anonymous
  • anonymous
Set theory operations from logical operations
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ParthKohli
  • ParthKohli
\[A \cap B = C\]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
  • anonymous
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.
anonymous
  • anonymous
It's not really a tutorial, it's more of a discussion. You can ask questions but there is no actual question.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[ A f(\circ) B = C\equiv \forall x:(x \in A) \circ (x\in B) \iff x \in C \]Hmmmmm
anonymous
  • anonymous
So \(f(\land) = \cap\).
anonymous
  • anonymous
so would set theory disallow this? Since it requires that there exist a set of all sets?
anonymous
  • anonymous
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.
perl
  • perl
How did you go from operating on a set to a set of all sets.
anonymous
  • anonymous
Well, what would you say is the domain and codomain of \(\cap\)?
perl
  • perl
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
perl
  • perl
https://en.wikipedia.org/wiki/Intersection_%28set_theory%29
anonymous
  • anonymous
Hmmm
anonymous
  • anonymous
power set still has 'all sets' as a domain, codomain
perl
  • perl
ok I think I understand you are saying. For intersection U X U -> U for union the same
anonymous
  • anonymous
Yep
perl
  • perl
but you are using the domain as U, not UXU ?
perl
  • perl
taking the cross product of such a large set can present its own problems
anonymous
  • anonymous
for a unary operation, you have U -> U
perl
  • perl
complement would be a unary operation
anonymous
  • anonymous
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
perl
  • perl
Why did it bother the early 20th century mathematicians so much, allowing a universal set.
anonymous
  • anonymous
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?
perl
  • perl
I would rather have a system that is clear and intuitive , that may have a paradox, than a complicated system that I cannot use.
anonymous
  • anonymous
Yeah, this is something we can do in programming, yet we can't do it in math. How bizarre!
perl
  • perl
you can allow self reference in programming, as you said with memory earlier?
anonymous
  • anonymous
yeah
perl
  • perl
what you described above with logic and set theory is whats called an isomorphism
anonymous
  • anonymous
You can have an array of arrays, where one of the elements is itself
perl
  • perl
with some qualifications , the logical *and* corresponds to intersection
perl
  • perl
can you find a logical operation that is equivalent to A subset B
anonymous
  • anonymous
Well \(\implies \) corresponds to \(\supseteq\)
perl
  • perl
with some tweaking it could work. note that you need for all x for the subset condition
anonymous
  • anonymous
Here is something I found that is interesting: https://en.wikipedia.org/wiki/Type_theory
perl
  • perl
"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
  • perl
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.
perl
  • perl
There is probably no way to avoid universe in some sense
anonymous
  • anonymous
Here is another idea: \[ f(\circ)A = y \equiv \forall x\in A: x \circ \bigg(f(\circ) (A\setminus x)\bigg) = y \]
anonymous
  • anonymous
wait, that isn't quite right...
anonymous
  • anonymous
Here is another idea: \[ f(\circ)A = y \equiv \bigg( f(\circ)\emptyset = I_{\circ} \bigg) \land x\in A \implies x \circ \bigg(f(\circ) (A\setminus x)\bigg) = y \]
anonymous
  • anonymous
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
anonymous
  • anonymous
The idea is \(f(+)\) corresponds to \(\sum\).
anonymous
  • anonymous
\[ f(\circ) = \begin{cases} \circ_{identity} & \emptyset \\ x\circ f(\circ) (A\setminus\{x\}) &\text{otherwise} \end{cases} \]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
  • anonymous
Also, for \(x\) to be an arbitrary element in \(A\), then I'm guessing this operation must be associative.
perl
  • perl
is your latter definition of \( f \circ \) different from your former definition above \( A f(\circ) B = C\equiv \forall x:(x \in A) \circ (x\in B) \iff x \in C\)
anonymous
  • anonymous
Yep, they are not the same \(f\).
anonymous
  • anonymous
Right now I'm doing something along the lines of: \[ g(+)A =\sum_{x \in A}x \]
anonymous
  • anonymous
Like \[ g(\cap) A =y \equiv \left(\bigcup_{x\in A}x \right)= y \]
anonymous
  • anonymous
an accumulator function
anonymous
  • anonymous
While \(f(\land) = \cap\) was more of a setwise function.
perl
  • perl
okay that looks valid, but your definition looks recursive
perl
  • perl
$$f(\circ) = \begin{cases} \circ_{identity} & \emptyset \\ x\circ \color{red}{f(\circ)} (A\setminus\{x\}) &\text{otherwise} \end{cases} $$
anonymous
  • anonymous
Eventually you will run out of elements to use and hit the base case of the empty set.
perl
  • perl
can you show me how it operates on \( \{1,2,3 \} \) $$\Large f(\circ) \{ 1,2,3\}$$
anonymous
  • anonymous
\[ f(\circ) \{1,2,3\} \equiv 1 \circ(2\circ 3) = 1 \circ(3\circ 2) = 2\circ (1\circ 3) = 2\circ(3\circ 1) = \ldots \]
anonymous
  • anonymous
Well, technically: \[ f(\circ)\{1,2,3\} \equiv 1\circ (2\circ (3\circ \text{idenity} )) \]
perl
  • perl
okay so this is defining an associative operation using a binary operation
anonymous
  • anonymous
Nope
anonymous
  • anonymous
It's like this: \[ f(+)\{1,2,3\} \equiv \sum_{k\in\{1,2,3\}}k = 1+ 2+ 3 \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.