[proofplan]
We compare the affine span of $S$ with the affine subspace obtained by translating the linear span of all displacement vectors from a chosen base point $p_0 \in S$. First we show that this translated subspace contains $S$, so the minimality of $\operatorname{Aff}(S)$ puts $\operatorname{Aff}(S)$ inside it. Then we show that every affine subspace containing $S$ must contain all those displacement directions, hence must contain the whole translate. Applying this to $\operatorname{Aff}(S)$ gives the reverse inclusion.
[/proofplan]
[step:Translate the displacement span to build an affine subspace containing $S$]
Define the linear subspace $W \subset V$ by
\begin{align*}
W:=\operatorname{span}_k\{p-p_0 : p \in S\}.
\end{align*}
Define the subset $C \subset A$ by
\begin{align*}
C:=p_0+W=\{p_0+w : w \in W\}.
\end{align*}
Since $W$ is a linear subspace of $V$, the set $C=p_0+W$ is an affine subspace of $A$. For every $p \in S$, the displacement vector $p-p_0$ belongs to $W$ by the definition of $W$, so
\begin{align*}
p=p_0+(p-p_0)\in p_0+W=C.
\end{align*}
Thus $S \subset C$. Since $\operatorname{Aff}(S)$ is the smallest affine subspace of $A$ containing $S$, and $C$ is an affine subspace containing $S$, we obtain
\begin{align*}
\operatorname{Aff}(S)\subset C.
\end{align*}
[/step]
[step:Show every affine subspace containing $S$ contains the translated span]
Let $B \subset A$ be an affine subspace such that $S \subset B$. Since $p_0 \in S$, we have $p_0 \in B$. Let $\operatorname{dir}(B) \subset V$ denote the direction space of $B$. By [citetheorem:9343], applied to the affine subspace $B$ and the point $p_0 \in B$, we have
\begin{align*}
B=p_0+\operatorname{dir}(B).
\end{align*}
For every $p \in S$, the inclusion $S \subset B$ gives $p \in B$, so the representation above implies
\begin{align*}
p-p_0\in \operatorname{dir}(B).
\end{align*}
Because $\operatorname{dir}(B)$ is a linear subspace of $V$ and contains every vector $p-p_0$ with $p \in S$, it contains their linear span:
\begin{align*}
W\subset \operatorname{dir}(B).
\end{align*}
Therefore
\begin{align*}
C=p_0+W\subset p_0+\operatorname{dir}(B)=B.
\end{align*}
[guided]
Let $B \subset A$ be any affine subspace containing $S$. The goal is to prove that the particular affine subspace $C=p_0+W$ constructed from displacement vectors is forced to lie inside $B$. Since $p_0 \in S$ and $S \subset B$, the base point $p_0$ belongs to $B$.
Let $\operatorname{dir}(B) \subset V$ denote the direction space of $B$. We apply [citetheorem:9343] to the affine subspace $B$ and the point $p_0 \in B$. Its hypotheses are satisfied because $B$ is an affine subspace and $p_0$ is a point of $B$. The conclusion gives
\begin{align*}
B=p_0+\operatorname{dir}(B).
\end{align*}
Now fix an arbitrary point $p \in S$. Since $S \subset B$, we have $p \in B$. The equality $B=p_0+\operatorname{dir}(B)$ means exactly that there exists a vector $u \in \operatorname{dir}(B)$ such that
\begin{align*}
p=p_0+u.
\end{align*}
By the defining property of displacement vectors in an [affine space](/page/Affine%20Space), this vector must be
\begin{align*}
u=p-p_0.
\end{align*}
Hence
\begin{align*}
p-p_0\in \operatorname{dir}(B).
\end{align*}
This holds for every $p \in S$. Since $\operatorname{dir}(B)$ is a linear subspace of $V$, it contains every finite $k$-linear combination of the vectors $p-p_0$ with $p \in S$. Therefore it contains their linear span:
\begin{align*}
W=\operatorname{span}_k\{p-p_0 : p \in S\}\subset \operatorname{dir}(B).
\end{align*}
Translating this inclusion by $p_0$ gives
\begin{align*}
p_0+W\subset p_0+\operatorname{dir}(B).
\end{align*}
Using again the equality $B=p_0+\operatorname{dir}(B)$, we conclude
\begin{align*}
C=p_0+W\subset B.
\end{align*}
[/guided]
[/step]
[step:Apply the containment to the affine span]
The affine span $\operatorname{Aff}(S)$ is itself an affine subspace of $A$ containing $S$. Applying the previous step with $B=\operatorname{Aff}(S)$ yields
\begin{align*}
C\subset \operatorname{Aff}(S).
\end{align*}
Together with the inclusion $\operatorname{Aff}(S)\subset C$ proved in the first step, this gives
\begin{align*}
\operatorname{Aff}(S)=C=p_0+W=p_0+\operatorname{span}_k\{p-p_0 : p \in S\}.
\end{align*}
This is the desired equality.
[/step]
