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!
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.
"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 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 :)
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 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!
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
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.
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% 🥳
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
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
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!
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.
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.
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 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.
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.
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).
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?
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 .
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.
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...?
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.
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?
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 :)
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.
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 ?
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.
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...
![[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)
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!
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😂😁
"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)?
"If your professor is sadistic" 😂😂
Too bad he is!
Bahahaha!!!!!! Right!!
Best username ever lol
power sets are wild! im rewatching that part so many times
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 :)
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
at 7:42 where he had to rework and count out 2^6 gives me hope for myself. thank you
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 ?
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
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
Sir.... Hats off to you, you are doing a great job. These videos are helping, we really appreacite that.
Thank you for making this chapter so interesting with your teaching style. I really appreciate it..
Thanks man I would be absolutely lost right now without your videos
"if your professor is a little bit sadistic " sounds about right
You're explanation of 10:50 was Fu****g awesome!! I was confused after the lecture but your videos are saving me :D
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
dude your handwriting is amazing
thank you
đón chờ những ca khúc tiếp theo của Phúc, càng nghe càng thích giọng ca của Phúc ❤️
Honestly. Thank you for make it so easy and fun to listen to.❤️
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.
Ur videos are awesome!! its saving me for my semester test, TYSM!!!
Wow man, I am very impressed with your teaching style.
I like how you draw brackets in 5:45 , it is somehow relatable.
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
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
Power sets very well explained. Especially the last |P(P(P(A)))| example, Thank You Trev :)
me and the boys enjoyed this
Chad
Thank you for becoming my teacher because my uni teacher isn't great.
This was the explanation for size of powersets I was looking for!
Thanks a lot man, You don't know how much you helped me.
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 😭💕
I found my professor. Thank you.
I am loving these videos i now have a better understanding on set theory thank yu very much trev
Hello
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
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 like your voice! Thank you so much for your videos!
Very well explained! Thank you
I’m reviewing this playlist for my theory of computation class
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
my respect is yours glad that you make these videos
thank you so much
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!
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.
thank you! your videos are really very helpful.
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}
its so interesting,before you explained the subset definition you quickly jump to the definition of p(A) .thanks for the video.
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.
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
Hi so do you never solve for the final product? Just leave it at 2^2^2*m?
yo thank you for this playlist honestly!
Really good content! I really wish your videos were recorded with a louder volume. 100% maxed vol and still have trouble hearing.
thank you so much for making the videos , it really helps a lot !
Your'e amazing, thankyou for all your videos!
A perfect complete course!!!
Great video, explained very well.
I wish the uni lecture notes can explain as well as u do XD Thank you for the awesome videos XD
I don’t understand what you are asking when you say “for any A?” Can you elaborate?
Thank you!! This was helpful, you made it easy to understand :)
i really wanna send this guy $10 and a hug!
Just got started learning discrete math yesterday because of boredom
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.
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.
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.
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.
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).
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
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 .
Take a shot every time he says "set" :D. Thanks for the video!
No
Thank you for this! Super helpful :)
something worth to watch
God bless you sir
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 👌❤
You're really good!! I'll be taking this course in my fall semester.
Thank you!
Thank you! This is extremely helpful. You are an excellent teacher!
great tutorail. thanks
I got some sadistic ass professors then.
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...?
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.
thank you
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?
thank you
8:15 thanks for this
Wow. Thank you. ❤
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 :)
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.
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?
This is amazing, thank you!
so any subset we find inside a power set is considered an element of this power set
correct me if am wrong
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 ?
It usually depends on what you want to prove. But the equals sign just means it’s a smaller set or the same set.
@@Trevtutor understood thanks
putting this vid on 1.5, way better for my focus
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.
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 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...
I have a question teacher:
In the set A={a,b}
Is a an subset of A? I think not, but is an element.