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!
"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."
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...
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)?
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.
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)
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.
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.
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)
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
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 :)
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!
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
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% 🥳
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.
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
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.
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?
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.
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
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 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.
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?
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!
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".
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).
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.
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.
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.
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:)
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.
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.
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 .
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.
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
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.
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?
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 :)
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!!
Wow you are amazing! I was wondering if you can clarify: if c is an element of S and d is an element of S, (c,d) implies that it is a subset of P(S)??? or Element of P(S) ? which one is always true? i am really confused with this.... would love to hear from you =)
"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
If the power set is the set containing all subset of a particular set as its elements. Then why isn’t the empty set written in curly brackets with the empty symbol inside it, but all the other subset are written in curly brackets?
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...
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
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 ?.
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.
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.
![[Discrete Mathematics] Subsets and Power Sets Examples](http://i.ytimg.com/vi/4hC8y9rVanw/mqdefault.jpg)
![[Discrete Mathematics] Subsets and Power Sets Examples](/img/tr.png)
![[UNCUT]"ตุลามรณะ"ย้อนฟังคำทำนาย หมอดูทุกสาย ดวงเมืองร้อน (หมอปลาย ฟองสนาน อ.โอเล่)I คนดังนั่งเคลียร์](http://i.ytimg.com/vi/lE5az5VvT9M/mqdefault.jpg)
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!
"If your professor is sadistic" 😂😂
Too bad he is!
Bahahaha!!!!!! Right!!
Best username ever lol
"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."
WHYYYY ISSS THISSS SOOOO TRUEEEE
Professor code for "Fuck if I know."
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...
@@nicholascunningham6936 Yeah that's a pretty common situation. Discrete Math seems to be a terribly taught class at a lot of universities.
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)?
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.
I agree
Saya minggu ke 4
damn... this is in chapter 2 of what we're doing
Damn I learnt this in week 3
So have you graduated yet😂😁
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)
hope this right cause I did understand it like this :D
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.
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.
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)
THANK YOU
power sets are wild! im rewatching that part so many times
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
great explanation i understood a math concept for first time in my life . great respect to you sir !! We need people like you !!
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 :)
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!
How was the exam ?
at 7:42 where he had to rework and count out 2^6 gives me hope for myself. thank you
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
"if your professor is a little bit sadistic " sounds about right
Sir.... Hats off to you, you are doing a great job. These videos are helping, we really appreacite that.
Perfect timing, i have discMath exam in 5 days
me too lol. mine is 5 days from today ;(
I start discrete math in 7 days. I'm just trying to learn what it's about.
Im doing mine right now
So weird. It' been 2 years
Thank you for making this chapter so interesting with your teaching style. I really appreciate it..
Great explanation but it would help if the sets are visually represented with circles and elements
Perfect timing, i have a discMath exam in 2 hours
So, how did it go? Did the video help?
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
Thanks man I would be absolutely lost right now without your videos
You're explanation of 10:50 was Fu****g awesome!! I was confused after the lecture but your videos are saving me :D
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% 🥳
so how was it?
@@Robotomy101 OH SHOOT.. I FORGOT TO PUT THE GRADE 😭
Thx for reminding me 😭💕
Ur videos are awesome!! its saving me for my semester test, TYSM!!!
thank you
Wow man, I am very impressed with your teaching style.
Honestly. Thank you for make it so easy and fun to listen to.❤️
đón chờ những ca khúc tiếp theo của Phúc, càng nghe càng thích giọng ca của Phúc ❤️
I got some sadistic ass professors then.
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.
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
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
This was the explanation for size of powersets I was looking for!
Power sets very well explained. Especially the last |P(P(P(A)))| example, Thank You Trev :)
I like how you draw brackets in 5:45 , it is somehow relatable.
I am loving these videos i now have a better understanding on set theory thank yu very much trev
Hello
dude your handwriting is amazing
Thanks a lot man, You don't know how much you helped me.
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.
I don’t understand what you are asking when you say “for any A?” Can you elaborate?
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?
I’m reviewing this playlist for my theory of computation class
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.
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
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.
Feel free to correct me if I'm wrong tho woops.
@@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.
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?
Exactly I don’t understand
Same, I don't understand
Very well explained! Thank you
Thank you for becoming my teacher because my uni teacher isn't great.
I wish the uni lecture notes can explain as well as u do XD Thank you for the awesome videos XD
me and the boys enjoyed this
Chad
my respect is yours glad that you make these videos
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!
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".
:(
same
Precisely
Same here, I don't understand
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).
its so interesting,before you explained the subset definition you quickly jump to the definition of p(A) .thanks for the video.
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.
Really good content! I really wish your videos were recorded with a louder volume. 100% maxed vol and still have trouble hearing.
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!
Yes, I totally forgot about that! The changes are there but I forgot to make them live.
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.
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.
thank you! your videos are really very helpful.
I found my professor. Thank you.
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:)
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.
You're really good!! I'll be taking this course in my fall semester.
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}
thank you so much
yo thank you for this playlist honestly!
thank you so much for making the videos , it really helps a lot !
Just got started learning discrete math yesterday because of boredom
I have a question teacher:
In the set A={a,b}
Is a an subset of A? I think not, but is an element.
Thank you!! This was helpful, you made it easy to understand :)
Your'e amazing, thankyou for all your videos!
Take a shot every time he says "set" :D. Thanks for the video!
No
that'd be better if the volume can be louder
Yeah. I'm not sure why the audio is lower on these ones. I'll make sure to fix that issue in future videos.
I like your voice! Thank you so much for your videos!
Thank you for this! Super helpful :)
A perfect complete course!!!
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.
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 .
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.
Great video, explained very well.
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
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.
i really wanna send this guy $10 and a hug!
Thank you! This is extremely helpful. You are an excellent teacher!
Thank you!
What are the differences between subsets & proper subsets? Thanks.
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?
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 :)
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!!
I'm talking about the second Q notthe third one :)
Wow you are amazing! I was wondering if you can clarify: if c is an element of S and d is an element of S, (c,d) implies that it is a subset of P(S)??? or Element of P(S) ? which one is always true? i am really confused with this.... would love to hear from you =)
"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
Sadistic and hilarious
The perfect professor
If the power set is the set containing all subset of a particular set as its elements. Then why isn’t the empty set written in curly brackets with the empty symbol inside it, but all the other subset are written in curly brackets?
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.
great tutorail. thanks
U give us really important thing
And good way to review this by the way
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...
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
I'm still confused but maybe I'll just rewatch this one
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 ?.
Sure, you could. It meets both definitions.
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.
Or I misunderstood Subsets? 😅😃
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.
Thanks a lot 👌❤
thank you
This is amazing, thank you!
Wow. Thank you. ❤