Is a vector space a Set?(Beginner's doubt)

[ask_LXXXVI]

A vector space is a collection of vectors. So can we say that it is a set (although with special properties) ? Just wanted to confirm this.

Is this also true for a vector space consisting of all functions from R to R , i.e can we say that it is also a set , where each member of the set is a function from R to R?

I mean can we say that "any vector space is a set, but any set may or may not be a vector space. "

 

[Fredrik]

The set of functions from ℝ into ℝ isn't a vector space, but the definitions

(f+g)(x)=f(x)+g(x)
(af)(x)=af(x)

turn it into one. It's the triple (set,addition operation,scalar multiplication operation) that should be called a vector space. However, we're only that careful with the details when we're stating the definition. After that, everyone gets sloppy and refers to the set as a vector space, even though this is technically incorrect. When the author of a book says that the set of functions from ℝ into ℝ is a vector space, he expects you to understand this.

 

[ask_LXXXVI]

It's the triple (set,addition operation,scalar multiplication operation) that should be called a vector space.

Thanks, that cleared the doubt!

 

[Studiot]

Thre is more to it than this.

A set is (by far) the more general object.

A set is a collection of uniquely identifiable or distinguishable objects.
There may be zero, one or more than one such object contained in the set.
The object, or elements, of a set may or may not have common properties, or even no properties at all except those by which we make the identification.
A set may contain all or only some of the possible objects with a particular property or collection of properties.

1)A vector space is a particular type of set where all the elements have a collection of common properties and all objects with these particular properties are elements of the set (vector space).
2)In addition the set itself has certain properties which exclude some collections that are sets but not vector spaces although they satisfy (1) above.

So yes

I mean can we say that "any vector space is a set, but any set may or may not be a vector space. "

The usual way to establish (prove) whether a given set is a vector space is to test is against all the axioms.
There are 9 in total, including the ones Frederik mentioned.
Others include the requirement for a zero vector and an identity vector to be in the set.

 

[Fredrik]

1)A vector space is a particular type of set where all the elements have a collection of common properties and all objects with these particular properties are elements of the set (vector space).

In all examples that I can think of, this is true (for example, they're all functions from X into Y, or they're all ordered pairs of real numbers), but the definition doesn't require anything like this.

2)In addition the set itself has certain properties which exclude some collections that are sets but not vector spaces although they satisfy (1) above.

This sounds like a reference to the vector space axioms. The axioms don't express properties of the set. (Let's call it V). They express properties of a function from V×V into V called addition and a function from ℝ×V into V called scalar multiplication. (This is assuming that we're talking about a vector space over the real numbers; if not, replace ℝ with some other field).

The usual way to establish (prove) whether a given set is a vector space is to test is against all the axioms.
There are 9 in total, including the ones Frederik mentioned.

I didn't mention any axioms. ask_LXXXVI gave us a set, and I supplied the other two members of the triple (set,addition,scalar multiplication). We would test if such a triple is a vector space by verifying all the vector space axioms, one at a time. The way I like to write them, there are 8.

Others include the requirement for a zero vector and an identity vector to be in the set.

The zero vector is the identity vector.

(It's the identity element of addition, which is the only binary operation in the triple).

 

[Studiot]

In all examples that I can think of, this is true (for example, they're all functions from X into Y, or they're all ordered pairs of real numbers), but the definition doesn't require anything like this.

Another way to put this is

if vector a is in the set and vector b is in the set then vector c = (a + b) is in the set.

This provides the basis () for the theorems about spanning and bases and many existence and uniquness theorems in applications.

My terminology on the rest of the axioms could certainly be brushed up - I didn't bother to check the list but will post a full set if the OP asks, although he should certainly look them up. I have seen 8, 9 or 10 set out depending upon how a particular author feels. Some are inherited from simpler (more general) mathematical structures.

However my point is that the way to test is to test (against the axioms).


