Deos this question make sense to anyone?

[Emspak]

Homework Statement

Let S be a nonempty set and K a field. Let C(S,K) denote the set of all functions $f \in C(S,K)$ such that $f(s) = 0$ for all but a finite number of elements of S.

Prove that C(S,K) is a vector space.

OK, I can't make the slightest sense of this. What in the name of all that is holy does the value of f(0) have to do with C(S,K) being a vector space? Can anyone at least hep me figure out what the heck I am supposed to even start with?

 

[Dick]

Homework Statement

Let S be a nonempty set and K a field. Let C(S,K) denote the set of all functions $f \in C(S,K)$ such that $f(s) = 0$ for all but a finite number of elements of S.

Prove that C(S,K) is a vector space.

OK, I can't make the slightest sense of this. What in the name of all that is holy does the value of f(0) have to do with C(S,K) being a vector space? Can anyone at least hep me figure out what the heck I am supposed to even start with?

I don't think there's to much to prove. Function spaces are pretty generally vector spaces. I think you just want to prove that if f and g vanish for all but a finite number of elements of S then so do f+g and k*f for k in K. Just show it's closed under addition and scalar multiplication. Though you didn't say what the scalars are supposed to be for this vector space, I'm just guessing.

 

[Emspak]

Where in the world does g come from?

I can see that f(s) = 0 should be a part of C(S,K). But that's an intuition more than anything else. I know that in order to be a vector space you have to be able to add and do scalar multiplications; there are several axioms in the book that define vector spaces. Well and good. But I am completely, utterly, totally baffled by this problem because I can't get a handle on what they are asking.

 

[Dick]

Where in the world does g come from?

I can see that f(s) = 0 should be a part of C(S,K). But that's an intuition more than anything else. I know that in order to be a vector space you have to be able to add and do scalar multiplications; there are several axioms in the book that define vector spaces. Well and good. But I am completely, utterly, totally baffled by this problem because I can't get a handle on what they are asking.

Well, if f(s) is zero for all but a finite number of elements of S and ditto for g(s), then does f(s)+g(s) vanish for all but a finite number of elements of S? Like I said, they may just want an argument for closure. Vector spaces have to be closed under addition and scalar multiplication. Isn't that one of your axioms?

 

[Emspak]

The axioms for the definition of vector space we have are:

(V is a Vector space, F is Field)

1. for all x,y in vector space V x+y = y+ x
2. for all x, y, z in V (x + y) + z = x + (y + z)
3. there exists an element in V denoted by 0 s.t. x + 0 = x for each x in v
4. for each element x in V there exits an element y in V s.t. x + y = 0
5. for each element x in V 1x = x
6. for each pair of elements a, b, in F and each element x in V, (ab)x = a(bx)
7. for each element a in F and each pair of elements x,y in V, a(x+y) = ax + ay
8. for each pair of elements a, b in F and each element x in V (a+b)x = ax + ab

we did do a whole exercise of showing closure under addition in certain cases, I think, but I am not entirely sure what this has to do with anything here. Closure under addition and multiplication is required for somethign to be a subspace, no?

I did see a proof like this one with the problem stated as:

That one is stated like this: Let S be a nonempty set and F a field. Let C (S,F) denote the set of all functions f ∈ F (S,F) such that f(s) = 0 for all but a finite number of elements of S . Prove that C(S,F) is a subspace of F(S,F).

Proof (as stated in the book)
For f(s) = 0, the ”zero” element of F(S,F) is clearly in C (S, F).

(It is not clear to me at all, really, but I'll take it as it is for the moment)

Let f(s) = c_i, for c_i ≠ 0 , s_i ∈ S, i = 1, ..., n, and g(t_i) = d_i for d_i ≠ 0 , t_i ∈S, i = 1 , .., m
and for all other s ∈ S let f(s) = 0 , g(s) = 0.
Now, ( f + g )(s) = 0 for all s ∈ S except for s = s_i, i = 1 , .., n and s = t i , i = 1 , ..., m . Since f + g is again a function from S to F, that vanishes at all but a finite number of elements of S . Similarly, one can show αf ∈ C(S, F).

