What I thought is, if I labelled subset as {a1,a2,...,ap}, and defined
ai∗aj=a(i×pj)
then I think, I get a cyclic group. I assume the operation is associative, but it follows from the associativity of multiplication modulo.
Then, if I define f(ai)=i, I get that
f(ai∗aj)=f(a(i×pj))=i×pj=f(ai)×pf(aj)
and this is clearly bijective, so it is an isomorphism to Zp.
In this case I get a cyclic group, for all sets with a prime number of elements. However, I need this to be true for every number of elements, and if it helps, they will be naturals.
I see what you're doing. You need to be quite careful here: Zn is not (usually) constructed a subset of N, or even of Z, and it's not a subgroup of either under any conventional operations. It is normally either defined abstractly or defined concretely a quotient of Z by the subgroup nZ.
Now, Zn is not a group under multiplication modulo n, even when n is prime (for instance, p has no multiplicative inverse modulo p). And it is always a group under addition, even when n isn't prime.
However, Zp∗is a group under multiplication; its elements are {1,2,⋯,p−1} with multiplication defined by a×pb=ab(modp). But Zp∗≅Zp−1, and the correspondence is not as simple as mapping, say, 1 to 1 and 2 to 2 and 3 to 3 and so on... the isomorphism will mess up the ordering of the integers.
And on this basis it doesn't make a difference that your indices are natural numbers, since if you want them to act multiplicatively then you'll need to screw up their ordering anyway in order to map them into Zp∗. So you can choose between making them act additively, so for instance ai∗aj=ai+nj, in which case it makes no difference whether or not n is prime; or you can choose another group and group operation and choose your isomorphism a different way.
So you can choose between making them act additively, so for instance ai∗aj=ai+nj, in which case it makes no difference whether or not n is prime; or you can choose another group and group operation and choose your isomorphism a different way.
I see. Well, it was the first one that I picked! So, if I choose them additively, then it seems plausible? I'll have a look at it now, and check the details to see whether it works. Thanks!
EDIT: I know what you mean by the above though, you didn't waste time to write it all out. I did it quickly, and slightly mispresented it.
I see. Well, it was the first one that I picked! So, if I choose them additively, then it seems plausible? I'll have a look at it now, and check the details to see whether it works. Thanks!
Yes. But you seem to be making it harder than you have to: by labelling the elements of your subset from 1 up to n you have already defined an isomorphism from your subset to Zn, by the method I described two posts ago, since the group structure is inherited from the identification with the elements of Zn.
A more critical issue is that not all subsets of N are finite, but I'm sure you can work out a way round that...
Yes. But you seem to be making it harder than you have to: by labelling the elements of your subset from 1 up to n you have already defined an isomorphism from your subset to Zn, by the method I described two posts ago, since the group structure is inherited from the identification with the elements of Zn.
A more critical issue is that not all subsets of N are finite, but I'm sure you can work out a way round that...
Yes, that is true, but I want to play a bit more with it.
It was just an idea, so I have to think through the details, and see how exactly I'm going to apply it. Well, more specifically, I'm looking at some periodic sequences, so that's not a problem for now.
More specifically, if A⊂N is any subset and (G,∗) is a group with ∣A∣=∣G∣ then any bijection f:A→G induces a group structure (A,⋅) on A by defining a⋅b=f−1(f(a)∗f(b)).
This is quite nice! I think I have seen it already used for the Isomorphism theorem, but obviously didn't process it well enough!
It's often the case that statements of the form "every set has [such and such a structure]" are equivalent to the axiom of choice (or a weaker variant of it), a famous example being "every set has a well-ordering". Most people accept the axiom of choice and get on with their lives, but it is quite interesting to know when it's being used I thought that the assertion that every set has a group structure depended on the axiom of choice, but I didn't know they were equivalent!
.. a famous example being "every set has a well-ordering".
So, that's the final ingredient of the argument! I have heard this axiom so many times that one may think I know what it means, but I find it interesting already!
Yup, that's fine, but the wording is slightly awkward and I can't put my finger on why. When dealing directly with the definition of continuity it's good to be more specific with ε and δs, so something like:
Spoiler
It sounds like more of a mouthful, but that way you're not skipping over details.
Why is this called the Axiom of ``Extensionality", expressed as: ∀x(x∈A≡x∈B)⇒A=B? It does not 'extend' anything.
It doesn't mean 'extension' as in 'make something bigger', it means extension as in 'the ability to include more things'. The reasoning goes a bit deeper into relations.
Background: A relation R on a set X can be thought of in a few ways (all are equivalent). One way is as a subset R⊆X×X, which means it's a collection of ordered pairs (x,y) where x,y∈X. (An example of a relation is the set X×X itself, which can be thought of as the set of all 'coordinates'.) Relations generalise functions: a function f is a relation for which for each x there is exactly one y such that (x,y)∈f, and we denote this y by f(x). In this sense, we can think of a relation as a 'multi-valued function'. (For instance, the set containing all (x,y) where y=x or y=−x is a relation, but not a function.) But the nitty-gritty aside, just think of a relation R as being a set containing ordered pairs of elements of some set.
