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Definition. A set $C \subseteq \mathbb{R}^n$ is affine if for any $x_1, x_2 \in C$ and $\theta \in \mathbb{R}$, $\theta x_1 + (1-\theta)x_2 \in C$.

The claim I wish to prove:

If $C$ is affine and $x_1, \dots, x_k \in C$ and $\theta_1, \dots, \theta_k \in \mathbb{R}$ satisfy $\sum_{i=1}^{k}\theta_i = 1$, then $\sum_{i=1}^{k}\theta_i x_i \in C$.

The base case is easy.

Now, suppose for some $m \geq 2$ that $$x_1, \dots, x_m \in C\text{ and } \sum_{i=1}^{m}\theta_i = 1 \implies \sum_{i=1}^{m}\theta_i x_i \in C\text{.}$$ Let $x_{m+1} \in C$ and $\theta_{m+1} \in \mathbb{R}$ be such that $\sum_{i=1}^{m+1}\theta_i = 1$.

Then, assuming $\theta_{m+1} \neq 1$, $$\sum_{i=1}^{m+1}\theta_i x_i = \sum_{i=1}^{m}\theta_ix_i+\theta_{m+1}x_{m+1}=(1-\theta_{m+1})\sum_{i=1}^{m}\dfrac{\theta_i}{1-\theta_{m+1}}x_i+\theta_{m+1}x_{m+1}\text{.}\tag{*}$$ Now observe $$\sum_{j=1}^{m}\dfrac{\theta_j}{1-\theta_{m+1}}=\dfrac{1}{1-\theta_{m+1}}\sum_{j=1}^{m}\theta_j=\dfrac{1}{\sum_{j=1}^{m}\theta_j}\sum_{j=1}^{m}\theta_j=1$$ so by the induction hypothesis, $$\sum_{i=1}^{m}\dfrac{\theta_i}{1-\theta_{m+1}}x_i \in C$$ and setting $\theta^{\prime} = 1-\theta_{m+1}$, since $C$ is affine, we have $$\theta^{\prime}\sum_{i=1}^{m}\dfrac{\theta_i}{1-\theta_{m+1}}x_i+(1-\theta^{\prime})x_{m+1} \in C$$ so the proof is done. $\square$

My question: I understand the proof above, but one thing annoys me: an implicit assumption in this proof is that $\theta_{m+1}\neq 1$ (because, if this were not the case, we would be dividing by $0$). Is there a way to avoid having to make this assumption? Is it reasonable to make the assumption that $\theta_{m+1}\neq 1$, or is this assumption actually not being made, even though I think it is?

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    $\begingroup$ If $\theta _{m+1}=1$, you may just change the label of the $x_i$'s to get back to the case where it is not $1$. Naturally, all the $\theta_i$ can not be equal to $1$ because $m\geq 2$. $\endgroup$
    – Suzet
    Commented Jun 6, 2018 at 12:29
  • $\begingroup$ @Kronnected The $\theta _i$'s are not assumed to be nonnegative. $\endgroup$
    – Suzet
    Commented Jun 6, 2018 at 12:32
  • $\begingroup$ @Suzet Could you please elaborate what you mean by "chang[ing] the label of the $x_i$s"? $\endgroup$ Commented Jun 7, 2018 at 1:17

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The same idea could be expressed as: since $\sum_{i=1}^{m+1}\theta_i = 1$, there exists $k\in \{1, \dots, m+1\}$ such that $\theta_k\ne 1$. Then $$ \sum_{i=1}^{m+1}\theta_i x_i = \sum_{i\ne k}\theta_ix_i+\theta_{k}x_{k}=(1-\theta_{k})\sum_{i\ne k} \dfrac{\theta_i}{1-\theta_{k}}x_i+\theta_{k}x_{k} $$ and so on. However, the author prefers to work with sums over consecutive indices. To this end, an observation is implicitly made that the numbering of terms $\theta_ix_i$ can be changed: we are not tied to any particular numbering. These terms can be renumbered so that the one we used to call $\theta_k$ is now $\theta_{m+1}$.

Compare to: "Let $x_1, x_2, x_3$ be real numbers. Without loss of generality we can assume $x_1\le x_2\le x_3$."

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  • $\begingroup$ So, if I'm understanding your answer, it has to be the case that one of the theta has to not be equal to 1? $\endgroup$ Commented Jun 7, 2018 at 4:43
  • $\begingroup$ It has to be the case, because the sum of $m+1$ numbers would not be equal to $1$ if each of them was equal to $1$. $\endgroup$
    – user357151
    Commented Jun 7, 2018 at 7:30
  • $\begingroup$ Wow, I realized how obvious this was earlier today. Appreciate your patience. For some bizarre reason, I had interpreted the proof to mean that there could not be a $\theta_j$ with $\theta_j = 1$, which is completely different than what is being stated here: that there has to be ONE $\theta_j$ with $\theta_j \neq 1$. This makes complete sense - you're arguing this by contradiction by saying that suppose that all $\theta_j$ are with $\theta_j = 1$, which obviously sums to $m+1 > 1$, so contradiction. Sorry for the confusion. $\endgroup$ Commented Jun 7, 2018 at 11:36

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