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$\newcommand{\span}{\operatorname{span}}$ $\newcommand{\aff}{\operatorname{aff}}$

Thm: Let $S\subseteq \mathbf{R}^n$. Then for any $m\in \aff S $ (in particular, for any $m\in S$) $\aff S=\left\{ m \right\}+ \span(S-S)$

Cor:Let $S\subseteq \mathbf{R}^n$. Then $x \in \aff S$ iff it is a linear combination of finitely many members of $S$ such that the coefficients are summed to $1$.

where $\aff S$ is the affine hull of $S$ which is the intersection of all affine subspaces that contains $S$, and $\span(S)$ is subspace generated by $S$(i.e., smallest subspace that contains $S$)

I am trying to prove this corrollary by using theorem, but have no idea at all. Any help would be greatly appreciated!

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Funny, I would normally expect the characterisation in the corolary to be the definition of the afffine hull and what you give as a definition a consequence. But anyway, using just the theorem to prove the corollary:

($\Rightarrow$) Suppose that $x \in \text{aff}(S)$. Take any $m \in S$. By the theorem, $x = m + \lambda_1 (s_1 - t_1) + \dots + \lambda_k (s_k - t_k)$ for certain $\lambda_i \in {\mathbb R}$ and certain $s_i, t_i \in S$. Note that this already expresses $x$ as a linear combination of finitely members of $S$ with the coefficients summing to 1: the 1 is in front of the $m$ and the $\lambda_i$ and $-\lambda_i$ in front of the $s_i$ and $t_i$ cancel.

($\Leftarrow$) Suppose that $x = \lambda_1 s_1 + \dots + \lambda_k s_k$ for certain $\lambda_i \in {\mathbb R}$ and certain $s_i \in S$ with $\lambda_1 + \dots + \lambda_k = 1$. Take $m = s_1$ and note that $x = m + \lambda_1 (s_1 - m) + \dots + \lambda_k (s_k - m)$: the first $m$ cancels with the $-\lambda_1 m - \dots - \lambda_k m_k$. Therefore $x \in \{m\} + \text{span}(S - S)$ and hence, by the theorem, $x \in \text{aff}(S)$.

Also note that ($\Rightarrow$) assumes $S \neq \emptyset$ to be able to take an $m \in S$ and apply the theorem. If $S = \emptyset$, $\text{aff}(S) = \{0\}$ by your definition; the theorem is then trivially true, but the corollary is false.

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