MHB Subsets of $\mathbb{R}^2$ Satisfying S2 and S3 but Not S1: Empty Set

[mathmari]

Hey! :giggle: The three axioms for a subspace are:

S1. The set must be not-empty.

S2. The sum of two elements of the set must be contained in the set.

S3. The scalar product of each element of the set must be again in the set.
I have shown that:

- $\displaystyle{X_1=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb{R}^2\mid 5x+3y=0_{\mathbb{R}}\right \}}$ is a subspace.

All axioms are satisfied. - $\displaystyle{X_2=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb{R}^2\mid 5x+3y=-2\right \}}$ is not a subspace.

S1 is satisfied, S2 and S3 are not satisfied. - $\displaystyle{X_3=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb{R}^2\mid x^2+y^2=0_{\mathbb{R}}\right \}=\left \{\begin{pmatrix}0 \\ 0 \end{pmatrix}\right \}}$ is a subspace.

All axioms are satisfied. - $\displaystyle{X_4=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb{R}^2\mid x^2+y^2=5\right \}}$ is not a subspace.

Only S1 is satisfied. - $\displaystyle{X_5=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb{R}^2\mid x,y\in \mathbb{Z}\right \}}$ is not a subspace.

S3 is not satisfied, S1 and S2 are satisfied. - $\displaystyle{X_6=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb{R}^2\mid xy\geq 0_{\mathbb{R}}\right \}}$ is not a subspace.

S3 is not satisfied, S1 and S2 are satisfied. - $\displaystyle{X_7=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb{R}^2\mid x> 0_{\mathbb{R}}<y\right \}}$ is not a subspace.

S3 is not satisfied, S1 and S2 are satisfied.

Give all subsets of $\mathbb{R}^2$ that

- satisfy S2 and S3, but S1
- satisfy S3, but S1 and S2

Since they shouldn't satisfy S1 do we not have only the empty set? :unsure:

 

[I like Serena]

- $\displaystyle{X_6=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb{R}^2\mid xy\geq 0_{\mathbb{R}}\right \}}$ is not a subspace.

S3 is not satisfied, S1 and S2 are satisfied.

Hi mathmari!

I don't think S2 is satisfied.

Give all subsets of $\mathbb{R}^2$ that
- satisfy S2 and S3, but S1
- satisfy S3, but S1 and S2
Since they shouldn't satisfy S1 do we not have only the empty set?

Indeed. (Nod)

 

[mathmari]

I don't think S2 is satisfied.

Ahh yes!

Counterexample: $\begin{pmatrix}2 \\ 2\end{pmatrix}, \begin{pmatrix}-1 \\ -5\end{pmatrix}$ then $\begin{pmatrix}2 \\ 2\end{pmatrix}+\begin{pmatrix}-1 \\ -5\end{pmatrix}=\begin{pmatrix}1 \\ -3\end{pmatrix}$.

:unsure:

 

[mathmari]

If we consider the following axioms:

S1. $0$ must be contained in the set.

S2. The sum of two elements of the set must be contained in the set.

S3. The scalar product of each element of the set must be again in the set. Then we have the following:

- $\displaystyle{X_1=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb{R}^2\mid 5x+3y=0_{\mathbb{R}}\right \}}$ is a subspace.

All axioms are satisfied.- $\displaystyle{X_2=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb{R}^2\mid 5x+3y=-2\right \}}$ is not a subspace.

S1, S2 and S3 are not satisfied.- $\displaystyle{X_3=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb{R}^2\mid x^2+y^2=0_{\mathbb{R}}\right \}=\left \{\begin{pmatrix}0 \\ 0 \end{pmatrix}\right \}}$ is a subspace.

All axioms are satisfied.- $\displaystyle{X_4=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb{R}^2\mid x^2+y^2=5\right \}}$ is not a subspace.

S1, S2, S3 are nont satisfied. - $\displaystyle{X_5=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb{R}^2\mid x,y\in \mathbb{Z}\right \}}$ is not a subspace.

S3 is not satisfied, S1 and S2 are satisfied.- $\displaystyle{X_6=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb{R}^2\mid xy\geq 0_{\mathbb{R}}\right \}}$ is not a subspace.

S2, S3 are not satisfied, S1 is satisfied.- $\displaystyle{X_7=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb{R}^2\mid x> 0_{\mathbb{R}}<y\right \}}$ is not a subspace.

S1, S3 are not satisfied, S2 is satisfied.:unsure:
What would we get for the following question, in this case?

Give all subsets of $\mathbb{R}^2$ that

- satisfy S2 and S3, but not S1
- satisfy S3, but not S1 and S2

For the first one we need a set that doesn't conatin the zero element but is closed by addition and scalar product.
For the second one we need a set that doesn't conatin the zero element and is not closed by addition but it is closed by scalar product.

How can we get something general? :unsure:

 

[I like Serena]

- $\displaystyle{X_6=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb{R}^2\mid xy\geq 0_{\mathbb{R}}\right \}}$ is not a subspace.

S2, S3 are not satisfied, S1 is satisfied.

Actually, what is wrong with S3?
Do you have a counter example?

What would we get for the following question, in this case?

Give all subsets of $\mathbb{R}^2$ that

- satisfy S2 and S3, but not S1

For the first one we need a set that doesn't contain the zero element but is closed by addition and scalar product.

Suppose the subset contains some element $x\ne 0$.
Then $-1\cdot x$ must also be in the set. And $x + -1\cdot x=0$ is then also in the set, which is a contradiction.

- satisfy S3, but not S1 and S2

For the second one we need a set that doesn't contain the zero element and is not closed by addition but it is closed by scalar product.

Can we use the same argument?

 

[mathmari]

Actually, what is wrong with S3?
Do you have a counter example?

Since $(\alpha x)\cdot (\alpha y)=\alpha^2(xy)\geq 0_{\mathbb{R}}$ S3 is satisfied, right? :unsure:

Suppose the subset contains some element $x\ne 0$.
Then $-1\cdot x$ must also be in the set. And $x + -1\cdot x=0$ is then also in the set, which is a contradiction.

So we get the empty set, right? :unsure:

Can we use the same argument?

Suppose the subset contains some element $x\ne 0$.
Then $0\cdot x$ must also be in the set, which is a contradiction, right? :unsure:

 

[I like Serena]

Since $(\alpha x)\cdot (\alpha y)=\alpha^2(xy)\geq 0_{\mathbb{R}}$ S3 is satisfied, right?

So we get the empty set, right?

Suppose the subset contains some element $x\ne 0$.
Then $0\cdot x$ must also be in the set, which is a contradiction, right?

All correct.