Exercise 1.3.31

Proof. If $v+W$ is a space, we have $0=v+(-v)\in v+W$ and thus $-v\in W$ and $v\in W$. Conversely, if $v\in W$ we have actually $v+W=W$, a space. □

Proof. We can proof that $v_1+W=v_2+W$ if and only if $(v_1-v_2)+W=W$. This is because $(-v_1)+(v_1+W)=\{-v+v+w:w\in W\}=W$ and $(-v_1)+(v_2+W)=\{-v_1+v_2+w:w\in W\}=(-v_1+v_2)+W$. So if $(v_1-v_2)+W=W$, a space, then we have $v_1-v_2\in W$ by the previous exercise. And if $v_1-v_2\in W$ we can conclude that $(v_1-v_2)+W=W$. □

Proof. We have $(v_1+W)+(v_2+W)=(v_1+v_2)+W=(v_1^{\prime }+v_2^{\prime })+W=(v_1^{\prime }+W)+(v_2^{\prime }+W)$ since by the previous exercise we have $v_1-v_1^{\prime }\in W$ and $v_2-v_2^{\prime }\in W$ and thus $(v_1+v_2)-(v_1^{\prime }+v_2^{\prime })\in W$. On the other hand, since $v_1-v_1^{\prime }\in W$ implies $a{v}_1-a{v}_1^{\prime }=a(v_1-v_1^{\prime })\in W$, we have $a(v_1+W)=a(v_1^{\prime }+W)$. □

Proof. It is closed because $V$ is closed. The commutativity and associativity of addition is also because $V$ is commutative and associative. For the zero element we have $(x+W)+W=x+W$. For the inverse element we have $(x+W)+(-x+W)=W$. For the identity element of multiplication, we have $1(x+W)=x+W$. The distribution law and combination law are also followed by the original propositions in $V$. But there are one more thing that should be checked, that is whether it is well-defined. But this is Exercise 1.3.31(c). □