@5:30 I believe you need a to be a nonzero integer for this proof to work, otherwise arithmetic sequences can be singletons and we get the trivial topology.
You’re right. Michael forgot to mention that. Otherwise constant integer sequences would be open sets which destroys the proof as it is necessary for nonempty finite sets to not be open in this topology
I came across the following variation which avoids the use of topological concepts but is still elegant. Define a subset A of Z to be periodic if there exists a > 0 such that x in A iff x + a in A. Note that the union of two periodic sets (with periods a and b) is again periodic (with period ab) and that the complement of a periodic set is again periodic. Also, a periodic set is either empty or infinite. Assume there are finitely many primes. We list them: p_1, ..., p_n Let A(p) be the set of all multiples of p. So A(p) is periodic. As the video explains, the union of A(p_1), ..., A(p_n) = Z \ { -1,1}. A finite union of periodic sets is periodic, so Z\{-1,1} is periodic, so its complement {-1,1} is periodic. However, it is clear that {-1,1} cannot be periodic since it is neither empty nor infinite. So our assumption leads to a contradiction. Q.E.D. Note that this is NOT Euclid's proof.
Every time I see primes, I think, ‘Okay, how’s this going to blow my mind?’ This video did not disappoint! Topology’s role here is fascinating, and SolutionInn has been my secret weapon for grasping these advanced concepts.
But this IS Euclid’s proof. There, you have a finite number of primes p1, p2, …, pk, and you generate a new one by considering the prime divisors of p1*p2*…*pk ± 1. This can’t be any of the original primes because it’s ±1 mod each of those, so it must be a new prime. Here, you consider a finite number of primes p1, p2, …, pk, and you look at the closed set A_{p1,0}∪A_{p2,0}∪…∪A_{pk,0} and it doesn’t contain ±1. Why is it closed? Because it’s a finite union of closed sets, because it’s the complement of a finite intersection of open sets, which is open because… why? Well, either such an intersection is empty, or there is a point that is in each set. But then the arithmetic difference of the numbers that define the open sets, when multiplied, give a difference that can define their total intersection. In other words, A_{x*y, a} ⊆ A_{x,a} ∩ A_{y,a}. This is the proof that this system is a topology, that Michael swept under the rug. And so if ±1 is not in the union of these closed sets, then ±1 IS in each of their complements. But then the entire arithmetic progression A_{p1*p2*…*pk, ±1} is in the intersection of these open sets, and in particular, p1*p2*…*pk±1 is not in A_{p1,0}∪A_{p2,0}∪…∪A_{pk,0} either. So it can’t be just {±1}, and thus any finite set of primes is insufficient and the set of primes is infinite. Notice how closely these mirror each other: in both cases, we proved that A_{p1,0}∪A_{p2,0}∪…∪A_{pk,0} was disjoint from A_{p1*p2*…*pk, ±1}; just in one place, we did it in not this much generality, and in the other, we wrapped it up in the proof that the system was a topology. This is the key to Euclid’s proof, so these are in fact the same!
But this IS Euclid’s proof. There, you have a finite number of primes p1, p2, …, pk, and you generate a new one by considering the prime divisors of p1*p2*…*pk ± 1. This can’t be any of the original primes because it’s ±1 mod each of those, so it must be a new prime.
Here, you consider a finite number of primes p1, p2, …, pk, and you look at the closed set A_{p1,0}∪A_{p2,0}∪…∪A_{pk,0} and it doesn’t contain ±1. Why is it closed? Because it’s a finite union of closed sets, because it’s the complement of a finite intersection of open sets, which is open because… why? Well, either such an intersection is empty, or there is a point that is in each set. But then the arithmetic difference of the numbers that define the open sets, when multiplied, give a difference that can define their total intersection. In other words, A_{x*y, a} ⊆ A_{x,a} ∩ A_{y,a}. This is the proof that this system is a topology, that Michael swept under the rug.
And so if ±1 is not in the union of these closed sets, then ±1 IS in each of their complements. But then the entire arithmetic progression A_{p1*p2*…*pk, ±1} is in the intersection of these open sets, and in particular, p1*p2*…*pk±1 is not in A_{p1,0}∪A_{p2,0}∪…∪A_{pk,0} either. So it can’t be just {±1}, and thus any finite set of primes is insufficient and the set of primes is infinite.