Again, none of this is as obvious as the text makes it out, I am looking at this and not sure if I can adapt this to what I see in front of me. I am feeling like this is mandarin right now.

 

[Dick]

The axioms for the definition of vector space we have are:

(V is a Vector space, F is Field)

1. for all x,y in vector space V x+y = y+ x
2. for all x, y, z in V (x + y) + z = x + (y + z)
3. there exists an element in V denoted by 0 s.t. x + 0 = x for each x in v
4. for each element x in V there exits an element y in V s.t. x + y = 0
5. for each element x in V 1x = x
6. for each pair of elements a, b, in F and each element x in V, (ab)x = a(bx)
7. for each element a in F and each pair of elements x,y in V, a(x+y) = ax + ay
8. for each pair of elements a, b in F and each element x in V (a+b)x = ax + ab

we did do a whole exercise of showing closure under addition in certain cases, I think, but I am not entirely sure what this has to do with anything here. Closure under addition and multiplication is required for somethign to be a subspace, no?

I did see a proof like this one with the problem stated as:

That one is stated like this: Let S be a nonempty set and F a field. Let C (S,F) denote the set of all functions f ∈ F (S,F) such that f(s) = 0 for all but a finite number of elements of S . Prove that C(S,F) is a subspace of F(S,F).

Proof (as stated in the book)
For f(s) = 0, the ”zero” element of F(S,F) is clearly in C (S, F).

(It is not clear to me at all, really, but I'll take it as it is for the moment)

Let f(s) = c_i, for c_i ≠ 0 , s_i ∈ S, i = 1, ..., n, and g(t_i) = d_i for d_i ≠ 0 , t_i ∈S, i = 1 , .., m
and for all other s ∈ S let f(s) = 0 , g(s) = 0.
Now, ( f + g )(s) = 0 for all s ∈ S except for s = s_i, i = 1 , .., n and s = t i , i = 1 , ..., m . Since f + g is again a function from S to F, that vanishes at all but a finite number of elements of S . Similarly, one can show αf ∈ C(S, F).

Again, none of this is as obvious as the text makes it out, I am looking at this and not sure if I can adapt this to what I see in front of me. I am feeling like this is mandarin right now.

That's fine. You are probably just thinking this is harder than it is. If the {di} are finite set and the {ti} are a finite set then their union (or intersection, which is what you really want) is a finite set. I think that's really all there to this problem. Same thing for scalar multiplication.

 

[Emspak]

so g is just an arbitrary function? It looks pulled out of nowhere.

Arrrgh. This is the problem. The proof I showed you looks like just magic right now and this is pretty frustrating.

Here's what I can sort of figure out. we have f(s) = 0 and we can say that it is in C(S,K) butI have o idea whether that's something we can take as given or not.

You say it vanishes. How? Because it is zero?

if we pick an arbitrary function g and set that = 0 yes, (f+g) = 0. But I am not getting the connection to proving that anything is a vector space. I just don't see what one has the slightest to do with the other. Unless it is something to do with axiom 3 and 4?

 

[Dick]

so g is just an arbitrary function? It looks pulled out of nowhere.

Arrrgh. This is the problem. The proof I showed you looks like just magic right now and this is pretty frustrating.

Here's what I can sort of figure out. we have f(s) = 0 and we can say that it is in C(S,K) butI have o idea whether that's something we can take as given or not.

You say it vanishes. How? Because it is zero?

if we pick an arbitrary function g and set that = 0 yes, (f+g) = 0. But I am not getting the connection to proving that anything is a vector space. I just don't see what one has the slightest to do with the other. Unless it is something to do with axiom 3 and 4?

Mmm. Your list of axioms omits the ones you really want to prove, the closure ones. Here's what you really need to prove. If f and g are in C(S,K) then f+g is in C(S,K). Is that true? If k is a scalar in K then is k*f in C(S,K). Just think about the definition of C(S,K).

