# Vector space - Prove or disprove

• MHB
• mathmari
In summary, the author has proved that the set $\{f\in \mathbb{R}^{\mathbb{R}} \mid \exists x\in \mathbb{R} : f(x)=0_{\mathbb{R}}\}$ is not a subspace of $\mathbb{R}^{\mathbb{R}}$.
mathmari
Gold Member
MHB
Hey!

Let $1\leq n\in \mathbb{N}$ and let $U_1, U_2$ be subspaces of the $\mathbb{R}$-vector space $\mathbb{R}^n$.

I want to prove or disprove the following:
• The set $\{f\in \mathbb{R}^{\mathbb{R}} \mid \exists x\in \mathbb{R} : f(x)=0_{\mathbb{R}}\}$ is a subspace of $\mathbb{R}^{\mathbb{R}}$.

What exactly is $\mathbb{R}^{\mathbb{R}}$ ?


• The set $U_1+U_2$ is a subspace of $\mathbb{Q}^n$.

I have shown that $U_1+U_2$ is a subspace of $\mathbb{R}^n$. I think that the sum $U_1+U_2$ doesn't have to be also a subspace of $\mathbb{Q}^n$. Is this correct?
(Wondering)

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Re: v=Vector space - Prove or disprove

mathmari said:
What exactly is $\mathbb{R}^{\mathbb{R}}$ ?
$A^B$ usually denotes the set of all functions from $B$ to $A$. This is because for finite $A$ and $B$ we have $|A^B|=|A|^{|B|}$.

mathmari said:
The set $U_1+U_2$ is a subspace of $\mathbb{Q}^n$.

I have shown that $U_1+U_2$ is a subspace of $\mathbb{R}^n$. I think that the sum $U_1+U_2$ doesn't have to be also a subspace of $\mathbb{Q}^n$. Is this correct?
$U_1\subseteq U_1+U_2$ does not even have to be a subset of $\mathbb{Q}^n$...

Re: v=Vector space - Prove or disprove

Evgeny.Makarov said:
$A^B$ usually denotes the set of all functions from $B$ to $A$. This is because for finite $A$ and $B$ we have $|A^B|=|A|^{|B|}$.

Ahh ok!

The set $\{f\in \mathbb{R}^{\mathbb{R}} \mid \exists x\in \mathbb{R} : f(x)=0_{\mathbb{R}}\}$ is non-empty, since it contains the function $\mathbf{0}$ defined by $\mathbf{0}(x)=0$.

Let $f,g$ be elements of the set, then there are $x,y\in \mathbb{R}$ such that $f(x)=g(y)=0$. If $x\neq y$ then there is no element $z$ such that $(f+g)(z)=0$, right?

That means that the set is not a subspace.

(Wondering)
Evgeny.Makarov said:
$U_1\subseteq U_1+U_2$ does not even have to be a subset of $\mathbb{Q}^n$...

Oh yes! And so $U_1+U_2$ is not (necessarily) a subspace of $\mathbb{Q}^n$.

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Re: v=Vector space - Prove or disprove

mathmari said:
Let $f,g$ be elements of the set, then there are $x,y\in \mathbb{R}$ such that $f(x)=g(y)=0$. If $x\neq y$ then there is no element $z$ such that $(f+g)(z)=0$, right?
Two words: in general.

mathmari said:
That means that the set is not a subspace.
Yes.

It's unusual that the problem says $0_{\mathbb{R}}$. Usually mathematicians don't make a distinction between 0 as, say, a natural and a real number. This may be a problem for programming languages where it is desirable for a single expression to have different types. Languages like Haskell have a whole mechanism called type classes for that, but this is rarely stressed in mathematics.

On second reading, the author may want to say that $f(x)$ is a value of $f$ at $x$ and is therefore a number as opposed to the whole function, in which case $0$ would denote a zero function. But this should be clear from $\exists x$.

Re: v=Vector space - Prove or disprove

mathmari said:
Let $f,g$ be elements of the set, then there are $x,y\in \mathbb{R}$ such that $f(x)=g(y)=0$. If $x\neq y$ then there is no element $z$ such that $(f+g)(z)=0$, right?

Your statement is not quite right. For example: if $f(x)=x$ and $g(x)=x-1$, then $f(0)=g(1)=0$ and $0\ne1$, but $\exists z$ (namely $z=\frac12$) such that $(f+g)(z)=2z-1=0$.

But this is neither here nor there. You want to show that the given set is not a subspace; thus it is enough to show a counterexample. Can you find one?

Hint: $f(x)=x$, $g(x)=1-x$.

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Thank you very much! (Smile)

Re: v=Vector space - Prove or disprove

mathmari said:
The set $\{f\in \mathbb{R}^{\mathbb{R}} \mid \exists x\in \mathbb{R} : f(x)=0_{\mathbb{R}}\}$ is non-empty, since it contains the function $\mathbf{0}$ defined by $\mathbf{0}(x)=0$.

Is this statement correct? (Wondering)

Re: v=Vector space - Prove or disprove

mathmari said:
Is this statement correct?

Sure it is. Why wouldn't it be? (Wondering)

The function $\mathbf 0: \mathbb R \to \mathbb R$ given by $\mathbf 0(x) = 0_{\mathbb R}$ is an element of $\mathbb R^{\mathbb R}$, and there is an $x \in \mathbb R$ such that $\mathbf 0(x) = 0_{\mathbb R}$, isn't it?

Re: v=Vector space - Prove or disprove

Klaas van Aarsen said:
Sure it is. Why wouldn't it be? (Wondering)

The function $\mathbf 0: \mathbb R \to \mathbb R$ given by $\mathbf 0(x) = 0_{\mathbb R}$ is an element of $\mathbb R^{\mathbb R}$, and there is an $x \in \mathbb R$ such that $\mathbf 0(x) = 0_{\mathbb R}$, isn't it?

Yes, you're right! Thank you! (Yes)

## What is a vector space?

A vector space is a mathematical structure that consists of a set of vectors and a set of operations that can be performed on those vectors. These operations include addition and scalar multiplication, and the set of vectors must satisfy certain properties such as closure and associativity.

## What does it mean to prove a vector space?

To prove a vector space means to show that a given set of vectors and operations satisfy all the necessary properties to be considered a vector space. This involves demonstrating that the operations are well-defined and that the set of vectors satisfies the required properties.

## How do you prove that a set of vectors is a vector space?

To prove that a set of vectors is a vector space, you must show that the set satisfies all the necessary properties of a vector space, such as closure under addition and scalar multiplication, associativity, and existence of a zero vector. You can also prove a set of vectors is a vector space by showing that it is a subspace of a larger vector space.

## Can a set of vectors be both a vector space and not a vector space?

No, a set of vectors cannot be both a vector space and not a vector space. A set either satisfies all the necessary properties to be considered a vector space or it does not. However, a set of vectors can be a vector space under one set of operations and not a vector space under another set of operations.

## What is the importance of proving a vector space?

The importance of proving a vector space lies in the fact that vector spaces are fundamental structures in mathematics and have numerous applications in fields such as physics, engineering, and computer science. Proving a vector space ensures that the operations and properties associated with it are well-defined and can be used reliably in mathematical and scientific calculations.

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