Proof: A is a Subset of B' iff A intersect B is Empty | Set Theory, Set Complements

A is a subset of the complement of B if and only if A intersect B is the empty set! We prove this simple set theory result in today's set theory video lesson!
Remember that the complement of a set A is the set of all elements not in A, typically this is taken relative to some "universal set", so the complement of A is the set of all elements in the universal set that are not in A. The proof follows simply from a few applications of definitions!
Lesson on set complements: • What is a Set Complement?
I hope you find this video helpful, and be sure to ask any questions down in the comments!
+WRATH OF MATH+
◆ Support Wrath of Math on Patreon: / wrathofmathlessons
Follow Wrath of Math on...
● Instagram: / wrathofmathedu
● Facebook: / wrathofmath
● Twitter: / wrathofmathedu
My Music Channel: / seanemusic

[@WhenCatPlayz]

thanks a lot sir...this helped me so much 👍🏻👍🏻

[@WrathofMath]

Glad to hear that!

[@projectiocon]

Thankyou so much ❣️

[@WrathofMath]

My pleasure, thanks for watching!

[@tijahhappy912]

Sir, I want to ask one question. How to proof, If A and B are sets, then A n (B-A) =empty set?

[@WrathofMath]

Thanks for watching and you just have to use what is often called a "double inclusion proof". If you look up some of my set equality proofs you'll see many examples of this. You need to show the empty set is a subset of A n (B-A) and then show that A n (B-A) is a subset of the empty set. The first part is immediate since the empty set is a subset of every set. Then, to show A n (B-A) is a subset of the empty set, you would need to show that it doesn't have any elements. I would use a contradiction argument. Suppose A n (B-A) has an element x. Then you will be able to show a contradiction since x must be in A, but x also must be in B-A.

[@tijahhappy912]

@@WrathofMath so, i need to proof using contradiction method. Thank you sir.

[@azizhani33]

A Π (B - A) = A Π A' = {}. Explanation: B - A is the complement of A in B, that is A'. So A Π A' is every element that belongs to both A and its complement, and of course there is none.

[@tenvolt]

Well explained

[@WrathofMath]

Thank you!

[@mairasayyed161]

Thanks sir for making this helpful video

[@WrathofMath]

My pleasure, thanks for watching!

[@maxpercer7119]

1:05 I think you meant A is a subset of B *complement* , (not A is a subset of B)

[@dalya.mohammed]

Thank you so much 🌹🌹🌹🌹

[@AnishKumar-uy7bu]

Thanks for this solution ☺☺

[@WrathofMath]

No problem! Thanks for watching!

[@AnishKumar-uy7bu]

@@WrathofMath ❤

[@ajmailsingh2806]

Thank you so much 😊😊

[@WrathofMath]

My pleasure! Thanks a lot for watching!

[@pramilak.]

Tq so much for solutions.

[@WrathofMath]

My pleasure, thanks for watching and let me know if you ever have any video requests!

[@Kau351]

Thanks a lot bro❤️❤️

[@WrathofMath]

No problem! Thanks a lot for watching and let me know if you ever have any questions!

[@dalya.mohammed]

You are the best teacher ❤️❤️

[@WrathofMath]

Thanks a lot - I do my best! Thanks for watching and let me know if you have any questions!

[@mairasayyed161]

Sir I have a question
Prove A subset of B if and only if A intersection B'= phi

[@WrathofMath]

That result is actually the same thing as what we prove in this video. Remember that the complement of a complement is the original set. As in, (B')' = B.
To try to make it more clear why what you stated is the same as what we prove in the video, imagine I relabel the sets used in this video. Perhaps I call the sets X and Y. Then, in this video we prove that X is a subset of Y' iff X intersect Y is empty. We could then decide to rename these sets. Let's set A = X and B = Y', and thus B' = Y. This renaming then gives the result you stated, A is a subset of B iff A intersect B' is empty. Does that help?

[@mairasayyed161]

@@WrathofMath yeah thanks

[@mairasayyed161]

@@WrathofMath
A subset of B iff A interaction B'=phi
A subset of B iff A' union B=U
A subset of B iff A\B=phi
Sir please upload a lecture at these questions if it can be possible today or tomorrow

[@azizhani33]

On one hand, A\B = A - B consists of all elements of A that are not elements of B, that is none, because A is a subset of B and therefore every element of A is by definition an element of B. As a result A\B = {}. On the other, A\B = {} implies that there is no element of A which is not an element of B. It stems from this that every element of A belongs to B and therefore A is a subset of B.

[@angelranapaheli9723]

let A,B subset of U. Prove that A subset of B A intersection B compliment= phi.
Sir please solve this question for me

[@WrathofMath]

Thanks for watching, Angel, and this video is a proof of that result. B' and B relate to each other in the same way, so we can interchange their names if we want, since they're arbitrary complementary sets. So the title of this video, the result we prove, could be stated as "A is a subset of B iff A intersect B' is empty". I just called A a subset of B' instead of B, but the result and logic behind the proof is exactly the same whether you state it as "A is a subset of B iff A intersect B' is empty", or if you state it as "A is a subset of B' iff A intersect B is empty". Does that make sense?

[@sudevpaul7530]

Sir can u do

[@chrisrainville6324]

4 out of 3 people struggle with math!

[@WrathofMath]

I am trying to create the antidote

[@arpitaghosh7214]

😂