If you're looking for more, I've got about a million other videos on power sets and subsets. Here is a fun one, the power set of the power set of the power set of the empty set: th-cam.com/video/d6k-qSbys4g/w-d-xo.html Thanks for watching and share the video if you enjoy the Christmas lessons!
Your videos are very helpful, thank you so much!!! You speak slowly and clearly and are easy to follow along with. I just wanted to say thank you again!!!
Thanks for watching and for the request! I'll be making some lessons about cardinality, rationals and irrationals and so on, to go at the beginning of my real analysis playlist. I'm not sure when it will be though, I'll see what I can do this month!
I have understood it ! Thanks.. I have a question .. Number of subsets of set A= 2 n(A) ( 2 n to the power A) Number of subsets of set A=32 Then n(A)=?? Can you say how to do this...
I think you are missing one part of provement for the total count of subsets that includes a(k+1) is 2^k. I understand for every subset of {a1, a2, ... ak}, if we throw a a(k+1), it would definitely be a subset of the total set S, but why are you sure the total count of subsets that includes a(k+1) is just 2^k? Why could not be another subset which is not coming from throwing a a(k + 1) into the existing subsets of {a1, a2, ... ak}? I think for this part, we can prove it using contradiction. I mean this contradiction might seem to be simple, but it needs this contradiction provement to rigorously prove the statement. In another words, there is a hidden theorem or something in this statement, which is for the total subsets of {a1, a2, ... a(k+1)}, there are exactly half would contains a(k+1) and exactly another half would not. The distribution of it should be right on 50/50. But there should be a provement for it. I would think using contradiction can prove it very easily.
Thanks for watching and good question! This is a proof by induction, so if you aren’t familiar with that technique I’d watch a couple videos on it with other examples too. I have plenty. The idea is that we prove it holds for n=0 (and maybe n=1? i don’t remember the video exactly). The idea is that we’re trying to prove an infinite “ladder” of statements (the rungs of the ladders are cardinalities of sets: n=0,1,2,3,…). When we prove the n=0 case, we’re proving that we can get on the ladder. What remains to be proven is that if we are on any rung of the ladder, we can move to the next one. Thus, we assume the statement is true for some n (which is valid because we proved it was true for n=0), then show it must be true for the next n [that is, a set with n+1 elements has 2^(n+1) subsets. EDIT: The short answer to your question is that the “it” you refer to is a set with n+1 elements.
If you're looking for more, I've got about a million other videos on power sets and subsets. Here is a fun one, the power set of the power set of the power set of the empty set: th-cam.com/video/d6k-qSbys4g/w-d-xo.html
Thanks for watching and share the video if you enjoy the Christmas lessons!
Your videos are very helpful, thank you so much!!!
You speak slowly and clearly and are easy to follow along with.
I just wanted to say thank you again!!!
Thanks so much, I'm glad you've found my lessons helpful! Let me know if you ever have any questions!
im confused wouldn't {a vk+1} also be counted in the cardinality of S? So wouldn't | P(s) | = 2 ^k+1 + 1?
Question;prove by method of induction that if A has n elements, then |P(A)|=2the power n.
answer please sir
Thank you so much sir
Glad to help, thanks for watching!
Awesome lecture..... So very clear!!!!! Thankyou sir!
My pleasure, thanks for watching!
Thank you for this video! I was just wondering, when you combine the subsets from k and k + 1, what happens to the empty set? Is it counted twice?
5:03 how did you know to even think of union?
haha fr, but it feels so obvious after he shows it
Please, make video about cardinality of real irrational numbers.
Also thank you very much for this great video...since it contains cool proof.
Thanks for watching and for the request! I'll be making some lessons about cardinality, rationals and irrationals and so on, to go at the beginning of my real analysis playlist. I'm not sure when it will be though, I'll see what I can do this month!
Thanks bro
Glad to help - thanks for watching!
Brilliant!!
Thank you so much
My pleasure, thanks for watching!
I have understood it ! Thanks..
I have a question ..
Number of subsets of set A= 2 n(A)
( 2 n to the power A)
Number of subsets of set A=32
Then n(A)=??
Can you say how to do this...
2^5
So 5 elements
I think you are missing one part of provement for the total count of subsets that includes a(k+1) is 2^k. I understand for every subset of {a1, a2, ... ak}, if we throw a a(k+1), it would definitely be a subset of the total set S, but why are you sure the total count of subsets that includes a(k+1) is just 2^k? Why could not be another subset which is not coming from throwing a a(k + 1) into the existing subsets of {a1, a2, ... ak}? I think for this part, we can prove it using contradiction. I mean this contradiction might seem to be simple, but it needs this contradiction provement to rigorously prove the statement.
In another words, there is a hidden theorem or something in this statement, which is for the total subsets of {a1, a2, ... a(k+1)}, there are exactly half would contains a(k+1) and exactly another half would not. The distribution of it should be right on 50/50. But there should be a provement for it. I would think using contradiction can prove it very easily.
I do not understand if we are proving the a set a will havd 2^n subsets why do you prove that it has 2^(n+1)? I don't get it. Thanks
Thanks for watching and good question! This is a proof by induction, so if you aren’t familiar with that technique I’d watch a couple videos on it with other examples too. I have plenty. The idea is that we prove it holds for n=0 (and maybe n=1? i don’t remember the video exactly). The idea is that we’re trying to prove an infinite “ladder” of statements (the rungs of the ladders are cardinalities of sets: n=0,1,2,3,…). When we prove the n=0 case, we’re proving that we can get on the ladder.
What remains to be proven is that if we are on any rung of the ladder, we can move to the next one. Thus, we assume the statement is true for some n (which is valid because we proved it was true for n=0), then show it must be true for the next n [that is, a set with n+1 elements has 2^(n+1) subsets.
EDIT: The short answer to your question is that the “it” you refer to is a set with n+1 elements.
@@WrathofMath you are a teacher every student would dream to have🙏
thanks boss
You're very welcome!