Notice how closely these mirror each other: in both cases, we proved that A_{p1,0}∪A_{p2,0}∪…∪A_{pk,0} was disjoint from A_{p1*p2*…*pk, ±1}; just in one place, we did it in not this much generality, and in the other, we wrapped it up in the proof that the system was a topology. This is the key to Euclid’s proof, so these are in fact the same!
An intuitive explanation of open/closed from continuous metric spaces (with the topology induced by the metric and e.g. the real number line): open sets = contain none of their boundary points closed sets = contain all of their boundary points neither open nor closed = contain some but not all BPs both open and closed ("clopen") = contain both none and all BPs = therefore have no BPs Doesn't really seem to help in this context though. 😅
You need to show that the arithmetic sequences and their unions can even constitute a topology, and proving this is probably more complicated than the rest of this proof...
What you need to show is that any element of the intersection of two basic open sets U1, U2 is a member of some third basic open set U3 which is fully contained in the intersection of U1 and U2, i.e. U3 is a subset of U1 and U3 is a subset of U2. Let's work this out for two sets U1 = A(a,b) and U2 = A(c,d). U1 = A(a,b) = { an+b : n in Z} = all elements in Z which are equal to b mod a. U2 = A(c,d) = { cn+d : n in Z} = all elements in Z which are equal to d mod c. Let s be an element in the intersection of U1 and U2. Then pick U3 = A(ac, s). Clearly, s is in U3. Since s is in U1, we know that s is equal to b mod a. All elements in U3 are equal to s mod ac and therefore equal to s mod a and therefore equal to b mod a. So U3 is contained in U1. By the same argument, U3 is contained in U2. Q.E.D.
Equiivalent, but a bit more elegant, is to note that if s is an element of the intersection of A(a,b) and A(c,d), then A(a,b) = A(a,s) and A(c,d) = A(c,s), and both contain A(ac,s).
@@expl0s10n Yeah, but this point is really the whole ballgame. You need number theoretic input equivalent to the infinitude of primes in order to show it, so leaving it as an exercise and then appealing to it to show the infinitude of primes as a consequence seems like question begging. I don't think dressing all of this up in the language of topology is really adding anything.
@@TonyCox-fh1po"number theoretic input equivalent to the infinitude of primes". What? No, not at all. It's just saying that the intersection of A(x,b) and A(x,c) is A(x,bc). That's extremely simple.
It would be fun to do this not by contradiction, just because you can construct a union of closed sets without making a claim about whether it is infinite or not, and then say that a non-closed union of closed sets is infinite. It's a neat trick to use an operation you can actually perform on a potentially infinite set of integers and get a result with an informative property.
Interesting! and Interesting? Open interval zero to three eg excluding 0 and 3, has compliment closed interval infinity to zero union closed set 3 to infinity. If so I likeee very muchlee! Trying in symbol-speak compliment(0,3) in reals ℝ is [-∞,0] union [3, ∞] And a hmmm @ 9:27 definitions? A(5,4) is a 'naively' open set and may be a 'conditionally' closed set. Equivalently the chained disjoint union of A(5,0) thru A(5,3) is both naively open and conditionally closed as the compliment of A(5,4). This is how math should be: a voyage of discovery If so I like this very much. Too often in math conditional definitions and conditionality seems second place to absolute definitions and absoluteness . So I propose: naive definitions and conditional definitions as part of axiomatic structure and pedagogical presentations. Absolute definitions, conditional definitions, relative definitions = yummy!
The number of likes on the video before I hit the button was 314, I almost felt bad about hitting it. And the number of comments right now is 31, I guess it's just my day to destroy π occurrences.
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
It is worth mentioning that this proof was provided by Hillel Furstenberg while he was still an undergraduate student in 1955.
@@yuan-jiafan9998 indeed. It would be nice if the videos in this channel included a reference to the relevant paper where applicable.
@5:30 I believe you need a to be a nonzero integer for this proof to work, otherwise arithmetic sequences can be singletons and we get the trivial topology.