 

[Simon Bridge]

so g is just an arbitrary function? It looks pulled out of nowhere.

Arrrgh. This is the problem. The proof I showed you looks like just magic right now and this is pretty frustrating.

Here's what I can sort of figure out. we have f(s) = 0 and we can say that it is in C(S,K) butI have o idea whether that's something we can take as given or not.

You say it vanishes. How? Because it is zero?

if we pick an arbitrary function g and set that = 0 yes, (f+g) = 0. But I am not getting the connection to proving that anything is a vector space. I just don't see what one has the slightest to do with the other. Unless it is something to do with axiom 3 and 4?

g comes from the same place as f and g(s)=0 under the same conditions that f(s)=0.
In order to prove something is a vector space, you need two arbitrary vectors in the space.

 

[Emspak]

OK, that makes more sense. And I can just pull g out of nowhere as an arbitrary function, right? I say that g is in C(S,K) along with f, and prove they are both in C(S,K). Is that about right? Because k*f being in C(S,K) would be axiomatic it seems to me (from the list I gave you).

Now, thinking about it, if I say that f(s) = 0 for all but a certain number of elements of S, and say that we're adding it to a random, arbitrary function g, then we get f + g = g

and f becomes the 0, per axiom 3.

am i getting there?

 

[Simon Bridge]

I'm going to leave the actual math part to Dick ... you also appear to have a conceptual issue, I'll address that:

I can just pull g out of nowhere as an arbitrary function, right?

You are not pulling g "out of nowhere", you are pulling it out of C(S,K). There are a lot of functions that would fit the definition you have, you pull two of them out, call one f and the other one g. You can call them Bazza and Trev if you like - it's just a label.

 

[Emspak]

I think I am just tired. Most of the time you have to justify that kid of thing somehow, and that was what was confusing me i think.

 

[Emspak]

OK. Here's a version of the proof.

Let S be a nonempty set and K a field. Let C(S,K) denote the set of all functions f∈C(S,K) such that f(s)=0 for all but a finite number of elements of S. Prove that C(S, K) is a vector space.

Vector spaces are defined by the following axioms:
1. for all x,y in vector space V x+y = y+ x
2. for all x, y, z in V (x + y) + z = x + (y + z)
3. there exists an element in V denoted by 0 s.t. x + 0 = x for each x in v
4. for each element x in V there exits an element y in V s.t. x + y = 0
5. for each element x in V 1x = x
6. for each pair of elements a, b, in F and each element x in V, (ab)x = a(bx)
7. for each element a in F and each pair of elements x,y in V, a(x+y) = ax + ay
8. for each pair of elements a, b in F and each element x in V (a+b)x = ax + ab

Proof:

Let g(s) be an arbitrary function. ƒ, g ∈ C(S,K)
ƒ + g = g when ƒ(s) = 0. So axiom (3) is satisfied.

Let scalars λ1, λ2 ∈ K
With the definition of addition: (f+g)(s) = f(s)+g(s) ∀s ∈ S

By definition the addition is commutative and satisfies axiom (1)

(λ1+ λ2)f(s)= λ1f(s) + λ2f(s) ∀x ∈ S satisfies axiom (8)

λ1ƒ,+ λ2g ∈ C(S,K) and satisfies axiom (7)

(λ1λ2)f(s) = λ1 (λ2f(s)) ∀s ∈ S satisfies axiom (6)

1ƒ(s) = ƒ(s) ∀s ∈ S and ƒ(s) is in C(S,K) so axiom (5) is satisfied

-11ƒ(s) = -ƒ(s) ∀s ∈ S and -ƒ(s) + ƒ(s) = 0 and -ƒ(s) ∈ C(S,K) satisfying axiom (4)

By the definition of the problem there is a finite set S1 ⊆ S such that ƒ(s) ≠ 0 and another finite set S2 ⊆ S such that g(s) ≠ 0. Therefore the only space where λ1ƒ,+ λ2g ≠ 0 is inside S1 ∪ S2 and (S1 ∪ S2) ⊆ S.

