ago
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    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

  21. perl
    • one year ago
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    Why did it bother the early 20th century mathematicians so much, allowing a universal set.

  22. anonymous
    • one year ago
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    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?

  23. perl
    • one year ago
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    I would rather have a system that is clear and intuitive , that may have a paradox, than a complicated system that I cannot use.

  24. anonymous
    • one year ago
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    Yeah, this is something we can do in programming, yet we can't do it in math. How bizarre!

  25. perl
    • one year ago
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    you can allow self reference in programming, as you said with memory earlier?

  26. anonymous
    • one year ago
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    yeah

  27. perl
    • one year ago
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    what you described above with logic and set theory is whats called an isomorphism

  28. anonymous
    • one year ago
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    You can have an array of arrays, where one of the elements is itself

  29. perl
    • one year ago
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    with some qualifications , the logical *and* corresponds to intersection

  30. perl
    • one year ago
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    can you find a logical operation that is equivalent to A subset B

  31. anonymous
    • one year ago
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    Well \(\implies \) corresponds to \(\supseteq\)

  32. perl
    • one year ago
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    with some tweaking it could work. note that you need for all x for the subset condition

  33. anonymous
    • one year ago
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    Here is something I found that is interesting: https://en.wikipedia.org/wiki/Type_theory

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

  35. perl
    • one year ago
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    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.

  36. perl
    • one year ago
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    There is probably no way to avoid universe in some sense

  37. anonymous
    • one year ago
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    Here is another idea: \[ f(\circ)A = y \equiv \forall x\in A: x \circ \bigg(f(\circ) (A\setminus x)\bigg) = y \]

  38. anonymous
    • one year ago
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    wait, that isn't quite right...

  39. anonymous
    • one year ago
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    Here is another idea: \[ f(\circ)A = y \equiv \bigg( f(\circ)\emptyset = I_{\circ} \bigg) \land x\in A \implies x \circ \bigg(f(\circ) (A\setminus x)\bigg) = y \]

  40. anonymous
    • one year ago
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    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

  41. anonymous
    • one year ago
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    The idea is \(f(+)\) corresponds to \(\sum\).

  42. anonymous
    • one year ago
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    \[ f(\circ) = \begin{cases} \circ_{identity} & \emptyset \\ x\circ f(\circ) (A\setminus\{x\}) &\text{otherwise} \end{cases} \]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?

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

  44. perl
    • one year ago
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    is your latter definition of \( f \circ \) different from your former definition above \( A f(\circ) B = C\equiv \forall x:(x \in A) \circ (x\in B) \iff x \in C\)

  45. anonymous
    • one year ago
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    Yep, they are not the same \(f\).

  46. anonymous
    • one year ago
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    Right now I'm doing something along the lines of: \[ g(+)A =\sum_{x \in A}x \]

  47. anonymous
    • one year ago
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    Like \[ g(\cap) A =y \equiv \left(\bigcup_{x\in A}x \right)= y \]

  48. anonymous
    • one year ago
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    an accumulator function

  49. anonymous
    • one year ago
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    While \(f(\land) = \cap\) was more of a setwise function.

  50. perl
    • one year ago
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    okay that looks valid, but your definition looks recursive

  51. perl
    • one year ago
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    $$f(\circ) = \begin{cases} \circ_{identity} & \emptyset \\ x\circ \color{red}{f(\circ)} (A\setminus\{x\}) &\text{otherwise} \end{cases} $$

  52. anonymous
    • one year ago
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    Eventually you will run out of elements to use and hit the base case of the empty set.

  53. perl
    • one year ago
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    can you show me how it operates on \( \{1,2,3 \} \) $$\Large f(\circ) \{ 1,2,3\}$$

  54. anonymous
    • one year ago
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    \[ f(\circ) \{1,2,3\} \equiv 1 \circ(2\circ 3) = 1 \circ(3\circ 2) = 2\circ (1\circ 3) = 2\circ(3\circ 1) = \ldots \]

  55. anonymous
    • one year ago
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    Well, technically: \[ f(\circ)\{1,2,3\} \equiv 1\circ (2\circ (3\circ \text{idenity} )) \]

  56. perl
    • one year ago
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    okay so this is defining an associative operation using a binary operation

  57. anonymous
    • one year ago
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    Nope

  58. anonymous
    • one year ago
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    It's like this: \[ f(+)\{1,2,3\} \equiv \sum_{k\in\{1,2,3\}}k = 1+ 2+ 3 \]

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