I meant that

There is a vector 0 in the set such that a + 0 = a

 

[ask_LXXXVI]

Thanks , the axioms are readily available , so I can look them up.
But I am a bit confused again after the last two posts.

Probably one more example should clear this.
Lets say we have the set of all ordered pairs ℝxℝ .Can we say that this set in itself is a vector space ?

Or do we say that the triple consisting of (ℝxℝ,addition,scalar multiplication) is the vector space?

 

[Rasalhague]

Thanks , the axioms are readily available , so I can look them up.
But I am a bit confused again after the last two posts.

Probably one more example should clear this.
Lets say we have the set of all ordered pairs ℝxℝ .Can we say that this set in itsielf is a vector space ?

Or do we say that the triple consisting of (ℝxℝ,addition,scalar multiplication) is the vector space?

This is something that confused (and niggled) me too. You could define a vector space as a triple, e.g. V = ((vectors V, +:V-->V), (scalars S, +:S-->S, x:S-->S), scalar multiplication: SxV-->V), or perhaps a 6-tuple (V, +, S, +, x, scalar multiplication), and something like this is considered the correct way to think of it. The set, V, containing the vectors, in your example ℝxℝ, can be called the "underlying set" of the vector space.

In practice, though, as Fredrik has explained, it's normal usage to refer sloppily to the underlying set, ℝxℝ, itself as the vector space, even in formal mathematical writing. I suppose that's because, very often, there's one particular standard scalar multiplication function, vector addition function etc. conventionally associated with a particular set, such as componentwise addition for ℝxℝ, so that there's not much risk of ambiguity. Maybe it's also because the sloppy idiom fits better with the intuitive associations of the word "space", whereas the correct definition clashes with them a bit.

 

[Studiot]

Or do we say that the triple consisting of (ℝxℝ,addition,scalar multiplication) is the vector space?

If you are going down that road, where do you stop?

I regard a vector space as a set of elements that satisfy the axioms or possesses the properties.
If you like it is an Abelian group with some additional properties : It 'inherits' some (of the axioms on the list) properties from this.
Remember that some vector spaces possesses properties additional to the axiomatic list, such as vector multiplication.

Pure mathematicians love to squash definitions down so a vector space is

A Module over a Field.

 

[disregardthat]

You could model a vector space in set theory as an ordered triple ((V,+),(F,+',.),*) where (V,+) is an abelian group, (F,+',.) a field and * the group action of F on V satisfying the vector space axioms. Here V and F are sets, + a commutative operator, satisfying the group axioms, and +' and . satisfying the field axioms as addition and multiplication on F respectively.

The + operator is a function from V x V to V, and similarly for . and +'. which have well-known models in set theory, namely sets of ordered pairs (subset of Domain x Codomain) in which each first coordinate have a unique second-coordinate. So all you have is sets.

You will need a similar model in order to make sense of vector spaces in set theory. Sets are very generic in set theory; the set of integers by itself does not know that 1+1 = 2. It doesn't make sense unless addition is a defined function. In order to "reach it" and prove things in formal syntactical notation you will need to order your information properly (e.g. as ordered tuples), and provide all information needed in such a way.

Outside of formal set theory however we are all satisfied by calling a vector space a set for which we have given some extra structure.

 

[Fredrik]

I regard a vector space as a set of elements that satisfy the axioms or possesses the properties.

Again, those are properties of functions associated with the set, not properties of the set.

I agree with the other things you said in this post and the one before it.

Lets say we have the set of all ordered pairs ℝxℝ .Can we say that this set in itself is a vector space ?

No. When you say "the set in itself", you're making it clear that you're referring to the set alone, not the set and the two functions I mentioned. The set isn't a vector space without those functions. Another way of looking at it is that there are infinitely many vector space structures on that set, one for each way to choose those functions. If addition is defined by

(a,b)+(c,d)=(a+c,b+d)

and we define two scalar multiplication functions, S and S' by

S(a,(b,c))=(ab,ac)
S'(a,(b,c))=(2ab,2ac),

then (ℝxℝ,addition,S) and (ℝxℝ,addition,S') are two different vector spaces, with the same underlying set.

