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Subset that satisfies all but one axioms of subspaces

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
Hey!! :eek:

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

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Hey!! :eek:

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

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
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

MHB Oldtimer
Staff member
Feb 7, 2012
2,678
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.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
Hey!! :eek:

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

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
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 $\forall \lambda\in\mathbb R, \forall \mathbf v\in S: \lambda\mathbf v\in S$.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
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

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MHB Site Helper
Apr 14, 2013
4,004
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)
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
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

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MHB Site Helper
Apr 14, 2013
4,004
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)
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
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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

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MHB Site Helper
Apr 14, 2013
4,004
$$\mathbb Z \cdot \left\{ \begin{pmatrix}1\\0\end{pmatrix} \right\}$$
(Thinking)
Ah and if we multiply the vector by a negative scalar, the result will not be in the set, right? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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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

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Apr 14, 2013
4,004
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)
 

Klaas van Aarsen

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Mar 5, 2012
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mathmari

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MHB Site Helper
Apr 14, 2013
4,004
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
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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)