Since both sets are finite and both sets are inside S.

Therefore C(S,K) is a vector space

anyone willing to tell me if I messed it up again?

 

[Dick]

OK. Here's a version of the proof.

Let S be a nonempty set and K a field. Let C(S,K) denote the set of all functions f∈C(S,K) such that f(s)=0 for all but a finite number of elements of S. Prove that C(S, K) is a vector space.

Vector spaces are defined by the following axioms:
1. for all x,y in vector space V x+y = y+ x
2. for all x, y, z in V (x + y) + z = x + (y + z)
3. there exists an element in V denoted by 0 s.t. x + 0 = x for each x in v
4. for each element x in V there exits an element y in V s.t. x + y = 0
5. for each element x in V 1x = x
6. for each pair of elements a, b, in F and each element x in V, (ab)x = a(bx)
7. for each element a in F and each pair of elements x,y in V, a(x+y) = ax + ay
8. for each pair of elements a, b in F and each element x in V (a+b)x = ax + ab

Proof:

Let g(s) be an arbitrary function. ƒ, g ∈ C(S,K)
ƒ + g = g when ƒ(s) = 0. So axiom (3) is satisfied.

Let scalars λ1, λ2 ∈ K
With the definition of addition: (f+g)(s) = f(s)+g(s) ∀s ∈ S

By definition the addition is commutative and satisfies axiom (1)

(λ1+ λ2)f(s)= λ1f(s) + λ2f(s) ∀x ∈ S satisfies axiom (8)

λ1ƒ,+ λ2g ∈ C(S,K) and satisfies axiom (7)

(λ1λ2)f(s) = λ1 (λ2f(s)) ∀s ∈ S satisfies axiom (6)

1ƒ(s) = ƒ(s) ∀s ∈ S and ƒ(s) is in C(S,K) so axiom (5) is satisfied

-11ƒ(s) = -ƒ(s) ∀s ∈ S and -ƒ(s) + ƒ(s) = 0 and -ƒ(s) ∈ C(S,K) satisfying axiom (4)

By the definition of the problem there is a finite set S1 ⊆ S such that ƒ(s) ≠ 0 and another finite set S2 ⊆ S such that g(s) ≠ 0. Therefore the only space where λ1ƒ,+ λ2g ≠ 0 is inside S1 ∪ S2 and (S1 ∪ S2) ⊆ S.

Since both sets are finite and both sets are inside S.

Therefore C(S,K) is a vector space

anyone willing to tell me if I messed it up again?

I think that's pretty much it but most of it's pretty generic. The only really important part for your problem is that if S1 and S2 are finite, then S1 U S2 if finite. Right?

 

[Emspak]

I suppose so, and I wonder if the reason for the problem (as in: why it's assigned that way is to illustrate that point.

 

[Dick]

I suppose so, and I wonder if the reason for the problem (as in: why it's assigned that way is to illustrate that point.

Hard to say. But if the problem had been define D(S,K) to be the set of all functions f:S->K such that f(s)=0 for all but AT MOST ONE value of s, then it would NOT be a vector space. If you see why then you probably understand the problem well enough.

 

[Emspak]

I am not sure I understand why it isn't a vector space with exactly one f(s) = 0 is 0 for just one s as opposed to a finite set of s values. I mean I can think of a few vector functions that are presumably in a vector space that do just that, unless I am reading it wrong. I get why S1 U S2 would be finite.

 

[Ray Vickson]

I am not sure I understand why it isn't a vector space with exactly one f(s) = 0 is 0 for just one s as opposed to a finite set of s values. I mean I can think of a few vector functions that are presumably in a vector space that do just that, unless I am reading it wrong. I get why S1 U S2 would be finite.

You are reading it wrong---exactly backwards. For example, we might have f(1) ≠ 0 but f(x) = 0 for all x≠1. We might have g(2) ≠ 0 but g(x) = 0 for all x≠2. Then f+g would be nonzero at two points, so would not be a function that is nonzero at only one point.