Also note that we can define other structures on ℝxℝ. Let's define addition by

(a,b)+(c,d)=(a+c,b+d),

scalar multiplication by

a(b,c)=(ab,ac),

multiplication by

(a,b)(c,d)=(ac-bd,ad+bc),

an inner product by

(a,b)·(c,d)=ab+cd,

and a norm by

$$\|(a,b)\|=\sqrt{a^2+b^2}.$$ℝxℝ is a set.
(ℝxℝ,addition) is an abelian group.
(ℝxℝ,addition,scalar multiplication) is a vector space over ℝ.
(ℝxℝ,addition,multiplication) is a field. (To be more specific, it's the field of complex numbers).
(ℝxℝ,addition,scalar multiplication,inner product) is a Hilbert space over ℝ.
(ℝxℝ,addition,scalar multiplication,norm) is a Banach space over ℝ.
(ℝxℝ,addition,scalar multiplication,norm,multiplication) is a Banach algebra over ℝ.

It's also fine to say that the pair (G,scalar multiplication), where G is the abelian group (ℝxℝ,addition), is a vector space over ℝ. No one cares if what you have in mind when you say "the vector space ℝxℝ" is the triple (ℝxℝ,addition,scalar multiplication) or the pair ((ℝxℝ,addition),scalar multiplication).

Note that if you say e.g. that "ℝxℝ with the standard definitions of addition and scalar multiplication is a vector space" you're not committing yourself to either of the choices (G,scalar multiplication) or (ℝxℝ,addition,scalar multiplication), but you're showing that you understand that those functions need to be a part of it.

 

[ask_LXXXVI]

Thanks Fredrik, your latest post made it comprehensively clear now!

 

[Studiot]

Again, those are properties of functions associated with the set, not properties of the set.

Again I say that it depends upon the set.

If you choose your set only from the elements of the set of all sets with the particular function then the function comes automatically and you don't need to repeat it in the axiom list.

But again my point is that this is really playing with words either way.

To actually establish that a particular set is a vector space you have to work through the axiom tests either by saying the set is a member of the set of all sets with (say) a zero element or you specifically display a zero element.

Practical applications over more than a century now have established which are the most useful list of axioms.

calling a vector space a set for which we have given some extra structure.

Hear hear.

 

[disregardthat]

Again I say that it depends upon the set.

What do you mean by this?

 

[Studiot]

I hoped to have explained this statement in the three paragraphs immediately following it.

Was there something confusing in these paragraphs?

 

[disregardthat]

I hoped to have explained this statement in the three paragraphs immediately following it.

Was there something confusing in these paragraphs?

Yes, it didn't make any sense at all.

If you choose your set only from the elements of the set of all sets with the particular function then the function comes automatically and you don't need to repeat it in the axiom list.

There is no "set of all sets". A set doesn't come with a function.

To actually establish that a particular set is a vector space you have to work through the axiom tests either by saying the set is a member of the set of all sets with (say) a zero element or you specifically display a zero element.

A set by itself doesn't have an operation associated with it. So you can't work through the axioms to check whether it is a vector space. It doesn't have a field associated with it either.

$$\mathbb{R}$$ can be an abelian group, a field, a vector space over $$\mathbb{Q}$$ and $$\mathbb{R}$$, a topological space in various topologies, a linear continuum, etc. It doesn't come with any of these properties as a set.

 

[Studiot]

Yes, it didn't make any sense at all.

I don't know whether you are extracting the Michael or what

Let me provide a few examples.

The set of all continuous functions

The set of all colours obtainable by mixing blue and red

Both of these come with built in functions.

Let us consider the second as it is simpler.

Let p% be the proportion of blue then (100-p)%) is the proportion of red

Thus the set contains all possible values of

p/100B + (100-p)/100R...F1

Now suppose we were to consider a second set with p <= 35.

