Subset that satisfies all but one axioms of subspaces

[mathmari]

Hey!

I want to find subsets $S$ of $\mathbb{R}^2$ such that $S$ satisfies all but one axioms of subspaces. A subset that doesn't satisfy the first axiom: We have to find a subset that doesn't contain the zero vector. Is this for example $\left \{\begin{pmatrix}x \\ y\end{pmatrix} : x,y>0\right \}$ ?

A subset that doesn't satisfy the second axiom: We have to find a subset that doesn't contain the sum of two vectors of $S$. Could you give me an example for that?

A subset that doesn't satisfy the third axiom: We have to find a subset that doesn't contain the scalar product of a vector of $S$. Is this for example $\left \{\begin{pmatrix}x \\ y\end{pmatrix} : x\geq 0\right \}$ since $\begin{pmatrix}1 \\ 0\end{pmatrix}\in S$ but $(-1)\cdot \begin{pmatrix}1 \\ 0\end{pmatrix}\notin S$.

(Wondering)

 

[HallsofIvy]

Hey!

I want to find subsets $S$ of $\mathbb{R}^2$ such that $S$ satisfies all but one axioms of subspaces. A subset that doesn't satisfy the first axiom: We have to find a subset that doesn't contain the zero vector. Is this for example $\left \{\begin{pmatrix}x \\ y\end{pmatrix} : x,y>0\right \}$ ?

Yes, that is correct.

A subset that doesn't satisfy the second axiom: We have to find a subset that doesn't contain the sum of two vectors of $S$. Could you give me an example for that?

What about $\{\begin{pmatrix}x \\ y\end{pmatrix}| y= 1\}$?

A subset that doesn't satisfy the third axiom: We have to find a subset that doesn't contain the scalar product of a vector of $S$. Is this for example $\left \{\begin{pmatrix}x \\ y\end{pmatrix} : x\geq 0\right \}$ since $\begin{pmatrix}1 \\ 0\end{pmatrix}\in S$ but $(-1)\cdot \begin{pmatrix}1 \\ 0\end{pmatrix}\notin S$.

(Wondering)

Yes, that is a good example. Another would be, again,

$\{\begin{pmatrix}x \\ y\end{pmatrix}| y= 1\}$.​

 

[mathmari]

Yes, that is correct.

For this set the third axiom is not satisfied, is it? (Wondering)

What set could we take so that only the first axiom is not satisfied? (Wondering)

 

[Opalg]

For this set the third axiom is not satisfied, is it? (Wondering)

What set could we take so that only the first axiom is not satisfied? (Wondering)

If a nonempty subset satisfies the second and third properties then it automatically also satisfies the first property. In fact, if a vector $\mathbf{x}$ is in the subset then so is $-\mathbf{x} = (-1)\mathbf{x}$ and therefore so is $\mathbf{0} = \mathbf{x} + (-\mathbf{x})$.

So the only example satisfying the second and third properties but not the first property is the empty set.

 

[I like Serena]

Hey!

I want to find subsets $S$ of $\mathbb{R}^2$ such that $S$ satisfies all but one axioms of subspaces.

Hey mathmari!

What are your axioms exactly? (Wondering)

Note that all axioms of associativity, commutativity, distributivity, and multiplicative identity are automatically satisfied since the subspace inherits them.

If I'm not mistaken, that leaves:

• $\forall \mathbf v,\mathbf w\in S: \mathbf v+\mathbf w\in S$
• $\forall \lambda\in\mathbb R, \forall \mathbf v\in S: \lambda\mathbf v\in S$
• $\mathbf 0 \in S$
• $\forall \mathbf v\in S: -\mathbf v\in S$

How did you number them? (Wondering)

 

[mathmari]

If a nonempty subset satisfies the second and third properties then it automatically also satisfies the first property. In fact, if a vector $\mathbf{x}$ is in the subset then so is $-\mathbf{x} = (-1)\mathbf{x}$ and therefore so is $\mathbf{0} = \mathbf{x} + (-\mathbf{x})$.

So the only example satisfying the second and third properties but not the first property is the empty set.

So, in every case is it like that? So if all axioms are satisfied except of one then this can just be the empty set? (Wondering)

- - - Updated - - -

What are your axioms exactly? (Wondering)

Note that all axioms of associativity, commutativity, distributivity, and multiplicative identity are automatically satisfied since the subspace inherits them.

If I'm not mistaken, that leaves:

• $\forall \mathbf v,\mathbf w\in S: \mathbf v+\mathbf w\in S$
• $\forall \lambda\in\mathbb R, \forall \mathbf v\in S: \lambda\mathbf v\in S$
• $\mathbf 0 \in S$
• $\forall \mathbf v\in S: -\mathbf v\in S$

How did you number them? (Wondering)

Property 1 is $\mathbf 0 \in S$.
Property 2 is $\forall \mathbf v,\mathbf w\in S: \mathbf v+\mathbf w\in S$.
Property 3 is $\forall \lambda\in\mathbb R, \forall \mathbf v\in S: \lambda\mathbf v\in S$.

 

[I like Serena]

So, in every case is it like that? So if all axioms are satisfied except of one then this can just be the empty set? (Wondering)

- - - Updated - - -

Property 1 is $\mathbf 0 \in S$.
Property 2 is $\forall \mathbf v,\mathbf w\in S: \mathbf v+\mathbf w\in S$.
Property 3 is $\forall \lambda\in\mathbb R, \forall \mathbf v\in S: \lambda\mathbf v\in S$.

