Are the sets subspaces?

We have the following subsets: \begin{align*}&U_1:=\left \{\begin{pmatrix}x \\ y\end{pmatrix} \mid x^2+y^2\leq 4\right \} \subseteq \mathbb{R}^2\\ &U_2:=\left \{\begin{pmatrix}2a \\ -a\end{pmatrix} \mid a\in \mathbb{R}\right \} \subseteq \mathbb{R}^2 \\ &U_3:=\left \{\begin{pmatrix}x \\ y \\ z\end{pmatrix} \mid y=0\right \}\subseteq \mathbb{R}^3 \\ &U_4:=\left \{\begin{pmatrix}x \\ y \\ z\end{pmatrix} \mid y=1\right \}\subseteq \mathbb{R}^3\end{align*}

I want to check if these are subspaces.

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I have done the following:

• $U_1$ :

  The set is non-empty, $(0,0)^T\in U_1$.

  Let $(x_1, y_1)^T, (x_2, y_2)^T\in U_1$. That means that $x_1^2+y_1^2\leq 4$ and $x_2^2+y_2^2\leq 4$. For the sum $(x_1, y_1)^T+ (x_2, y_2)^T=(x_1+x_2, y_1+y_2)^T$ we have \begin{align*}(x_1+x_2)^2+(y_1+y_2)^2&=x_1^2+2x_1x_2+x_2^2+y_1^2+2y_1y_2+y_2^2 \\ & =\left (x_1^2+y_1^2\right )+\left (x_2^2+y_2^2\right )+2x_1x_2+2y_1y_2 \\ & \leq 4+4+2x_1x_2+2y_1y_2\end{align*} This is not necessarily less than $4$ and so this is not a subspace. Is that correct?

• $U_2$ :

  The set is non-empty, $(0,0)^T\in U_2$.

  Let $(x_1, y_1)^T, (x_2, y_2)^T\in U_2$. That means that $y_1=-2x_1$ and $y_2=-2x_2$. For the sum $(x_1, y_1)^T+ (x_2, y_2)^T=(x_1+x_2, y_1+y_2)^T$ we have $y_1+y_2=-2x_1-2x_2=-2(x_1+x_2)$ and that means that the sum is also in $U_2$.

  Let $r\in \mathbb{R}$ and $(x_1, y_1)^T\in U_2$. That means that $y_1=-2x_1$. For the scalar multiplication $r(x_1, y_1)^T=(rx_1, ry_1)^T=(rx_1, ry_1)^T$ we have $ry_1=r\cdot (-2x_1)=-2(rx_1)$ and that means that the result is also in $U_2$.

  Therefore $U_2$ is a subspace.

• $U_3$ :

  The set is non-empty, $(0,0,0)^T\in U_3$.

  Let $(x_1, y_1, z_1)^T, (x_2, y_2, z_2)^T\in U_3$. That means that $y_1=y_2=0$. For the sum $(x_1, y_1, z_1)^T+ (x_2, y_2, z_2)^T=(x_1+x_2, y_1+y_2, z_1+z_2)^T$ we have $y_1+y_2=0+0=0$ and that means that the sum is also in $U_3$.

  Let $r\in \mathbb{R}$ and $(x_1, y_1, z_1)^T\in U_3$. That means that $y_1=0$. For the scalar multiplication $r(x_1, y_1, z_1)^T=(rx_1, ry_1, rz_1)^T$ we have $ry_1=r\cdot 0=0$ and that means that the result is also in $U_3$.

  Therefore $U_3$ is a subspace.

• $U_4$ :

  Since the zero vector is nott included, $U_4$ is not a subspace.

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Is everything correct and complete?

For every one want to know:

$U_1: \qquad $ There is no bounded subspace but $\{0\}$

$U_2 = \langle \binom{2}{-1} \rangle: \qquad $ Multiples of any vector produce a subspace of degree $1$.

$U_2' = \langle v_i \rangle_i: \qquad $ linear combinations of any set of vectors produce a subspace of degree equals to the max number of independent vectors in generating set.

$U_{3,4} = y=y_0: \qquad $ Any linear equation produces a subspace of degree $n-1$ where your space is of degree $n$.

$U'_{3,4} : \qquad $ A system of linear equations produce a subspace of degree $n-k$ where you have $k$ independent equations.