SUBSETS AND POWER SETS - DISCRETE MATHEMATICS

[@Trevtutor]

Check out my new course in Set Theory: trevtutor.com/p/master-discrete-mathematics-set-theory It comes with video lectures, text lectures, practice problems, solutions, and a practice final exam!

[@nataliesagnoy3579]

"If your professor is sadistic" 😂😂 Too bad he is!

[@addieadds1264]

Bahahaha!!!!!! Right!!

[@onegrapefruitlover]

Best username ever lol

[@jackmenirons4989]

"Is A a subset of the p(A)?" Professor at the college you're paying thousands of dollars for: "Well it's trivial so I'll leave it to you to think about." Random youtube playlist for free: "Well this is tricky at first so let me explain it in detail so everyone understands."

[@RH-hv4ir]

WHYYYY ISSS THISSS SOOOO TRUEEEE

[@christianfranklin7043]

Professor code for "Fuck if I know."

[@nicholascunningham6936]

Better yet, we're paying thousands of dollars to a college for an education, and yet we still have to come to some random youtube playlist and teach ourselves because our professors suck at teaching. At least that's the situation in my case. A classmate recommended TrevTutor. Thank God he did...

[@jackmenirons4989]

@@nicholascunningham6936 Yeah that's a pretty common situation. Discrete Math seems to be a terribly taught class at a lot of universities.

[@MikeRosoftJH]

For example, let A={1,2}. P(A)={{}, {1}, {2}, {1,2}}. Neither 1 nor 2 is an element of P(A) - {1} and {2} is. (In set theory the usual construction of natural numbers as sets is: 0 is the empty set, 1 is {0}, 2 is {0,1}, 3 is {0,1,2}, and so on; but not even under this definition is 1 or 2 an element of P(A).) So A is not a subset of P(A). (A is an element of P(A).) Can you find an example when A *is* a subset of P(A)?

[@yurilegs1]

I'm taking Discrete Math this semester. We are on the fifth week now and we're just learning this. We learned proofs last week. I think the ordered of your videos are much easier to follow :) thank you for the awesome videos! You make Discrete Math easier.

[@zeromaster8666]

I agree

[@sports6872]

Saya minggu ke 4

[@michaelwest2451]

damn... this is in chapter 2 of what we're doing

[@Jerry-yq8vt]

Damn I learnt this in week 3

[@DJK001]

So have you graduated yet😂😁

[@justingolden21]

12:01 so A is not a subset of p(A) because the elements a and b aren't in p(A), only {a} and {b} which are different. but A is an element of p(A) because {a,b} is in p(A)

[@YBBY1928]

hope this right cause I did understand it like this :D

[@Nichama70]

This wasn't clear to me at first, but after watching it a few times and looking up the differences between an element and a subset, I've come to the same conclusion.

[@ritesh890]

Yes. While considering the subset of P(A) when A={a, b} , one of the subsets involving A will be { {a, b} } which is clearly not A. But in subset of P(A) where A={phi}, one subset will be {phi}, hence A is a subset here.

[@pokerwithoutknowing]

I suck at proofs, could you guys confirm a formal version would look something like this: Let A={a,b}, then p(A)={∅, {a}, {b}, {a,b}} => p(A)={∅, {a}, {b}, A} => A ∈ p(A) and p(A)≠{∅, {a}, {b}, {A}} => some A ⊄ p(A)

[@DarvinJ]

THANK YOU

[@The_double_side]

great explanation i understood a math concept for first time in my life . great respect to you sir !! We need people like you !!

[@CoreyBass-t9b]

at 7:42 where he had to rework and count out 2^6 gives me hope for myself. thank you

[@Shiroyashasama]

power sets are wild! im rewatching that part so many times

[@timxhx1832]

i just want to thank you, we need more videos like these on youtube rather than people playing with fidget spinners or slime. Thank you for the help :)

[@marcelleivyzabat9913]

I have an exam tomorrow and I'm quite confused about power sets, the rest of the coverage is okay for me. The way you explained it is so much easier to understand. Thank you!

[@kenzom2424]

How was the exam ?

[@wirito]

You're explanation of 10:50 was Fu****g awesome!! I was confused after the lecture but your videos are saving me :D

[@S4mm1ch]

thank you for the videos, i just started this course and was feeling so lost, i am starting to grasp the concepts a bit easier watching your videos

[@naomifinkilegouwe7798]

Sir.... Hats off to you, you are doing a great job. These videos are helping, we really appreacite that.