 

[Emspak]

Ah, what you are showing me looks a lot like the delta function. There's a relation there, no?

 

[Dick]

I am not sure I understand why it isn't a vector space with exactly one f(s) = 0 is 0 for just one s as opposed to a finite set of s values. I mean I can think of a few vector functions that are presumably in a vector space that do just that, unless I am reading it wrong. I get why S1 U S2 would be finite.

The problem says f(s) is only NONZERO at a finite number of points. If f is nonzero at one point and g is nonzero at one point is f+g nonzero at one point?

 

[Emspak]

well, of course they are nonzero at more than one point. But I am still not sure what the connection is to being a vector space or not if it just so happens that there is only one such function as opposed to more than one.

EDIT: You have all officially made me despair of understanding any of this :-)

 

[Dick]

well, of course they are nonzero at more than one point. But I am still not sure what the connection is to being a vector space or not if it just so happens that there is only one such function as opposed to more than one.

EDIT: You have all officially made me despair of understanding any of this :-)

If you define a space V to consist of all functions which are nonzero at at most one point, then that shows that if f and g are in V then f+g doesn't have to be in V. It's not 'closed under addition'. Hence not a vector space.

 

[Emspak]

OK, but take this function: z^3=y+(x-3)^2

That's nonzero at lots of points and zero at lots of points. Is that function NOT in a vector space? I get that we're defining V as a vector space consisting of all functions that are nonzero at one point or none. OK, but that doesn't seem to match up with what we do with vectors all the time. That's what is confusing as all hell.

 

[Dick]

OK, but take this function: z^3=y+(x-3)^2

That's nonzero at lots of points and zero at lots of points. Is that function NOT in a vector space? I get that we're defining V as a vector space consisting of all functions that are nonzero at one point or none. OK, but that doesn't seem to match up with what we do with vectors all the time. That's what is confusing as all hell.

That's NOT a function. Review what f being a function from S->K means.

 

[Emspak]

well, what about f(x,y) = y+ (x-3)^2 ? That's a function, isn't it? It's 0 at (x,y) = (0,0) or (0,3), so at least two points. You're saying it's not in a vector space. But those are 2 vectors right there.

Looking up the definition of function, it's when you do an operation on a set, right? Or not? What is it then if not that?

 

[Dick]

well, what about f(x,y) = y+ (x-3)^2 ? That's a function, isn't it? It's 0 at (x,y) = (0,0) or (0,3), so at least two points. You're saying it's not in a vector space. But those are 2 vectors right there.

Looking up the definition of function, it's when you do an operation on a set, right? Or not? What is it then if not that?

f is a function from S->K if f maps every element of S to an element of K. And in your problem the functions themselves ARE the vectors. You are saying some very confused things.

 

[Emspak]

of course they are confused. Because I AM CONFUSED.

 

[Ray Vickson]

well, what about f(x,y) = y+ (x-3)^2 ? That's a function, isn't it? It's 0 at (x,y) = (0,0) or (0,3), so at least two points. You're saying it's not in a vector space. But those are 2 vectors right there.

Looking up the definition of function, it's when you do an operation on a set, right? Or not? What is it then if not that?

Yes, of course it is a function, But it is NOT a function in the set C(S,K). What is it you are not seeing?

 

[Emspak]

I was referring to a generic function and sort of picking one at random, trying to understand why the example I used -- f(x,y) = whatever, can be 0 at more than one point and describe a vector, and not be in a vector space at the same time. Do you understand why what you just posted here confuses me? You are saying something is and isn't a function at the same time!

I am trying to understand why if (in the original problem) f(s)=0, and there were another function g(s) = 0, and adding them gets you something equal to zero at two points, that it isn't a vector space anymore. Is it because in a vector space the zero element has to be unique?

I asked if it was connected to the delta function concept. But remember, I may not understand enough of this to express that as precisely as you might.

Look, maybe I am having a conceptual problem. As it is I feel lost again, completely.