This is obviously a subset.

But it inherits F1

 

[disregardthat]

Both of these come with built in functions.

Not in set theory. As I said, sets does not "come with" built in functions on them. They may be defined afterwards.

 

[Fredrik]

The set of all continuous functions

The set of all colours obtainable by mixing blue and red

Both of these come with built in functions.

Let us consider the second as it is simpler.

Let p% be the proportion of blue then (100-p)%) is the proportion of red

Thus the set contains all possible values of

p/100B + (100-p)/100R...F1

I'm not sure I understand what you mean by F1. It looks like it's just a member of the set. I'll call the set S. Scalar multiplication is a function from ℝxS into S. What such function "comes with" the set S?

 

[Rasalhague]

You will need a similar model in order to make sense of vector spaces in set theory.

We often have cases where there exists a mathematical idea, such as "vector space" or "the field of real numbers" or "function" or "counting", and to check that our intuitions about this idea are correct, we need to formalise it in some way. So we define the idea in terms of set theory. But there's usually more than one way to define the idea in terms of sets. E.g. I've seen the idea of a function defined in terms of tuples, which in turn can be defined as embedded sets (in various ways), but also the idea of tuples defined in terms of a function. So a particular choice of formalism seems more like a representation of the idea, or as you called it a model, than the idea itself. In some sense, we consider certain models equivalent ways to formally characterise an idea; but I suppose the lack of a set of sets might make it impossible to define the idea literally as equivalence classes of structures that behave in a certain way. I wonder if maybe, if the idea of sets hadn't been chosen for this role, some other idea could have been taken as fundamental, and sets defined in terms of that.

 

[sponsoredwalk]

You could model a vector space in set theory as an ordered triple ((V,+),(F,+',.),*) where (V,+) is an abelian group, (F,+',.) a field and * the group action of F on V satisfying the vector space axioms. Here V and F are sets, + a commutative operator, satisfying the group axioms, and +' and . satisfying the field axioms as addition and multiplication on F respectively.

The + operator is a function from V x V to V, and similarly for . and +'. which have well-known models in set theory, namely sets of ordered pairs (subset of Domain x Codomain) in which each first coordinate have a unique second-coordinate. So all you have is sets.

This is very interesting, ((V,+),(F,+',°),•) clears up a lot of conceptual dissonance I've been
feeling & I'd like to ask you about it if you don't mind.

In set theory you can define a set (X,Y,+) where + is a relation. A relation is a subset of
the cartesian product of the sets X & Y so + ⊆ X × Y := { (x,y) | (x ∈ X) ⋀ (y ∈ Y) }.

I'm trying to figure out how to think about the set ((V,+),(F,+',°),•) along those lines
clearly if you'd help me with this. In the set (V,+), + : V × V → V correct? In the set
(F,+',.) we have (+' : F × F → F) & ( ° : F × F → F) right? I justify this by saying that
the sets +,+' & ° are just subsets of the cartesian product (i.e. the set of ordered pairs
of all possible configurations & the sets +,+' & ° are simply a rule that determines what
particular pairs are chosen from the cartesian product). As for •, you called it
"the group action of F on V", now I have never come across anything like this, could
you could explain this? Maybe what you said is just another way to talk about relations
on sets?

I can't put my finger on it, I am thinking that • : F × V → V but if I mimic the language
that I used when I defined (X,Y,+) I would write the definition of • as • : (V,+) × (F,+',°) → (V,+).
Whatever the technical details, I see that • : F × V → V represents scalar multiplication
of vectors, i.e. that • : (x,u) ↦ (xu) & this notation really codifies that idea for me.

I think this way of looking at vector spaces is clearer as (V,+) is an abelian group, (F,+',°)
is a field & you need not write a big list of axioms each time you write the definition of a
vector space & with • I think what I've written above makes sense, it explains why
those particular axioms are chosen when scalars act on vectors where before it seemed
a bit contrived.

Let me know