the formula for the powerset is because we have 2 choices for each element: be included or not, in a subset. it's a combinatory problem, how many ways there is to form a subset with "n" elements. 2 possibilities for the first element (be included or not), times 2 for the second, times 2 for the third, and so on, resulting in the formula given in the video.
Thank you for these videos. I prepared for my Lin A class this semester by doing a few playlists from native English speakers (as I'm a native speaker) but it has turned out to be completely useless for my LA class, which is in German. The German class is a lot more abstract and covers things none of the English playlists covered; only your playlist does (and I see you're German, so maybe that's why). It seems like the English playlists are about actually solving things, whereas my German class is just about writing definitions and then proving them... (hope that changes...).
6:20 The description of Capital Union: you use "there exists" i in I. Can we say "for each" instead of "there exists"? Do they mean the same? Or, is there a different notation for "for each"? Or "for every" or "for any" ?
was having the same question myself.. then I figured out something say, I = {1,2,3....} if Capital Union = {x | "for each" i in I : x in A_i} would mean that x has to be in A_1, A_2, A_3... at the same time. which would mean -> we are describing intersection instead
@@ganglinwu6835 Yes, right. It seems the word "exists" bother me. As I said in my first comment, the right word should be "for any". So, the fittest word must be "there exists".
Your video was so nice! I'm a college student interested in mathematics, and I have a question about implications involving the universal quantifier. (1) [∀x: x ∈ A ∪ B ↔ x ∈ A or x ∈ B] is true. (2) ∀x: x ∈ A ∪ B ⇔ x ∈ A or x ∈ B. (3) x ∈ A ∪ B ↔ x ∈ A or x ∈ B is true. (4) x ∈ A ∪ B ⇔ x ∈ A or x ∈ B. I believe all of these statements are the same. Is that correct? (Note: In the case of (3) and (4), 'x ∈ A ∪ B ↔ x ∈ A or x ∈ B' is not only a predicate but also a logical statement itself. For example, even if the predicate 'x or ~x' has no quantifiers, it still serves as a logical statement.)
Hi there! Your videos give such clarity with the examples and explanation! Thank you so much! I have a question.. you have mentioned that we can always build a new set from two sets A & B. But what if A and B are empty sets? Are we really building a new set from the union of two empty sets?
Thank you very much! You are absolutely right. The word "new" was not meant in the literal sense. It can happen that you get the same set back, for example, in the case that A = B.
Im not sure I understand this new big U notation. What do sets I and A represent here? I think set A represents a set with all other sets as subsets? And set I is effectively an itterator containing all positions in set A? But I'm not perfectly sure
@@gustavaggeboe7014 In his example it ended up being the set of natural numbers after the big union, but take this as an example: Define A_i := {2 * i} Then we can take the Big Union of A_i for i in the natural numbers so our index set, I, is just the set of natural numbers N = {1, 2, 3, ...}, therefore: A_1 = {2 * 1} = {2} A_2 = {2 * 2} = {4} A_3 = {2 * 3} = {6} ect... so the big union of A_i for i in the naturals = {2, 4, 6, ...} = the set of all positive even numbers notice how this union does not equal our index set N? Also Big union doesn't have to be an infinite union either, we could have: A_1 := {1, 2, 3, 4} A_2 := {3, 4, 7, 9} A_3 := {9, 1, 0 7} Now if we say our index set, I, is I := {1, 2, 3} (since we have A_1, A_2, A_3) then: Big Union of A_i for i in I = {x | x is an element of AT LEAST ONE of the A_1, A_2, A_3 sets} = A_1 U A_2 U A_3 = {1, 2, 3, 4} U {3, 4, 7, 9} U {9, 1, 0 7} = {1, 2 ,3, 4, 7, 9, 1, 0, 7} I think we could also have said Big union of A_i's is {1, 2, 3, 4, 3, 4, 7, 9, 9, 1, 0, 7} and it would be equivalent, but conventionally we don't put duplicate items in sets since it is redundant. We could also instead define our index set as I := {1, 2}, then the Big union of A_i's would have just been A_1 U A_2 = {1, 2, 3, 4, 7, 9} since we just aren't considering A_3 in that union anymore.
Hello! I'm not quite sure if I got the big union part down 100%, so could you check if this is correct? Let's say I := {1, 2, 3, 4} U Ai = A1 ∪ A2 ∪ A3 ∪ A4 where A1, A2, A3, A4 are different sets? i∈I Thank you!
the formula for the powerset is because we have 2 choices for each element: be included or not, in a subset.
it's a combinatory problem, how many ways there is to form a subset with "n" elements.
2 possibilities for the first element (be included or not), times 2 for the second, times 2 for the third, and so on, resulting in the formula given in the video.
Thanks, I had exactly this question in mind. Makes perfect sense!
Hi, why is the empty set not written in a set?
Thanks for your videos.
Such a good time.
The very important notion of a power set is really relevant down the road
yes
i am dyslexic, it so clear that i am learning v fast thank you..
I love your videos! I'm using them to help teach my sister discrete math :D
Keep up the great work!
That sounds great and you are very welcome!
Wonderful video, clear and nice explanations.
