SUBSETS AND POWER SETS - DISCRETE MATHEMATICS

Today we look at subsets and power sets. This includes the empty set, and the power set of the empty set.
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-Playlists-
Discrete Mathematics 1: • Discrete Math (Sets, L...
Discrete Mathematics 2: • Discrete Math (Countin...
-Recommended Textbooks-
Discrete and Combinatorial Mathematics (Grimaldi): amzn.to/2T0iC53
Discrete Mathematics (Johnsonbaugh): amzn.to/2Hh7H41
Discrete Mathematics and Its Applications (Rosen): amzn.to/3lUgrMI
Book of Proof (Hammack): amzn.to/35eEbVg
Hello, welcome to TheTrevTutor. I'm here to help you learn your college courses in an easy, efficient manner. If you like what you see, feel free to subscribe and follow me for updates. If you have any questions, leave them below. I try to answer as many questions as possible. If something isn't quite clear or needs more explanation, I can easily make additional videos to satisfy your need for knowledge and understanding.

[@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😂😁

[@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 :)

[@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)

[@yb7535]

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 !!

[@Shiroyashasama]

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

[@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 ?

[@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

[@user-kx2cq5gz4y]

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

[@slyanna3688]

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

[@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.

[@Towhid-ze3rw]

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

[@wirito]

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

[@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

[@ivasto6182]

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

[@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 😭💕

[@louisdiaz4179]

I got some sadistic ass professors then.

[@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.

[@kateye6513]

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

[@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.

[@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.

[@MrRobot-gm9cv]

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

[@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

[@SquirtIeSquad]

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

[@kayleenicholelopez]

I have a midterm tonight and I’m absolutely lost on the symbols and what their functions are. I appreciate these videos and the explanations. I’m probably going to bomb the test, but it’s because of my brain lol

[@aboutthereality179]

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

[@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

[@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.

[@drakeultima9146]

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

[@JVenom_]

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

[@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 ❤️

[@shubhrogupta9409]

me and the boys enjoyed this

[@agastyasingh3764]

Chad

[@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?

[@Tortuex_]

thank you

[@jordantaylor5552]

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

[@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}

[@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.

[@walternyarukuhwe3180]

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

[@paulcarello6390]

Given: A = {a,b} then P(A) = { { },{a},{b},{a,b} } Note: A is a set, and P(A) is a set of sets ... this is important . Now, any subset of P(A) must also be a set of sets - eg { { } } ⊆ P(A), { {a} } ⊆ P(A), { {b} } ⊆ P(A), { {a,b} } ⊆ P(A), { { },{a} } ⊆ P(A) ... etc. But {a,b} is an ∈ P(A), it is not in a set

[@ozgurkuzu2202]

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

[@user-ke2rp4gz2c]

dude your handwriting is amazing

[@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!

[@martijn130370]

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

[@detectivedoom5175]

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

[@padayaomartinjanc.3958]

Just got started learning discrete math yesterday because of boredom

[@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

[@Jeff-on7ug]

Same, I don't understand

[@Eis461]

thank you so much

[@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.

[@rchtchauhan]

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

[@isurusandaruwan4769]

I found my professor. Thank you.

[@jayant9151]

My friend, a math professor, puts an empty plastic bag on the table when he teaches that. - What's on the table? - An empty bag. - What's in the bag? - Nothing. That's the same thing. The set \{ \emptyset \} contains one element, a set which happens to be empty. The set \emptyset contains nothing at all.

[@Fabius11k]

Very well explained! Thank you

[@yuriibech6691]

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

[@CrisRorens]

my respect is yours glad that you make these videos

[@Lena-of7wd]

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

[@benjaminwest4852]

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

[@somedatussr4323]

No

[@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.

[@justozo3130]

putting this vid on 1.5, way better for my focus

[@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

[@jackpnguinne9956]

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

[@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 .

[@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.

[@carmenx3789]

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

[@yuehernkang]

thank you! your videos are really very helpful.

[@sunsonny9132]

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

[@yoitslemonboy6988]

yo thank you for this playlist honestly!

[@pranaymarella2260]

Great video, explained very well.

[@nayeem.j.i]

A perfect complete course!!!

[@SebHaugeto]

Thank you for this! Super helpful :)

[@umutberkorhan9579]

God bless you sir

[@sben125]

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

[@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...?

[@salmasoul8628]

8:15 thanks for this

[@caloz.3656]

"if your professor is sadistic" um he made us watch this video and in the exercise section question 2 he made us do p(p(p(p(phi)))) which has 16 elements instead of the normal question bruh

[@kylegarcia3639]

Sadistic and hilarious The perfect professor

[@zaylo9273]

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

[@KmInthu-m1i]

thank you

[@samaiatraforti9060]

Your'e amazing, thankyou for all your videos!

[@materialknight]

I find those trees a bit confusing, I prefer them as a truth tree, the following way (you'll have to imagine the branches connecting the truth values): Assume the set {a, b, c}; the members of its powerset would be those sets that lie at the end of each branch of the following truth tree: a b c T : {a, b, c} T F : {a, b} T F T : {a, c} F : {a} T : {b, c} T F : {b} F F T : {c} F : {} This shows why in every thruth table, the number of horizontal lines (rows or valuations) is always 2^n, where n is the number of letters; in turn, it also explains why |P(A)| = 2^|A|, since it's the same principle (every element can be or not be, and for every instance where one is, the next one can be or not be); in this case, we have 3 letters, so their truth table should have 2^3 = 8 rows, and it does: a b c T T T T T F T F T T F F F T T F T F F F T F F F This shows the straightforward procedure to make truth tables: Calculate the number of rows, and for the first letter, mark half the rows as T, half as F, and for each half, make sure the next letter has half the rows as T and half as F, rinse and repeat until the table is complete.

[@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 👌❤

[@azatgaripov2380]

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.

[@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

[@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.

[@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.

[@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.

[@vector2-074]

I have a question teacher: In the set A={a,b} Is a an subset of A? I think not, but is an element.

[@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).

[@Hacklberry18]

Thank you!

[@soundcodetutorials1825]

great tutorail. thanks

[@janniklovesicetea]

This is amazing, thank you!

[@mcity6043]

something worth to watch

[@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?

[@manarelbashier5944]

Thank you very much

[@Watch519]

I think there is a mistake in the third Q in Tricky Questions : the empty set cannot be an ELEMENT of the power set of the set containing the empty set or any other set , the empty set is a subset of every set but never an elemnt. Please correct me if am wrong!!

[@Watch519]

I'm talking about the second Q notthe third one :)

[@aboitoo]

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

[@GitaGhasemi]

Wow. Thank you. ❤

[@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...