Set theory operations from logical operations

- anonymous

Set theory operations from logical operations

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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

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

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

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

- anonymous

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

- anonymous

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

- 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

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

- anonymous

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

- 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

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

- anonymous

Hmmm

- anonymous

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

- perl

ok I think I understand you are saying.
For intersection
U X U -> U
for union the same

- anonymous

Yep

- perl

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

- perl

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

- anonymous

for a unary operation, you have U -> U

- perl

complement would be a unary operation

- 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

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

- 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

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

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

- perl

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

- anonymous

yeah

- perl

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

- anonymous

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

- perl

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

- perl

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

- anonymous

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

- perl

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

- anonymous

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

- 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

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 â€“ being able to say things about all types â€“ still exist. So many type theories have a "universe type", which contains all other types.

- perl

There is probably no way to avoid universe in some sense

- 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

wait, that isn't quite right...

- 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

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

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

- 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

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

- 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

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

- anonymous

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

- anonymous

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

- anonymous

an accumulator function

- anonymous

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

- perl

okay that looks valid, but your definition looks recursive

- perl

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

- anonymous

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

- perl

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

- 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

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

- perl

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

- anonymous

Nope

- 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.