Affine hull smallest affine space

Question

I would like to prove the following statement:

"The affine hull is the smallest affine space containing $S$, where $S$ is an arbitrary set".

I think the proof is rather trivial, but I cannot find it.

Answer 1

With the affine hull defined as $$\text{aff}(S) = \{ \sum_{i=1}^k a_ix_i | x_i \in S, a_i \in R, \sum_{i=1}^k a_i = 1, k\geq1\}$$ we can show it as follows.

First we need to show that $\text{aff}(S)$ is an affine space, then we show it is the smallest. To show that $\text{aff}(S)$ is an affine space we need only show it is closed under affine combinations. This is simply because an affine combination of affine combinations is still an affine combination. But I'll provide full details here.

Suppose that for $i=1\ldots k$ we have elements $s_i\in S$ say $$s_i = \sum_j a_{i,j}x_{i,j}$$ Where the $x_{i,j}\in S$ and for fixed $i$ we have $\sum_j a_{i,j}=1$. and we have $a_i\in \mathbb R$ such that $\sum_i a_i = 1$. Then $$\begin{align} \sum_i a_is_i &= \sum_{i=1}^ka_i\left(\sum_j a_{i,j}x_{i,j}\right)\\ &= \sum_{i,j} (a_ia_{i,j})x_{i,j}$ \end{align}$$ This is an element of $\text{aff}(S)$ since $$\begin{align} \sum_{i,j} (a_ia_{i,j})&= \sum_i a_i(\sum_j a_{i,j}) \\ &= \sum_i a_i\cdot 1 \\ &= \sum_i a_i = 1 \end{align}$$ So we have established that $\text{aff}(S)$ is an affine space.

Now to show that $\text{aff}(S)$ is the smallest affine space suppose that $A$ is an affine space containing $S$. Let $s=\sum_i a_ix_i\in \text{aff}(S)$. Since $A$ is closed under affine combinations and $x_i\in S\subseteq A$ it follws that $s\in A$. Therefore $\text{aff}(S)\subseteq A$ and we're done.