Now, if (x,y)∈R then we say x is an 'R-predecessor' of y. So for instance if f is a function then you could say (although most people don't) that x is an f-predecessor of f(x) for each x in the domain of f.
The relevant bit: A relation R is called 'extensional' if each element y∈X is determined by its R-predecessors. That is, if for all x∈X we have (x,y)∈R⇔(x,y′)∈R, then y=y′. The axiom of extensionality, therefore, is just the assertion that ∈ is an extensional relation.
It doesn't mean 'extension' as in 'make something bigger', it means extension as in 'the ability to include more things'. The reasoning goes a bit deeper into relations.
Background: A relation R on a set X can be thought of in a few ways (all are equivalent). One way is as a subset R⊆X×X, which means it's a collection of ordered pairs (x,y) where x,y∈X. (An example of a relation is the set X×X itself, which can be thought of as the set of all 'coordinates'.) Relations generalise functions: a function f is a relation for which for each x there is exactly one y such that (x,y)∈f, and we denote this y by f(x). In this sense, we can think of a relation as a 'multi-valued function'. (For instance, the set containing all (x,y) where y=x or y=−x is a relation, but not a function.) But the nitty-gritty aside, just think of a relation R as being a set containing ordered pairs of elements of some set.
Now, if (x,y)∈R then we say x is an 'R-predecessor' of y. So for instance if f is a function then you could say (although most people don't) that x is an f-predecessor of f(x) for each x in the domain of f.
The relevant bit: A relation R is called 'extensional' if each element y∈X is determined by its R-predecessors. That is, if for all x∈X(x,y)∈R⇒(x,y′)∈R, then y=y′. The axiom of extensionality, therefore, is just the assertion that ∈ is an extensional relation.
Most textbooks introduce it as you go along; the notation I used would all be in a typical first year textbook. I can't really think of any to suggest though. The table on Wikipedia isn't very useful but might help a little bit.
The relevant bit: A relation R is called 'extensional' if each element y∈X is determined by its R-predecessors. That is, if for all x∈X(x,y)∈R⇒(x,y′)∈R, then y=y′. The axiom of extensionality, therefore, is just the assertion that ∈ is an extensional relation.
I see. Just for clarity, this notion of a ``relation" is as general as it can be. e.g. I don't attribute it any properties other than being a subset of a Cartesian product?
I am still trying to decipher the last bit though.
Is x∈X(x,y)∈R⇒(x,y′)∈R the same as
If for all x∈X, we have xRy⇒xRy′, then y=y′?
In this case, I get that the axiom of extensionality would be: R is ∈ and y=A,y′=B.
Following this logic, would = and ≡ be extensional relations?
I see. Just for clarity, this notion of a ``relation" is as general as it can be. e.g. I don't attribute it any properties other than being a subset of a Cartesian product?
Well it's nothing more than that - a subset of a Cartesian product (of a set with itself). Relations are extremely useful (equivalence relations, partial orders and preorders are all types of relation). They can be generalised further, say to n-ary relations, which are subsets of Xn or to subsets of X×Y, or even more general products.
Following this logic, would = and ≡ be extensional relations?
Yes. But, for instance, if we write H≤G to mean that H is a subgroup of a group G (where we consider the groups as being 'up to isomorphism'); then ≤ is not extensional (for instance, D6 and C6 have the same subgroups).
Yes. But, for instance, if we write H≤G to mean that H is a subgroup of a group G (where we consider the groups as being 'up to isomorphism'); then ≤ is not extensional (for instance, D6 and C6 have the same subgroups).
Oh, yes. I wasn't very sure about the meaning of = in that case (definition), but took it as being the same object.
Would you recommend Naive Set Theory? Or, perhaps, any other book that might be useful -- I notice most of these have no prerequisites. Comments suggest it is a well structured introduction that covers all axioms of ZF? (It's funny how naive is this book with this axiomatic nature. )
Oh, yes. I wasn't very sure about the meaning of = in that case, but took it as being the same object.
Would you recommend Naive Set Theory? Or, perhaps, any other book that might be useful -- I notice most of these have no prerequisites. Comments suggest it is a well structured introduction that covers all axioms of ZF? (It's funny how naive is this book with this axiomatic nature. )
I didn't learn naive set theory from a book, and as far as I know there isn't a copy of the lecture notes online, so I can't really say. I've heard of that book, though, which probably means it's not too bad.
Promise me you won't just do maths this summer? I'm a bit worried.
Would you recommend Naive Set Theory? Or, perhaps, any other book that might be useful -- I notice most of these have no prerequisites. Comments suggest it is a well structured introduction that covers all axioms of ZF? (It's funny how naive is this book with this axiomatic nature. )