What happened to the property of the additive inverse? (Wondering)

Suppose we 'break' only property 2.
So we have some $\mathbf v,\mathbf w \in S$ such that $\mathbf v+\mathbf w\not\in S$.
Simplest possible example is $\mathbf v=\binom 1 0, \mathbf w=\binom 0 1$.
Which set would we get if it still satisfies the other properties? (Thinking)

 

[mathmari]

Suppose we 'break' only property 2.
So we have some $\mathbf v,\mathbf w \in S$ such that $\mathbf v+\mathbf w\not\in S$.
Simplest possible example is $\mathbf v=\binom 1 0, \mathbf w=\binom 0 1$.
Which set would we get if it still satisfies the other properties? (Thinking)

The set $\left \{\begin{pmatrix} x \\ y\end{pmatrix} : x\neq y\right \}\cup \left \{\begin{pmatrix} 0 \\ 0\end{pmatrix} \right \}$ ? (Wondering)

If we want that just the axiom 3 is broken then we have $\mathbf{v} \in S$ and $\lambda\in \mathbb{R}$ such that $\lambda \mathbf{v}\not\in S$.

For example $\mathbf{v}=\binom{1}{0}$ and $\lambda= 5$.

A set that satisfies the other twoaxioms is for example $\left \{\begin{pmatrix} x \\ y\end{pmatrix} : x= y+1\right \}$.

If we want that just the axiom 1 is broken then we consider the empty set. Is everything correct? (Wondering)

 

[I like Serena]

What happened to the property of the additive inverse?

Just realized that the additive inverse is covered by property 3 since $-1\cdot \mathbf v=-\mathbf v$. (Blush)

The set $\left \{\begin{pmatrix} x \\ y\end{pmatrix} : x\neq y\right \}\cup \left \{\begin{pmatrix} 0 \\ 0\end{pmatrix} \right \}$ ?

That works yes. (Nod)

Alternatively we could have constructed the smallest set that satisfies the remaining properties.
Due to property 1 we must include $\mathbf 0$.
And due to property 3 we must include $\mathbb R\cdot\binom 10$ and $\mathbb R \cdot \binom 01$.
So a minimal set is:
$$\mathbb R\left \{\begin{pmatrix} 1 \\ 0 \end{pmatrix} \right\}\cup \mathbb R\left \{\begin{pmatrix} 0 \\ 1 \end{pmatrix} \right\}$$
(Nerd)

If we want that just the axiom 3 is broken then we have $\mathbf{v} \in S$ and $\lambda\in \mathbb{R}$ such that $\lambda \mathbf{v}\not\in S$.

For example $\mathbf{v}=\binom{1}{0}$ and $\lambda= 5$.

A set that satisfies the other twoaxioms is for example $\left \{\begin{pmatrix} x \\ y\end{pmatrix} : x= y+1\right \}$.

That does not work since $\mathbf 0$ in not in the set now. (Worried)
Furthermore, if $\binom 10$ is in the set, then according to property 2 so is $\binom 10+\binom 10+\binom 10+\binom 10+\binom 10=\binom 50$, which is a contradiction. (Sweating)

We can start with $\mathbf{v}=\binom{1}{0}$ though.
From property 2 it follows that every integer multiple must be in the set as well.
And that includes $\mathbf 0$.

That is the minimum we must have according to properties 1 and 2.
If we pick this minimal set, does it satisfy property 3? (Wondering)

If we want that just the axiom 1 is broken then we consider the empty set.

Yep. (Nod)

 

[mathmari]

That does not work since $\mathbf 0$ in not in the set now. (Worried)
Furthermore, if $\binom 10$ is in the set, then according to property 2 so is $\binom 10+\binom 10+\binom 10+\binom 10+\binom 10=\binom 50$, which is a contradiction. (Sweating)

We can start with $\mathbf{v}=\binom{1}{0}$ though.
From property 2 it follows that every integer multiple must be in the set as well.
And that includes $\mathbf 0$.

That is the minimum we must have according to properties 1 and 2.
If we pick this minimal set, does it satisfy property 3? (Wondering)

I haven't really understood that part. What set could we take? (Wondering)

 

[I like Serena]

I haven't really understood that part. What set could we take?

$$\mathbb Z \cdot \left\{ \begin{pmatrix}1\\0\end{pmatrix} \right\}$$
(Thinking)

 

[mathmari]

$$\mathbb Z \cdot \left\{ \begin{pmatrix}1\\0\end{pmatrix} \right\}$$
(Thinking)

Ah and if we multiply the vector by a negative scalar, the result will not be in the set, right? (Wondering)

 

[I like Serena]

Ah and if we multiply the vector by a negative scalar, the result will not be in the set, right?

No... $-2\cdot\binom 10=\binom {-2}0$ will be in the set as well... (Sweating)
After all, $-2$ is an element of $\mathbb Z$, isn't it?

 

[mathmari]

No... $-2\cdot\binom 10=\binom {-2}0$ will be in the set as well... (Sweating)
After all, $-2$ is an element of $\mathbb Z$, isn't it?

Oh yes... But then why is the axiom 3 not satisfied? (Wondering)

 

[I like Serena]

Oh yes... But then why is the axiom 3 not satisfied?

How about $\binom{1.5}0$? (Wondering)

 

[mathmari]

How about $\binom{1.5}0$? (Wondering)

Ahh so the if we multiply the vector of the set by a non-integer real number, the result will not be in the set and so this is set is not closed under scalar multipliation.

Have I understood that correctly? (Wondering)

 

[I like Serena]

Ahh so the if we multiply the vector of the set by a non-integer real number, the result will not be in the set and so this is set is not closed under scalar multiplication.

Have I understood that correctly?

Yup. (Nod)