You’re right. Michael forgot to mention that. Otherwise constant integer sequences would be open sets which destroys the proof as it is necessary for nonempty finite sets to not be open in this topology
13:46 : Missed opportunity to say clopen set
15:40
He's back
Can I suggest a problem? Prove that no group of order 2*p^2 is simple. I personally found two really nice proofs, one that uses linear algebra.
I came across the following variation which avoids the use of topological concepts but is still elegant.
Define a subset A of Z to be periodic if there exists a > 0 such that x in A iff x + a in A.
Note that the union of two periodic sets (with periods a and b) is again periodic (with period ab) and that the complement of a periodic set is again periodic.
Also, a periodic set is either empty or infinite.
Assume there are finitely many primes. We list them: p_1, ..., p_n
Let A(p) be the set of all multiples of p. So A(p) is periodic.
As the video explains, the union of A(p_1), ..., A(p_n) = Z \ { -1,1}.
A finite union of periodic sets is periodic, so Z\{-1,1} is periodic, so its complement {-1,1} is periodic. However, it is clear that {-1,1} cannot be periodic since it is neither empty nor infinite.
So our assumption leads to a contradiction. Q.E.D.
Note that this is NOT Euclid's proof.
Every time I see primes, I think, ‘Okay, how’s this going to blow my mind?’ This video did not disappoint! Topology’s role here is fascinating, and SolutionInn has been my secret weapon for grasping these advanced concepts.
It is really a nice idea, but nothing can beat the greatness of Euclid's.
But this IS Euclid’s proof. There, you have a finite number of primes p1, p2, …, pk, and you generate a new one by considering the prime divisors of p1*p2*…*pk ± 1. This can’t be any of the original primes because it’s ±1 mod each of those, so it must be a new prime.
Here, you consider a finite number of primes p1, p2, …, pk, and you look at the closed set A_{p1,0}∪A_{p2,0}∪…∪A_{pk,0} and it doesn’t contain ±1. Why is it closed? Because it’s a finite union of closed sets, because it’s the complement of a finite intersection of open sets, which is open because… why? Well, either such an intersection is empty, or there is a point that is in each set. But then the arithmetic difference of the numbers that define the open sets, when multiplied, give a difference that can define their total intersection. In other words, A_{x*y, a} ⊆ A_{x,a} ∩ A_{y,a}. This is the proof that this system is a topology, that Michael swept under the rug.
And so if ±1 is not in the union of these closed sets, then ±1 IS in each of their complements. But then the entire arithmetic progression A_{p1*p2*…*pk, ±1} is in the intersection of these open sets, and in particular, p1*p2*…*pk±1 is not in A_{p1,0}∪A_{p2,0}∪…∪A_{pk,0} either. So it can’t be just {±1}, and thus any finite set of primes is insufficient and the set of primes is infinite.
Notice how closely these mirror each other: in both cases, we proved that A_{p1,0}∪A_{p2,0}∪…∪A_{pk,0} was disjoint from A_{p1*p2*…*pk, ±1}; just in one place, we did it in not this much generality, and in the other, we wrapped it up in the proof that the system was a topology. This is the key to Euclid’s proof, so these are in fact the same!
But this IS Euclid’s proof. There, you have a finite number of primes p1, p2, …, pk, and you generate a new one by considering the prime divisors of p1*p2*…*pk ± 1. This can’t be any of the original primes because it’s ±1 mod each of those, so it must be a new prime.
Here, you consider a finite number of primes p1, p2, …, pk, and you look at the closed set A_{p1,0}∪A_{p2,0}∪…∪A_{pk,0} and it doesn’t contain ±1. Why is it closed? Because it’s a finite union of closed sets, because it’s the complement of a finite intersection of open sets, which is open because… why? Well, either such an intersection is empty, or there is a point that is in each set. But then the arithmetic difference of the numbers that define the open sets, when multiplied, give a difference that can define their total intersection. In other words, A_{x*y, a} ⊆ A_{x,a} ∩ A_{y,a}. This is the proof that this system is a topology, that Michael swept under the rug.
And so if ±1 is not in the union of these closed sets, then ±1 IS in each of their complements. But then the entire arithmetic progression A_{p1*p2*…*pk, ±1} is in the intersection of these open sets, and in particular, p1*p2*…*pk±1 is not in A_{p1,0}∪A_{p2,0}∪…∪A_{pk,0} either. So it can’t be just {±1}, and thus any finite set of primes is insufficient and the set of primes is infinite.
