# Subset that satisfies all but one axioms of subspaces

• MHB
• mathmari
In summary, if a subset satisfies the second and third properties but not the first, it is the empty set.
mathmari
Gold Member
MHB
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)

mathmari said:
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,
$\{\begin{pmatrix}x \\ y\end{pmatrix}| y= 1\}$.

HallsofIvy said:
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)

mathmari said:
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.

mathmari said:
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)

Opalg said:
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 - - -

Klaas van Aarsen said:
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 $\forall \lambda\in\mathbb R, \forall \mathbf v\in S: \lambda\mathbf v\in S$.

mathmari said:
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)

Klaas van Aarsen said:
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 by a moderator:
Klaas van Aarsen said:
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)

mathmari said:
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)

mathmari said:
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)

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

Yep. (Nod)

Klaas van Aarsen said:
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)

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

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

Klaas van Aarsen said:
$$\mathbb Z \cdot \left\{ \begin{pmatrix}1\\0\end{pmatrix} \right\}$$
(Thinking)

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

mathmari said:
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?

Klaas van Aarsen said:
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)

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

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

Klaas van Aarsen said:
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)

mathmari said:
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)

## 1. What is a subset that satisfies all but one axioms of subspaces?

A subset that satisfies all but one axioms of subspaces is a set of vectors that follows all of the properties of a subspace, except for one. This means that the subset may not be closed under addition or scalar multiplication, but still meets all other criteria for being a subspace.

## 2. Can a subset that satisfies all but one axioms of subspaces be considered a subspace?

No, a subset that satisfies all but one axioms of subspaces cannot be considered a subspace. In order for a set to be classified as a subspace, it must meet all of the required properties, including closure under addition and scalar multiplication.

## 3. What is an example of a subset that satisfies all but one axioms of subspaces?

An example of a subset that satisfies all but one axioms of subspaces is a set of vectors in R^2 that does not contain the zero vector. This subset would meet all of the other criteria for being a subspace, but would not be closed under addition since the zero vector is missing.

## 4. How are subsets that satisfy all but one axioms of subspaces useful in mathematics?

Subsets that satisfy all but one axioms of subspaces are useful in mathematics because they help to illustrate the importance of each axiom in defining a subspace. They also allow for the exploration of mathematical concepts and properties in a more nuanced way.

## 5. How can a subset that satisfies all but one axioms of subspaces be transformed into a subspace?

A subset that satisfies all but one axioms of subspaces can be transformed into a subspace by adding the missing element, such as the zero vector, and ensuring that all of the required properties are met. This may involve manipulating the vectors in the subset or adding additional vectors to the set.

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