I found this statement in these lecture notes: https://www1.se.cuhk.edu.hk/~manchoso/2021/engg5501/2-cvxanal.pdf. It is in the proof of Prop. 7, page 9.
Let $A \subset \mathbb{R}^{n}.$ $aff(A)$ is as defined here: https://en.wikipedia.org/wiki/Affine_hull. $cl(A)$ is not explicitly defined in the notes, but it is the closure of A wrt to its affine hull, so $cl(A) = \{ x \in aff(A): \forall \epsilon >0, \mathbb{B}_{\epsilon}(x) \cap A \neq \emptyset \}. $
Here $\mathbb{B}_{\epsilon}(x)$ denotes the Euclidean ball around an element $x \in \mathbb{R}^{n}. $
The statement that I want to prove or refute from the notes is:
"Let $\epsilon >0.$ If $x \in cl(A)$, then $x \in A + \mathbb{B}_{\epsilon}(0) \cap aff(A).$"
My reasoning: Suppose $x \in cl(A)$ and $\epsilon >0.$ There is $a \in \mathbb{B}_{\epsilon}(x),$ so that $x - a \in \mathbb{B}_{\epsilon}(0)$ and $x = a + x-a \in A + \mathbb{B}_{\epsilon}(0).$ However, by definition of $cl(A),$ $x \in aff(A),$ so $x \in ( A + \mathbb{B}_{\epsilon}(0) ) \cap aff(A).$ So, $x = a + y$ with $y \in \mathbb{B}_{\epsilon}(0)$ and $a = \sum_{i = 1}^{k} \lambda_{i}x_{i}$ where $k \in \mathbb{N}$ and $\lambda_{i} \in \mathbb{R}, i \in \{ 1, .., k\}$ and $\sum_{i = 1}^{k}\lambda_{i} = 1.$ So, if we write $y = x - a,$ $y$ is a weighted sum of elements of $A,$ whose weights sum to $0$ instead of $1,$ so it can't be in the affine hull of A.
Can someone find my mistake and explain why the statement holds, in case it does?