Vector space - Prove or disprove

[mathmari]

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)

 

[Evgeny.Makarov]

Re: v=Vector space - Prove or disprove

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 $|A^B|=|A|^{|B|}$.

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 $\mathbb{Q}^n$...

 

[mathmari]

Re: v=Vector space - Prove or disprove

$A^B$ 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)

$U_1\subseteq U_1+U_2$ 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 $\mathbb{Q}^n$.

 

[Evgeny.Makarov]

Re: v=Vector space - Prove or disprove

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.

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 $\exists x$.

 

[Olinguito]

Re: v=Vector space - Prove or disprove

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?

Spoiler

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

 

[mathmari]

Thank you very much! (Smile)

 

[mathmari]

Re: v=Vector space - Prove or disprove

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)

 

[I like Serena]

Re: v=Vector space - Prove or disprove

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?

 

[mathmari]

Re: v=Vector space - Prove or disprove

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)