[@slyanna3688]

"if your professor is a little bit sadistic " sounds about right

[@allissavalenzuela385]

I am taking mathematical economics this semester. My professor just kept saying 2^n is the powerset but never explained any of this! Discrete math is not a prereq for this class but my professor just acts like this stuff is common knowledge. I am so thankful for these videos because I have been feeling so lost in this class

[@Towhidchannel]

Thank you for making this chapter so interesting with your teaching style. I really appreciate it..

[@defaultimatum]

I like how you draw brackets in 5:45 , it is somehow relatable.

[@SquirtIeSquad]

Thanks man I would be absolutely lost right now without your videos

[@kateye6513]

Honestly. Thank you for make it so easy and fun to listen to.❤️

[@MrRobot-gm9cv]

Wow man, I am very impressed with your teaching style.

[@alebuser4933]

Perfect timing, i have discMath exam in 5 days

[@djjambo2821]

me too lol. mine is 5 days from today ;(

[@tarakhanal1864]

I start discrete math in 7 days. I'm just trying to learn what it's about.

[@SquirtIeSquad]

Im doing mine right now

[@gorkemaslan1858]

So weird. It' been 2 years

[@jarncherry4581]

It's funny when you talk you hold ur teeth together like you're getting angry aha. It's a little hard to wrap ur head around some of these, but there is no better person at explaining discrete mathematics than you. Props you did an EXTREMELY good job.

[@Kerotaroily]

Thank you so much for this! I’m taking a retake test tomorrow because I sorta.. failed the first one… It makes it seem a lot easier! (I’ll update the grade) Grade: 95% 🥳

[@Robotomy101]

so how was it?

[@Kerotaroily]

@@Robotomy101 OH SHOOT.. I FORGOT TO PUT THE GRADE 😭 Thx for reminding me 😭💕

[@hello-gt2su]

12:00 why A is a subset of the power set of A? I don't see the difference with the previous example of {a,b}

[@quanghuynguyen4222]

Ur videos are awesome!! its saving me for my semester test, TYSM!!!

[@jeffreyslater4416]

đón chờ những ca khúc tiếp theo của Phúc, càng nghe càng thích giọng ca của Phúc ❤️

[@user-ke2rp4gz2c]

dude your handwriting is amazing

[@swanhtet1]

The subset of powerset example confused me. 11:35 "we get A back as a possible subset" *Okey* 11:51 "the set {a,b} is not the subset of the powerset, it is just an element in this case." *WHY NOT?* 11:57 "if we compare it to this example, we would see that A ISSSSS a subset of the powerset." *WHY?* There was no explanation why it is not the subset or why it is a subset. You just point and said "it issss" and "it isn't".

[@CessPlays]

:(

[@CessPlays]

same

[@satyamprakash7030]

Precisely

[@dhruvipaprunia3279]

Same here, I don't understand

[@im-essi]

For A to be a subset of P(A), each element in A must be in P(A), in other words, since A = {a, b}, both a and b must be in P(A). But they aren't, {a} is, and {b} is, but those are not the same thing as a or b! a is just an element, and {a} is a set containing that element. A itself however *is* an *element* of P(A), because P(A) = { {}, {a}, {b}, {a, b} }, and since A = {a, b}, that is the same thing as { {}, {a}, {b}, A }. So A is *not a subset* of P(A), because not all elements in A are in P(A), since a ≠ {a} and b ≠ {b}, so none of them show up in P(A), but A is *an element* of P(A), because {a, b}, which is the same as A, *is* in P(A).

[@ivasto6182]

Great explanation but it would help if the sets are visually represented with circles and elements

[@walternyarukuhwe3180]

I am loving these videos i now have a better understanding on set theory thank yu very much trev

[@ikeikeikeikeikeikeikeike]

Perfect timing, i have a discMath exam in 2 hours

[@comfortkabanshi5048]

So, how did it go? Did the video help?

[@johnztech1651]

Hey this is a good lecture. I suggest to the author, please include in other means of payments in support functionality like PayPal. Also am software developer. How best do u think the content in this video can help me in the arena of programming

[@skylarngoi3262]

I can't believe it. I can do the questions that you show, but I can't solve the questions what my lecture gave me! Thanks for your video

[@XxThatGuyMysteryXx]

DUDE THIS VIDEO IS FUCKING AWESOME, YOU MADE EVERYTHING SO EASY TO UNDERSTAND! My professor is so dry and follows our textbook without any deviation. Thank you so much!

[@fatlirtopalli7641]

At 12:05, isn't it also true for {∅} that it is a subset of its power set, p({∅})? My reasoning is that ∅ is an element of both the {∅} and its power set, i.e. the condition for a subset?

