if A-B = B then A=∅

[peak scroll]

Can someone check if my proof is correct
Assume by contradiction that A-B = B and A≠∅
so there is an element x that is in the set A
x∈A∩U => x∈A∩(B∪B') => x∈A∩B or x∈A∩B' => x∈A∩B or x∈B => x∈B => x∈A-B => x∉B and that's a contradiction.

[random fern]

what exactly is U?

and B'?

[peak scroll]

universal set

and B complement

[late plaza]

"x∈A∩B or x∈A∩B' => x∈A∩B or x∈B"

not getting this part, where do you get x∈B?

[random fern]

your proof is correct bro

[late plaza]

oh i got it nvm

yes it's correct

although i would be more careful with using the => symbol at "x∈B => x∈A-B => x∉B"

[random fern]

i agree its kinda confusing

[peak scroll]

it's a conclusion from the assumption

[edgy crag]

How are you defining A-B?

[peak scroll]

A-B = A\B = { x | x∈A and x∉B }

[edgy crag]

If $A-B=B$, then $(A-B)\cup B=B\cup B$

[limber hearth]

SWR

[peak scroll]

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