Showing space closed under affine combinations is translation of vector space

Question

I'm struggling to reconcile two different definitions of an affine space. The definition in my course notes is:

An affine space in $\mathbb{R}^n$ is a non-empty subset closed under affine combinations; that is, $X$ is an affine space if, whenever $x_1, \ldots, x_n \in X$ and $\lambda_1 + \cdots + \lambda_n = 1$, we have that $\lambda_1 x_1 + \cdots \lambda_n x_n \in X$.

It is then claimed that it is easy to see that every affine space can be expressed as $u + W$, where $u \in \mathbb{R}^n$ and $W$ is a subspace of $\mathbb{R}^n$. (I can see that the converse is true).

I tried writing $x_i = u + w_i$ for each $x_i \in X$ and then taking $W = \textrm{span}\{w_i\}$, but since there might be infinitely many vectors $w_i$ I don't think this will work. Am I missing something obvious? Any help would be appreciated.

Answer 1

Let $u\in X$ and definte $W=X-u=\{w\in\mathbb{R}^n\colon \exists x\in X, w=x-u \}$. Then $W$ is a subspace of $\mathbb{R}^n$. Indeed,

1. $w_1,w_2\in W$ $\Rightarrow$ $w_1+w_2=\underbrace{x_1-u+x_2}_{\in X}-u\in W$.
2. $w\in W$ $\Rightarrow$ $\lambda w=\lambda(x-u)=\underbrace{\lambda x+(1-\lambda)u}_{\in X}-u\in W$.