Thank you for these videos. I prepared for my Lin A class this semester by doing a few playlists from native English speakers (as I'm a native speaker) but it has turned out to be completely useless for my LA class, which is in German. The German class is a lot more abstract and covers things none of the English playlists covered; only your playlist does (and I see you're German, so maybe that's why). It seems like the English playlists are about actually solving things, whereas my German class is just about writing definitions and then proving them... (hope that changes...).
Thank you very much. I am glad that the videos can help you :)
Waiting patiently for the next video. Thank you very much.
thank you
شكرا و جزاك الله خيرا
6:20 The description of Capital Union: you use "there exists" i in I.
Can we say "for each" instead of "there exists"? Do they mean the same? Or, is there a different notation for "for each"?
Or "for every" or "for any" ?
was having the same question myself.. then I figured out something
say, I = {1,2,3....}
if Capital Union = {x | "for each" i in I : x in A_i}
would mean that x has to be in A_1, A_2, A_3... at the same time.
which would mean -> we are describing intersection instead
@@ganglinwu6835 Yes, right. It seems the word "exists" bother me. As I said in my first comment, the right word should be "for any". So, the fittest word must be "there exists".
Your video was so nice!
I'm a college student interested in mathematics, and I have a question about implications involving the universal quantifier.
(1) [∀x: x ∈ A ∪ B ↔ x ∈ A or x ∈ B] is true.
(2) ∀x: x ∈ A ∪ B ⇔ x ∈ A or x ∈ B.
(3) x ∈ A ∪ B ↔ x ∈ A or x ∈ B is true.
(4) x ∈ A ∪ B ⇔ x ∈ A or x ∈ B.
I believe all of these statements are the same. Is that correct?
(Note: In the case of (3) and (4), 'x ∈ A ∪ B ↔ x ∈ A or x ∈ B' is not only a predicate but also a logical statement itself.
For example, even if the predicate 'x or ~x' has no quantifiers, it still serves as a logical statement.)
Hi there! Your videos give such clarity with the examples and explanation! Thank you so much!
I have a question.. you have mentioned that we can always build a new set from two sets A & B.
But what if A and B are empty sets? Are we really building a new set from the union of two empty sets?
Thank you very much! You are absolutely right. The word "new" was not meant in the literal sense. It can happen that you get the same set back, for example, in the case that A = B.
@@brightsideofmaths thank you! again with much clarity in your replies as well!
Do you have a series on Topology?
In my manifolds series: tbsom.de/s/mf
Im not sure I understand this new big U notation. What do sets I and A represent here? I think set A represents a set with all other sets as subsets? And set I is effectively an itterator containing all positions in set A? But I'm not perfectly sure
Yes, for each i in I, we have a set A_i. We take the union of all of them :)
But how is this union different from the original set, I?@@brightsideofmaths
@@gustavaggeboe7014 In his example it ended up being the set of natural numbers after the big union, but take this as an example:
Define A_i := {2 * i}
Then we can take the Big Union of A_i for i in the natural numbers
so our index set, I, is just the set of natural numbers N = {1, 2, 3, ...}, therefore:
A_1 = {2 * 1} = {2}
A_2 = {2 * 2} = {4}
A_3 = {2 * 3} = {6}
ect...
so the big union of A_i for i in the naturals = {2, 4, 6, ...} = the set of all positive even numbers
notice how this union does not equal our index set N?
Also Big union doesn't have to be an infinite union either, we could have:
A_1 := {1, 2, 3, 4}
A_2 := {3, 4, 7, 9}
A_3 := {9, 1, 0 7}
Now if we say our index set, I, is I := {1, 2, 3} (since we have A_1, A_2, A_3) then:
Big Union of A_i for i in I = {x | x is an element of AT LEAST ONE of the A_1, A_2, A_3 sets}
= A_1 U A_2 U A_3
= {1, 2, 3, 4} U {3, 4, 7, 9} U {9, 1, 0 7}
= {1, 2 ,3, 4, 7, 9, 1, 0, 7}
I think we could also have said Big union of A_i's is {1, 2, 3, 4, 3, 4, 7, 9, 9, 1, 0, 7} and it would be equivalent, but conventionally we don't put duplicate items in sets since it is redundant.
We could also instead define our index set as I := {1, 2}, then the Big union of A_i's would have just been A_1 U A_2 = {1, 2, 3, 4, 7, 9} since we just aren't considering A_3 in that union anymore.
Have you this video in a playlist?
Yeah, I have :)
ua-cam.com/play/PLBh2i93oe2qtbygdXz4u6Mkh7c_hMLBA8.html
The Bright Side Of Mathematics lovely. Thank you. I love all your videos. 🙂
Thanks
Thanks a lot!
You are welcome :)
Merely asserting that the empty set is a subset of every set by definition or axiom does not logically explain why such proposition is true.
Hello! I'm not quite sure if I got the big union part down 100%, so could you check if this is correct?
Let's say I := {1, 2, 3, 4}
U Ai = A1 ∪ A2 ∪ A3 ∪ A4 where A1, A2, A3, A4 are different sets?
i∈I
Thank you!
Yes, correct :)