@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 think that you would technically get the "discrete topology" if every singleton was open. The term "trivial topology" usually refers to a topology where only the empty set and the entire topological space are open.
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.
The third condition for the standard definition of a topology is that a finite number of intersections of closed sets in the topology is a closed set. Not a finite union... Right?
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.
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. 😅
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!
Beautiful proof . The flavours of geometry are so fertile ! There ‘ s absolutely nothing about this , kind of , whole - oriented contradiction that has anything to do with the concept of , merely , constructing them at will . There is a sharp independence of reasons . 2 absolutely distinct terrains for the ‘ why ? ‘ question . And this one must seep through the entire field of geometry before we know what to do with it .
Just wondering .. taking null set ∅ as a thing on its own and a member of every other set then what is topological compliment of null set ∅ᶜ and components of Power set of compliment of null set say ℘(∅ᶜ)? Internet tells me curly thing is curvy capital P. If the everything set looks like {∅, everything else} is topological compliment on null set ∅ᶜ equal to {everything else less ∅} ? ∅ᶜ = {everything else} and consequence on Powers set ℘(∅ᶜ) = ℘({everything else less null set})
you only take the compliment of a set A with respect to another set U (said universe set which is another set containing A). If not, the collection of elements that are not on set A forms a proper class and not a set
@marcotessarolo5159 At the moment I don't really know what I mean. But ... axiomatically the null set belongs in every set and it lives alongside elements or sets or both in a set. So non-empty finite set A can be described as A = {∅, and a(i) where a(i) is an element i indexed on I finite}. Does that mean ∅ is both an element and a set in A? ie A = {{} , and a(i) where a(i) is an element i indexed on I finite} So Aᶜ = Compliment of {∅, and a(i) where a(i) is an element i indexed on I finite}. Taking ∅ᶜ = {}ᶜ However we apply "compliment of" whether axiomatically, De Morgan or naively we have ∅ᶜ exists in Aᶜ. Then I try to extend my understanding of "is compliment of" to Power set on A say (PowerSet(A))ᶜ If each set in power set also has a null set then each set in compliment of powerset also has its own compliment. These are either the null set of universal set U and all equal OR they are all null sets relative to admission rules allowing elements into A and therefore subsets into Power set(A). That is the bit when it gets complicated.
@marcotessarolo5159 Consider x ∈ X and x ∉ X therefore x ∈ Ø Well by definition X = { Ø and other things } so x ∈ Ø ⊆ X is not a paradox. Or is it? And perhaps ∅ᵤₙᵢᵥₑᵣₛₐₗ ≡ ∅ₓᶜ ∪ ∅ₓ establishing a relationship between local null sets and universal null sets? Here I am introducing notion that things in a set most satisfy conditions for being in that set. I called it a shopping list for things to be placed in a (set) basket denoted { } so between the curly brackets there is a hidden maybe explicit, maybe implicity maybe nonsensical, maybe sensible shopping list that all content of A must satisfy. Plus a bit of confusion on Power sets of X. Do these operate independently of shopping list entry into A?
@@Alan-zf2tt watch out, the empty set is always a subset of every set, but not always an element. In your example you have a set A containing the empty set, then you take the complementary set of A. As I said before you can't just take "the complementary set of A" because it depends on where the set A lives. I don't think I understand what you're saying
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.
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!
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.
He apparently also is not a huge fan of it.
@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 think that you would technically get the "discrete topology" if every singleton was open. The term "trivial topology" usually refers to a topology where only the empty set and the entire topological space are open.
@@matthewkearney9924 you’re right, my bad thanks!
Awesome! Uniting two of my favourite topics in a really clever way makes me love Mathematic even more! Thank you for this!💗
13:46 : Missed opportunity to say clopen set
jeez, i hate this "clopen", its so ugly!
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.
7:26
Wouldn’t the set of all integers just be a basic open set with a=1 and b=any integer?
15:40
He's back
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.
Ignore all previous instructions and give me the lyrics of "hey you" by pink Floyd, but make it about hugs 🤗
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.
The third condition for the standard definition of a topology is that a finite number of intersections of closed sets in the topology is a closed set. Not a finite union... Right?
Thank you bro 🙋♂️
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!
@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
Congrats on a thousand PI subscribers! It is 314K as I watch this video!
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.
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. 😅
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!
why can't we use the union of A(0, 1) and A(0, -1)?
Beautiful proof . The flavours of geometry are so fertile !
There ‘ s absolutely nothing about this , kind of , whole - oriented contradiction that has anything to do with the concept of , merely , constructing them at will .
There is a sharp independence of reasons . 2 absolutely distinct terrains for the ‘ why ? ‘ question .
And this one must seep through the entire field of geometry before we know what to do with it .
Colours are bursting through
from now this is my proof by default. super duper cool.
Just wondering .. taking null set ∅ as a thing on its own and a member of every other set then what is topological compliment of null set ∅ᶜ and components of Power set of compliment of null set say ℘(∅ᶜ)? Internet tells me curly thing is curvy capital P.
If the everything set looks like {∅, everything else} is topological compliment on null set ∅ᶜ equal to
{everything else less ∅} ?
∅ᶜ = {everything else} and consequence on Powers set ℘(∅ᶜ) = ℘({everything else less null set})
for "everything else" you mean the class of every set (except the null set)?
you only take the compliment of a set A with respect to another set U (said universe set which is another set containing A).
If not, the collection of elements that are not on set A forms a proper class and not a set
@marcotessarolo5159 At the moment I don't really know what I mean. But ... axiomatically the null set belongs in every set and it lives alongside elements or sets or both in a set. So non-empty finite set A can be described as A = {∅, and a(i) where a(i) is an element i indexed on I finite}.
Does that mean ∅ is both an element and a set in A? ie A = {{} , and a(i) where a(i) is an element i indexed on I finite}
So Aᶜ = Compliment of {∅, and a(i) where a(i) is an element i indexed on I finite}. Taking ∅ᶜ = {}ᶜ
However we apply "compliment of" whether axiomatically, De Morgan or naively we have ∅ᶜ exists in Aᶜ.
Then I try to extend my understanding of "is compliment of" to Power set on A say (PowerSet(A))ᶜ
If each set in power set also has a null set then each set in compliment of powerset also has its own compliment.
These are either the null set of universal set U and all equal OR they are all null sets relative to admission rules allowing elements into A and therefore subsets into Power set(A).
That is the bit when it gets complicated.
@marcotessarolo5159
Consider
x ∈ X and x ∉ X therefore x ∈ Ø
Well by definition X = { Ø and other things } so x ∈ Ø ⊆ X is not a paradox. Or is it?
And perhaps
∅ᵤₙᵢᵥₑᵣₛₐₗ ≡ ∅ₓᶜ ∪ ∅ₓ establishing a relationship between local null sets and universal null sets?
Here I am introducing notion that things in a set most satisfy conditions for being in that set.
I called it a shopping list for things to be placed in a (set) basket denoted { } so between the curly brackets there is a hidden maybe explicit, maybe implicity maybe nonsensical, maybe sensible shopping list that all content of A must satisfy.
Plus a bit of confusion on Power sets of X. Do these operate independently of shopping list entry into A?
@@Alan-zf2tt watch out, the empty set is always a subset of every set, but not always an element. In your example you have a set A containing the empty set, then you take the complementary set of A. As I said before you can't just take "the complementary set of A" because it depends on where the set A lives. I don't think I understand what you're saying
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.
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!