# Are These Subsets of R^R Subspaces?

• karnten07
In summary, Dick said that the subset U is closed under vector addition and that the set of all odd functions is a subspace of RR.
karnten07

## Homework Statement

Which of the following subsets of the vector space R^R of all functions from R to R are subspaces? (proofs or counterexamples required)

U:= f $$\in$$R^R, f is differentiable and f'(0) = 0

V:= f$$\in$$R^R, f is polynomial of the form f=at^2 for some a$$\in$$R
= There exists a of the set R: for all s of R: f(s) = as^2

W:= " " f is polynomial of the form f=at^i for some aof the set R and i of the set N
= there exists i of N, there exists a of R: that for all s of R: f(s) = as^i

X:= " " f is odd
(f is odd such that f(-s) =-f(s) for all s of R

## The Attempt at a Solution

Okay so i want to start with , odd functions. I can use the sine function as a counterexample because i don't think X is a subspace. I think that it isn't closed under addtion because sin90 +sin90 = 2 which isn't a solution to any elements of the set X. So i can use this as a counterexample right?

I will start thinking about the other subsets.

Gack! The question doesn't ask you whether a single odd function forms a subspace, it asks you whether ALL odd functions form a subspace. I.e. you have to prove things like if f(s) and g(s) are odd, that f(s)+g(s) is odd and k*f(s) is odd, right?

Dick said:
Gack! The question doesn't ask you whether a single odd function forms a subspace, it asks you whether ALL odd functions form a subspace. I.e. you have to prove things like if f(s) and g(s) are odd, that f(s)+g(s) is odd and k*f(s) is odd, right?

yes, if i want to show the subset is a subspace. But if i want to show that it is not a subset as one function in the subset doesn't meet the criteria i can use a counterexample, right? So for this one, i can make it easy in doing it that way.

You are on the right track checking closure properties: you must have closure under scalar multiplication and closure under vector addition for a subset to be subspace. Your example of sin90 + sin90 = 2 didn't start with two vectors in X (sin90 = 1 is not an odd function).

Note: Here the vectors are functions, so testing closure under vector addition for this problem looks like this:

Let $$f(x)$$ and $$g(x)$$ be vectors (functions) in the subset X.
Then $$f(x)+g(x)$$ is (or is not) a vector in the subset X because...
<Put some stuff here>
Hence U is (or is not) closed under vector addition.

No! For one thing the odd functions ARE a subspace. Look, the real numbers are a vector space, right? Just because 1+1 is not equal to 1 doesn't prove they aren't. A subspace is a group of objects and closure just says that combinations of them remain in the same group. You can't pick one out and form a contradiction solely on that.

Maybe this is what is confusing you. sin(x)+sin(x)=2*sin(x) NOT sin(x+x). All of those are odd, and sin(90)+sin(90)=2*sin(90). There is absolutely nothing in the definition of subspace that says you have to be able to solve sin(90)+sin(90)=sin(y). Is there?

Last edited:
Dick said:
Maybe this is what is confusing you. sin(x)+sin(x)=2*sin(x) NOT sin(x+x). All of those are odd, and sin(90)+sin(90)=2*sin(90). There is absolutely nothing in the definition of subspace that says you have to be able to solve sin(90)+sin(90)=sin(y). Is there?

Oh yes, i see what you are saying now. These questions takes ages to write out ugh.

benorin said:
You are on the right track checking closure properties: you must have closure under scalar multiplication and closure under vector addition for a subset to be subspace. Your example of sin90 + sin90 = 2 didn't start with two vectors in X (sin90 = 1 is not an odd function).

Note: Here the vectors are functions, so testing closure under vector addition for this problem looks like this:

Let $$f(x)$$ and $$g(x)$$ be vectors (functions) in the subset X.
Then $$f(x)+g(x)$$ is (or is not) a vector in the subset X because...
<Put some stuff here>
Hence U is (or is not) closed under vector addition.

mmm I am confused. How do i show f(x) + g(x) is still odd? I don't get the vector addition part either??

The point Dick was making is that you are wrong. The sum of two odd functions is an odd function and any number times an odd function is an odd function. The set of all odd functions is a subspace of RR. What you need to do is to prove the two statements in the second sentence.

## 1. What is a subspace?

A subspace is a subset of a vector space that satisfies the three properties of closure under addition, closure under scalar multiplication, and the existence of a zero vector. It can also be thought of as a smaller space within a larger space that maintains the same operations and structure.

## 2. How can I determine if a subset is a subspace?

To determine if a subset is a subspace, you must check if it satisfies the three properties of closure under addition, closure under scalar multiplication, and the existence of a zero vector. If all three properties hold, then the subset is a subspace.

## 3. What does it mean for a subset to have closure under addition?

Closure under addition means that when two vectors in the subset are added together, the resulting vector is also in the subset. In other words, the subset contains all possible linear combinations of its vectors.

## 4. How do I check if a subset has closure under scalar multiplication?

To check for closure under scalar multiplication, you must multiply each vector in the subset by a scalar. If the resulting vectors are also in the subset, then the subset has closure under scalar multiplication.

## 5. Can a subset be a subspace of multiple vector spaces?

Yes, a subset can be a subspace of multiple vector spaces as long as it satisfies the three properties of a subspace in each vector space. The same subset may have different properties and operations in different vector spaces, but as long as it satisfies the three properties, it can be considered a subspace in each one.

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