[@Tortuex_]

thank you

[@asdgvdasadsssgdsad]

Just to clarify 12:10 (as I struggled to understand this concept initially). From what I've understood thus far from my reading: If B = {a, {a}} then P(B) = { Ø, {a}, {{a}}, {a,{a}} } and you will notice that the element {a} is present in both B and P(B), and thus B is a subset of P(B). On the other hand, if B = { a }, then P(B) = { Ø, {a} } and B behaves as an element of P(B) and not a subset as they share no common elements. Hope it helps.

[@asdgvdasadsssgdsad]

Feel free to correct me if I'm wrong tho woops.

[@Nichama70]

@@asdgvdasadsssgdsad In your example, the elements of B are a, {a}. From what I understand, for B to be a subset of P(B), all elements of B must also be elements of P(B). As you stated the elements of P(B) are Ø, {a}, {{a}}, {a,{a}} So as you can see, the element a is missing from list of elements in P(B), so I believe that is why B is not a subset of P(B). Hopefully I am understanding this correctly.

[@aboutthereality179]

Power sets very well explained. Especially the last |P(P(P(A)))| example, Thank You Trev :)

[@louisdiaz4179]

I got some sadistic ass professors then.

[@detectivedoom5175]

Thank you for becoming my teacher because my uni teacher isn't great.

[@rchtchauhan]

Thanks a lot man, You don't know how much you helped me.

[@sakshisharma4335]

How can the p(p()) be {()]} 14:03

[@TargetJEE-p6h]

in JEE, those crazy examiners expect us to find subsets of set with 7-8 elements and impose condition that the sum of elements should be odd, even, prime number............

[@mohamadtshehab8027]

03:57 First, sorry if my english was bad. The empty set cannot ever be equal to the set {a,b,c} because it is always empty so why don't we just call it a Proper subset of the other set ?.

[@Trevtutor]

Sure, you could. It meets both definitions.

[@mohamadtshehab8027]

Yep. But in this case I think we can only call it a proper subset since it is all the ways a part of the other set and it cannot be equal to it .. I wish I declared my idea well. And I really appreciate your quick reply.

[@mohamadtshehab8027]

Or I misunderstood Subsets? 😅😃

[@Trevtutor]

So would you say that 3 is not less than or equal to 5? It’s a similar concept. Which symbol/descriptor we use depends on context and what we want to prove.

[@mohamadtshehab8027]

Thanks a lot 👌❤

[@taizyakapula6328]

The subset symbol with the line underneath; should it only be used if the sets are equivalent that is when they contain all the same elements? If so in the first example at 1:58 shouldn’t that be false ?

[@Trevtutor]

It usually depends on what you want to prove. But the equals sign just means it’s a smaller set or the same set.

[@taizyakapula6328]

@@Trevtutor understood thanks

[@isurusandaruwan4769]

I found my professor. Thank you.

[@martijn130370]

This was the explanation for size of powersets I was looking for!

[@ekronb5287]

Honestly I was getting confused by my professors powerpoint and the book but this is really similar to logic. I took it last sem so it should help understand the concepts behind it.

[@theguildofsilence]

So if we think of constructing a power set via a decision tree, we always root it with the empty set, then on every right-hand branch of 'add nothing', we left-branch with adding a subsequent set element...until when? What determines what is a leaf node and therefore a unique subset? I guess I'm just a little confused about how you decide to create those branches. Unless I'm thinking too hard and it isn't supposed to be a generalizable way to find subsets.

[@mustafamalik4211]

I literally just understood the definition of a subset and a proper subset in 1 minute here. I legit spent like 30 minutes trying to understand the definition from my class notes.

[@saumyojitdas4212]

for each element we can add it to a subset or we cannot add it to a subset ...means suppose A is a set {1,2} for 1 i can add it to a subset { 1 } or i add nothing to 1 that means it still remains {1} right? u mean by adding means we add to another element to make it a subset of two elements {1,2} same goes from 2 's perspective .

[@Lena-of7wd]

Thank you!! This was helpful, you made it easy to understand :)

[@carmenx3789]

thank you so much for making the videos , it really helps a lot !

[@RRatedT]

I've been watching your videos about Discrete and I keep on watching right now, I have to say you explain this subject so well and make it look so much easier than what it is. I hope my nightmares would stop by today :D

[@Fabius11k]

Very well explained! Thank you

[@yoitslemonboy6988]

yo thank you for this playlist honestly!

[@anassirelkhatim9846]

Bruv you are the definition of what I call a mad lad fam you the best lecturer out there m8t somebody please give this gentleman a place in the hall of fame.there is nothing I didn't understand the first time I watched all your videos your just the best man straight up.

[@CrisRorens]

my respect is yours glad that you make these videos

[@yuehernkang]

thank you! your videos are really very helpful.