Notice how closely these mirror each other: in both cases, we proved that A_{p1,0}∪A_{p2,0}∪…∪A_{pk,0} was disjoint from A_{p1*p2*…*pk, ±1}; just in one place, we did it in not this much generality, and in the other, we wrapped it up in the proof that the system was a topology. This is the key to Euclid’s proof, so these are in fact the same!
Colours are bursting through
After this, I understand more about sets than ever before!
Not sure what I do with this information, but at least I have it now, so thanks!
from now this is my proof by default. super duper cool.
An intuitive explanation of open/closed from continuous metric spaces (with the topology induced by the metric and e.g. the real number line):
open sets = contain none of their boundary points
closed sets = contain all of their boundary points
neither open nor closed = contain some but not all BPs
both open and closed ("clopen") = contain both none and all BPs = therefore have no BPs
Doesn't really seem to help in this context though. 😅
You need to show that the arithmetic sequences and their unions can even constitute a topology, and proving this is probably more complicated than the rest of this proof...
What you need to show is that any element of the intersection of two basic open sets U1, U2 is a member of some third basic open set U3 which is fully contained in the intersection of U1 and U2, i.e. U3 is a subset of U1 and U3 is a subset of U2.
Let's work this out for two sets U1 = A(a,b) and U2 = A(c,d).
U1 = A(a,b) = { an+b : n in Z} = all elements in Z which are equal to b mod a.
U2 = A(c,d) = { cn+d : n in Z} = all elements in Z which are equal to d mod c.
Let s be an element in the intersection of U1 and U2. Then pick U3 = A(ac, s). Clearly, s is in U3.
Since s is in U1, we know that s is equal to b mod a.
All elements in U3 are equal to s mod ac and therefore equal to s mod a and therefore equal to b mod a. So U3 is contained in U1.
By the same argument, U3 is contained in U2.
Q.E.D.
Equiivalent, but a bit more elegant, is to note that if s is an element of the intersection of A(a,b) and A(c,d), then A(a,b) = A(a,s) and A(c,d) = A(c,s), and both contain A(ac,s).
He mentioned this point at 5:50, he know he needs to check but this is left as an exercise to the readers
@@expl0s10n Yeah, but this point is really the whole ballgame. You need number theoretic input equivalent to the infinitude of primes in order to show it, so leaving it as an exercise and then appealing to it to show the infinitude of primes as a consequence seems like question begging. I don't think dressing all of this up in the language of topology is really adding anything.
@@TonyCox-fh1po"number theoretic input equivalent to the infinitude of primes". What? No, not at all. It's just saying that the intersection of A(x,b) and A(x,c) is A(x,bc). That's extremely simple.
It would be fun to do this not by contradiction, just because you can construct a union of closed sets without making a claim about whether it is infinite or not, and then say that a non-closed union of closed sets is infinite. It's a neat trick to use an operation you can actually perform on a potentially infinite set of integers and get a result with an informative property.
Thank you bro 🙋♂️
@9:30 - I can't quite see which open sets we can make that are not closed sets. Please could someone provide an example?
Sorry I asked this before I got through the whole video. I didn't realise that a=/=0 in the definition of the the basic sets
Interesting! and Interesting? Open interval zero to three eg excluding 0 and 3, has compliment closed interval infinity to zero union closed set 3 to infinity. If so I likeee very muchlee!
Trying in symbol-speak compliment(0,3) in reals ℝ is [-∞,0] union [3, ∞]
And a hmmm @ 9:27 definitions? A(5,4) is a 'naively' open set and may be a 'conditionally' closed set. Equivalently the chained disjoint union of A(5,0) thru A(5,3) is both naively open and conditionally closed as the compliment of A(5,4). This is how math should be: a voyage of discovery
If so I like this very much. Too often in math conditional definitions and conditionality seems second place to absolute definitions and absoluteness .
So I propose: naive definitions and conditional definitions as part of axiomatic structure and pedagogical presentations.
Absolute definitions, conditional definitions, relative definitions = yummy!
The number of likes on the video before I hit the button was 314, I almost felt bad about hitting it. And the number of comments right now is 31, I guess it's just my day to destroy π occurrences.