This is Theorem $3.2$ from Serg Lang's 'Introduction to Linear Algebra' and Theorem $5.2$ from Lang's 'Linear Algebra'; I was hoping to get some clarification on the proof of the theorem.
Let $P_1,...,P_n$ be elements of a vector space V. Any convex set $S$ which contains $P_1,...,P_n$ also contains all linear combinations $$t_1P_1+...+t_nP_n=\sum_{i=1}^{n}t_iP_i$$ with $0\leq t_i$ for all $i$ and $\sum_{i=1}^nt_i=1.$
Pf:
The proof proceeds nicely, (as usual) by induction. If $n=1$, then $t_1=1$ and clearly the assertion follows.
Now assume the theorem is proved for some $n-1\ge1$; we prove it for $n$. Let $t_1,...,t_n$ be such that $$\sum_{i=1}^nt_i=1$$
If $t_n=1,$ then the assertion follows again trivially from the base case since $t_1=...=t_{n-1}=0$. So assume $t_n\neq1$ so we may write $$t_1P_1+...+t_nP_n=(1-t_n)\left(\frac{t_1}{1-t_n}P_1+...+\frac{t_{n-1}}{1-t_n}P_{n-1}\right)+t_nP_n$$
The author then states the following in the first text: $$\text{Let $s_i=\frac{t_i}{1-t_i}$ for $i=1,...,n-1$}$$ and in the second: $$\text{Let $s_i=\frac{t_n}{1-t_n}$ for $i=1,...,n-1$}$$
then after these statements in each text (which I think are both wrong?), the author concludes $$\text{Then $s_i\geq0$ and $\sum_{i=1}^{n-1}s_i=1$}$$
I think both of these expressions for $s_i$ are wrong; shouldn't $s_i$ be $$s_i=\frac{t_i}{1-t_n}\text{ for $i=1,...,n-1$}$$ But then, I don't see how $$\frac{t_1}{1-t_n}+...+\frac{t_{n-1}}{1-t_n}=\sum_{i=1}^{n-1}\frac{t_i}{1-t_n}=^?1$$
I get the following by the induction hypothesis: $$\sum_{i=1}^{n-1}\frac{t_i}{1-t_n}=\frac{\sum_{i=1}^{n-1}t_i}{1-t_n}=\frac{1}{1-t_n}$$
But I don't see how this helps me.
Can anyone help me decipher this proof?
You're right about the $s_i$; it should indeed be $$s_i = \frac{t_i}{1 - t_n}.$$ I think what you're missing is that $\sum_{i=1}^{n-1} t_i$ need not be $1$; it is assumed, as part of the induction step that $\sum_{i=1}^{n} t_i = 1$, and so $$\sum_{i=1}^{n-1} t_i = \sum_{i=1}^{n} t_i - t_n = 1 - t_n \implies \sum_{i=1}^{n-1} \frac{t_i}{1 - t_n} = 1.$$