Been stuck on this claim for a while

[unborn loom]

Not sure how to do this exactly, I have been stuck on this for a while and just don't know what to do.

[buoyant iron]

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[hollow quarry]

what does "×" means?

[unborn loom]

cartesian product

[unborn loom]

@hollow quarryany idea on how to do it?

i have been looking at this since 11am

[hollow quarry]

so like for example if A={1,2,3} and B={4,5,6} then A×B={4,10,18}?

[unborn loom]

no

AxB = {(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)}

[stable cape]

let there exist an element x in a and an element y in B
so A x B will have an element (x,y)
now, we have AxB is subset of BxC
so, (x,y) is in BxC
so, x is in B and y is in C
but we know that x is in A
so A is subset of B
and we know that y is in B
B is subset of C
so A is subset of C
so it follows that A - C is the empty set

were we supposed to do that?

[unique moss]

now you need to prove that if A is subset of C then A - C = empty set

[stable cape]

i alr did lol

[stable cape]

no that part is logical
proving that is trivial, and I trust OP enough to fill in the blanks where necessary

[lime fern]

the sufficient and necessary condition for this is tush A is a subset of B and B is a subset of C, this is evident from the behavior of Cartesian products. Now due to the transitivity of inclusion we can say that A is a subset of C => A/C = null. [A/C is the same as A-C].

[stable cape]

bruh that's what I said

[lime fern]

did you?

[stable cape]

yeah lol

[lime fern]

ig so

didn't read it

[stable cape]

without the technical terms at least

[lime fern]

what's the technical term

[lime fern]

@stable cape

[stable cape]

idk what the technical terms are
ik cartesian product and all but not that transitivity of inclusion thingy

[lime fern]

A relation $R$ is transitive $\equiv (xRy \land yRz \implies xRz)$

[quick herald]

Comfey

[lime fern]

@stable cape

for example, ">" is transitive

because if 2>1 and 3>2 then 3>1

[stable cape]

oh like that

bruh

so

[lime fern]

and the $A \times B$ is the Cartesian product

[quick herald]

Comfey

[stable cape]

ik that

[lime fern]

this is also why the x-y plane is Called the Cartesian plane

it's the Cartesian product, R x R

[stable cape]

$A \times B = {(x,y)| x \in A \text{ and } y \in B}$

[lime fern]

@stable cape what's $A \times \phi$

[quick herald]

No Lifer (no lifer)

Comfey

[stable cape]

phi

there's nothing in phi

so

[lime fern]

[stable cape]

nothing in any cartesian product of phi

in other terms

[stable cape]

there is no element y for which $y \in \phi$

[quick herald]

No Lifer (no lifer)

[stable cape]

so A x phi = phi

[lime fern]

okay

[stable cape]

yee

[unique moss]

you need to prove it as a lemma otherwise your proof will be wrong

it's because the path you chosen

[lime fern]

Pardon me I forgot that part too

lemme fill that out real quick

[unique moss]

by contradiction or contrapositive

direct proof will not work, I think

contrapositive is better for this case

[lime fern]

claim : $A \subseteq B \implies A/B = \phi$ \ $A/B = {x \vert x \in A \land x \in ' B}$ \ Suppose, $A/B \neq \phi$. Thus, $\exists x : x \in A \land x \in ' B$, which is a contradiction of the given condition that $A \subseteq B$. So $A/B$ is an empty set.

[quick herald]

Comfey

[unique moss]

uhum

[lime fern]

is it incorrect?

[stable cape]

let there exist an element x in a and an element y in B
so A x B will have an element (x,y)
now, we have AxB is subset of BxC
so, (x,y) is in BxC
so, x is in B and y is in C
but we know that x is in A
so A is subset of B
and we know that y is in B
B is subset of C
so A is subset of C
so every element in A is in C
let's phrase this in technical terms
if x is in a, x is in C
now, we know that the definition of A - C is
{x| (x in A) and (x not in C)}
but for all x in A x is also in C
so there is no element x which satisfies the above condition
hence, A - C must be the empty set

[unique moss]

no

[unique moss]

you only need to put, suppose a subset of b and a/b =! empty set

because the way you wrote seems like contrapositive

but it's by contradiction, so you need the "suppose a subset of b"

[lime fern]

this gives me physical pain

[lime fern]

That's a part of the claim

[stable cape]

bruh I don't really want to write it in fancy terms

[lime fern]

I mean make it 3 lines?

and easier to read

[stable cape]

oh that I can do

wait a min

[unique moss]

the point is, you need to prove that "so there is no element x which satisfies the above condition", how you do that?

it's because that when proving things which involve empty set, you need to negate most of statemants

so you usually use contradiction or contrapositive

suppose there is a element

[stable cape]

oh...

[lime fern]

when doing set theory proves

[stable cape]

you have to logic everything huh

[lime fern]

think of it as you are explaining it to a -1 yr old

yes

[stable cape]

ah

so explain every little thing

someth I'm bad at

[unborn loom]

Thanks for all the help, I got it 👍

+close