 

[Dick]

I was referring to a generic function and sort of picking one at random, trying to understand why the example I used -- f(x,y) = whatever, can be 0 at more than one point and describe a vector, and not be in a vector space at the same time. Do you understand why what you just posted here confuses me? You are saying something is and isn't a function at the same time!

I am trying to understand why if (in the original problem) f(s)=0, and there were another function g(s) = 0, and adding them gets you something equal to zero at two points, that it isn't a vector space anymore. Is it because in a vector space the zero element has to be unique?

I asked if it was connected to the delta function concept. But remember, I may not understand enough of this to express that as precisely as you might.

Look, maybe I am having a conceptual problem. As it is I feel lost again, completely.

That's because you seem to have a bunch of extraneous thoughts in your head that are preventing you from concentrating. Look, define V to the set of all functions, say from R->R that are only nonzero at AT MOST ONE point. Define f such that f(1)=1 and f(x)=0 for all other values of x. That's only nonzero at one point, so it's in V. Define g such that g(2)=1 and g(x)=0 for all other values of x. That's also in V. f+g is nonzero at both x=1 and x=2, therefore f+g is NOT in V, because it's nonzero at TWO points. That means V is NOT a vector space because it's NOT closed under addition. If I substitute the definition "V is the set of all functions that are nonzero for a FINITE number of points", then that V is closed under addition and you can show it's a vector space. Please tell me if you understand this before launching off on another tangent.

 

[Emspak]

OK Let me see if I get this: the issue is that you're defining V (the vector space) from the get-go as the set of functions that are non zero at one point only. If you can put two functions together in that space V it isn't a vector space anymore. But if it isn't just one point then it's ok, it is still a vector space. As I understand it this starts <i>from the definition of V itself</i>.

(BTW I also discovered earlier this evening that the original problem as I presented it was wrong because the prof who wrote the assignment had a huge typo, the "real" problem was to prove a set was as subspace. Oh lord.)

Anyhow I think what was confusing me was that I got hung up on what functions do when you've already defined a space that they are in rather differently. In vector calculus for instance you don't get into the properties that make a vector space a vector space.

And I want to thank you for your patience. There are aspects of this that are really frustrating at first. I am not trying ot be flip or go off tangentwise, I was sort of following some reasoning in the text.

 

[Dick]

OK Let me see if I get this: the issue is that you're defining V (the vector space) from the get-go as the set of functions that are non zero at one point only. If you can put two functions together in that space V it isn't a vector space anymore. But if it isn't just one point then it's ok, it is still a vector space. As I understand it this starts <i>from the definition of V itself</i>.

(BTW I also discovered earlier this evening that the original problem as I presented it was wrong because the prof who wrote the assignment had a huge typo, the "real" problem was to prove a set was as subspace. Oh lord.)

Anyhow I think what was confusing me was that I got hung up on what functions do when you've already defined a space that they are in rather differently. In vector calculus for instance you don't get into the properties that make a vector space a vector space.

And I want to thank you for your patience. There are aspects of this that are really frustrating at first. I am not trying ot be flip or go off tangentwise, I was sort of following some reasoning in the text.

Proving it's a subspace of a vector space is certainly easier. You don't have to prove all of the axioms. Given the axioms are true for the larger space means most of them are automatically true for the subset. You just have to prove the closure ones. But it's not horribly different. And, no, in vector calculus you don't have to worry about whether it is a vector space. That's sort of implicit in the subject. This is more abstract. But do you see the difference between "at most one" and "a finite number" I was trying to spell out?

 

[Emspak]

Well, I see that if the set is defined in the first place as being that which only has functions that are nonzero at one point, and further reduce that (by definition) to only one. if you have any other functions that are zero then you aren't in the set anymore. Yes?

 

[Dick]

Well, I see that if the set is defined in the first place as being that which only has functions that are nonzero at one point, and further reduce that (by definition) to only one. if you have any other functions that are zero then you aren't in the set anymore. Yes?

I hope that means the same thing I tried to explain, it's not a very clear statement.