For any vector space $V$, subset $S \subseteq V $, and vector $\vec{v} \in V$, we define the set $$\vec{v}+S = \{\vec{v} + \vec{x} : \vec{x}\in S\} $$ Prove that $\vec{a}+W=\vec{b}+W$ iff $\vec{a} -\vec{b}\in W$.
What I have so far is:
Assume $\vec{a}+W=\vec{b}+W$. The a bases of the two are $\{\vec{a},\vec{a}+\vec{w_1}, ...,\vec{a}+\vec{w_n}\}$ and $\{\vec{b},\vec{b}+\vec{w_1}, ...,\vec{b}+\vec{w_n}\}$ respectively.
Let $\vec{w}\in \vec{a}+W$. It can be written as a linear combination of the bases such that $$ c_0(\vec{a}) + c_1(\vec{a}+\vec{w_1})+...c_n(\vec{a}+\vec{w_n}) = d_0(\vec{b}) + d_1(\vec{b}+\vec{w_1})+...d_n(\vec{b}+\vec{w_n})$$ Then by subtracting the sides $$ (c_0 + ... +c_n)\vec{a} - (d_0 + ... +d_n)\vec{b} + (c_1 - d_1)\vec{w_1}+...+(c_n-d_n)\vec{w_n}=0$$ Then isolating the first two terms $$ (c_0 + ... +c_n)\vec{a} - (d_0 + ... +d_n)\vec{b} =(d_1 - c_1)\vec{w_1}+...+(d_n-c_n)\vec{w_n}$$
So clearly, some linear combination of $\vec{a}$ and $\vec{b}$ are in $W$, but how can I conclude that $\vec{a}-\vec{b}$ is also in $W$ since the two terms have different coefficients so I can't factor it out?
Follow up to Martin's comment:
If $\vec{a} \in \vec{b} +W$, then for any $\vec{x} \in \vec{a}+W$, it can be written as $\vec{x} = \vec{a} + \vec{w}, \vec{w} \in W$ by definition of adding a vector to a subspace. Since $\vec{a}, \vec{w} \in \vec{b}+W$, $\vec{x} \in \vec{b}+W$.
You are making it too complicated. If $a+W=b +W$, then there exist $u,v\in W$ such that $a+u=b+v$. So $a-b=v-u\in W$.
Conversely, if $a-b\in W$, there exists $w\in W$ with $a-b=w$. Then $a=b+w\in b+W$, so $a+W\in b+W$; and the reverse inclusion follows but reversing roles.