[@KineHjeldnes]

Thank you for your videos! But i dont understand, at 2:11 you say if A then B, right? But isn't it the other way around since A is contained within B? Because if you have A you don't have B if A is a,b and B is a,b,c.

[@YuriiBech]

I like your voice! Thank you so much for your videos!

[@JVenom_]

I’m reviewing this playlist for my theory of computation class

[@ozgurkuzu2202]

its so interesting,before you explained the subset definition you quickly jump to the definition of p(A) .thanks for the video.

[@MarlitaPangda]

Dont listen to those haters, btw i liked and sub😊❤

[@shubhrogupta9409]

me and the boys enjoyed this

[@agastyasingh3764]

Chad

[@sben125]

i really wanna send this guy $10 and a hug!

[@jordantaylor5552]

I don’t understand what you are asking when you say “for any A?” Can you elaborate?

[@zaylo9273]

I wish the uni lecture notes can explain as well as u do XD Thank you for the awesome videos XD

[@SebHaugeto]

Thank you for this! Super helpful :)

[@kaizenalive]

12:15 A can be a proper subset of p(A)?

[@hieudoan6532]

Hi for the last example don't you have to solve from the right to left since P(A) is inside another P and so on?

[@aboitoo]

Hi so do you never solve for the final product? Just leave it at 2^2^2*m?

[@pranaymarella2260]

Great video, explained very well.

[@ishanshmalviya1089]

hello in the 12th minute third question states for "any" A and not for "every" A, and since it is true in the case stated above, shouldnt the answer be yes...?

[@martindeveloper4856]

maybe just as a easier technique to get around with nested sets, so sets that contains another sets and so on. Since @TrevTutor has made somewhere in the past videos an visual example with boxes you open like amazon boxes containing another boxes, we can just simplify the things with substitution or in real world "just not opening the nested boxes". That works because neither in the cardinality nor in power sets are we interested in the nested sets, but simply on the outer sets in the set we're looking at. Meaning p({{a}}) can be just substituted as p({Z}), where Z = {a} thus p({Z)} = {{}, Z} so {{}, {a}} works also if the element inside Z is the empty set. So p({{}}) is the same as p({Z}) where Z = {} so {{}, Z} = {{}, {{}}}. I hope this is clear, typing with the computer and the sets is confusing.

[@tabolihk1]

that'd be better if the volume can be louder

[@Trevtutor]

Yeah. I'm not sure why the audio is lower on these ones. I'll make sure to fix that issue in future videos.

[@benjaminwest4852]

Take a shot every time he says "set" :D. Thanks for the video!

[@somedatussr4323]

No

[@venom830]

I have a question. You said, If a set "A" is inside a set "B" then "A" is a subset of "B". But in the case of power sets If set "A" is inside the power set "p(A)" then why is "A" not a subset?

[@NEMOBANDZBEATS]

Exactly I don’t understand

[@piastristan]

Same, I don't understand

[@coryanders6328]

I'm a little confused by the idea that A is not always a subset of the superset of A. The explanation shown on here just states that it's only an element (just because). Why?

[@nayeem.j.i]

A perfect complete course!!!

[@samaiatraforti9060]

Your'e amazing, thankyou for all your videos!

[@TrueCyberian]

1:00 There's a clear contradiction in the 1st example with what you say at the very beginning. {a, b} < {a, b, c}. It's strictly less. Therefore, A cannot be less or EQUAL to B. So the statement is False.

[@alittlebyte]

On 1:02 you said that {a,b} is a subset of {a,b,c}. Shouldn't it be a proper subset, as it contains less elements? Or do both subset and proper subset apply? Very nice videos btw :)

[@meikogatdula2770]

2:43 i think the first one will be a subset not a proper subset, because there is one element on Set B that is not on Set A. I am right? I’m sorry😐😶

[@temwanimutale3760]

Firstly your videos are awesome. In the last question of the exercise where did the 2 come from if the size of A is m. Why were you raising it to the power of two. And can't I write the final answer as 2^4m

[@john_paul]

I am confused with your example at 10:50. You define a subset relationship "A is a subset of a power set p(A)" as one where B contains every element which can be found in set A, and even with the example that you work on here it seems that it is actually the case, right? A = {Empty} and p(A) = {Empty, {Empty}}, where Empty is an element of both sets, so wouldn't A in this case be a subset of p(A)? It confused me because I believe it is but you wrote otherwise. Thanks!

[@john_paul]

Edit: You define a subset relationship "A is a subset of B"... not p(A).

