I am trying to prove the following statement: given two sets $A, B \subseteq \mathbb{R}^n$ with $A \subseteq B$ and $\dim\aff(A)=\dim\aff(B)$ it follows that $\aff(A)=\aff(B)$.
So first of all, is that statement correct?
Secondly, I try to prove it exploiting the corresponding property for linear subspaces but I cant quite finish the proof.
My initial try is: We know that $\aff(A)=a_0+\aff(A−a_0)=a_0+\span(A−a_0)$, for some $a_0 \in A$ and $\span(A−a_0)$ is the linear subspace parallel to $\aff(A)$. Likewise, $\aff(B)=b_0+\span(B−b_0)$. Due to $\aff(\cdot)$ being a monotone transformation, from $A \subseteq B$ follows $\aff(A)\subseteq \aff(B)$ and hence $a_0+\span(A−a_0)\subseteq b_0+\span(B−b_0)$. I also have $\dim(a_0+\span(A−a_0))=\dim(b_0+\span(B−b_0))$. However, in all these statements I still have the translation vectors $b_0$ and $c_0$ so I never work with the actual linear subspace $\span(A-a_0)$ and $\span(B-b_0)$ but only with its translations which are affine subspaces. How to complete this proof or am I on the wrong track anyway?