[@simplyexplained875]

p(A) = { {a} }, A = { a }, therefore, A is not a subset of p(A)

[@supmethods]

You mentioned "@@simplyexplained875 p(A) = { {a} }, A = { a }, therefore, A is not a subset of p(A)". The video mentioned there was an exception and he seemed to have pointed out where p(Empty set) = {Empty set}, which seems incorrect to me because A is just the "empty set". The "empty set" does not equal to {"empty set"}. It seems he might have been talking about A = {Empty} and p(A) = {Empty, {Empty}}. Here the element "Empty" is in both p(A) and A, so A is a subset of p(A). Can you clarify that he was pointing this out or can you elaborate a bit more on why it's not the case?

[@jackpnguinne9956]

Really good content! I really wish your videos were recorded with a louder volume. 100% maxed vol and still have trouble hearing.

[@lunarbeats_sa]

I have a question about the empty set being in all set's. So could you say that a set is a box that for example can theoretically hold 5(subsets) and you only put 4(subsets) inside therefore the last space is technically filled by an empty set which would be 1 of the subsets? 🤔I may have just confused myself in that question but ok...

[@blablaman9]

Thank you for making these tutorials they really help, just if you could update the links on the website to these newer version tutorials. Thanks!

[@Trevtutor]

Yes, I totally forgot about that! The changes are there but I forgot to make them live.

[@sunsonny9132]

At 9:48, the answer for p({empty set}) should be {empty set, {{empty set}}}

[@zerocool1032]

Where can we practice these discrete mathematics problems?

[@mochhisyamtanzil9741]

3:50 i am confused why he seperate a and set containing a. arent those two have the same element which is a so A should be a subset of B

[@sukalyanroy765]

As far as I understand, "a" is an element and the set containing "a" is a separate and distinct another element. So that's why they are different.

[@salmasoul8628]

8:15 thanks for this

[@saumyojitdas4212]

@TheTevTutor in the tree part why do we start with an empty set ? because we know that it will be there for any set

[@Trevtutor]

We’re building all possible subsets, so we start with a set with nothing in it. Then we add an element or we don’t add an element in each step.

[@saumyojitdas4212]

@@Trevtutor my explanation and doubt to formation of the 2to the power n is kinda like this : We have a original set A={2,3} Now we have to create a powerset, we know that by definition a powerset will contain all the subsets of original set. For 2 I can add it to the power set or leave it . If I add it then power set becomes {{2}} If I leave it then power set has zero sets . Assume if I added 2 For 3 also if I add 3 then power set is {{3},{2}} If not then power set remains {{2}} If I haven't added any thing then power set is null . Then I have another option of adding the 2 elements at the same time which will make the power set {{2,3}} Now my question is are all the options mandatory means if u think there are total 4 options each 2 of 2 elements ; I have to take into account of every combination of any 2 options (at most atleast )taking one option at a time of each element . The picture will be 2 3 adding 2 |\ not add adding 3| ot Total I have 4 combination Why I have to take account of all the 4 options (2 at a time) Because I know subsets of 2,3 will be {2},{3},{2,3} and empty set will be automatically added. Thats why from this info we have created that 4 options algorithm.

[@saumyojitdas4212]

@TheTrevTutor please understand what i am trying to say above . My tree has the same concept i am not beginning from empty set as i know it is a subset of every set but my question is why and how we have created that 4 options algo just like i told above ..please reply

[@amahlesmahle7597]

Can you please help me with writing down a set of all odd numbers

[@yvng4697]

am literally confused yesterday till today, at 7:29 why is the answer is 64? What if in p|A| there was 4 or 5 elements?

[@DJK001]

To get it you do 2 power 5 you get 64!

[@yvng4697]

@@DJK001 thanks Jesse but i stopped learning it, but i will def return to learn it in the future!

[@DJK001]

@@yvng4697 oooh

[@husseinmatar4156]

is this playlist with discrete math 2 playlist enough for a discrete math university course ?

[@janniklovesicetea]

This is amazing, thank you!

[@sing759]

11:40 - isn't contradict your definition of power set? A power set contains all possible subsets. NOT element.. i am bit confused

[@RafaelRabinovich]

Calculating 2 ^ 6 by hand is excessively tedious, but programming it in Python is fairly simple. Then you get to see all the results listed.

[@sabinabeisembayeva8873]

I didn't understand the part, when you said, that A is not a subset of the power set of A, but A is an element of power set of A. Please, answer me, if it is possible:)

[@vishnuvikash5194]

Lets say A = empty set, and power set of A = {empty set}, Here A is not a subset of p(A) since empty set is an element of p(A) and not an subset. If p(A) = {empty set, {empty set}}, then A is a subset of p(A). But that does not apply here.