Induction step in showing a convex set is convex iff it contains all convex combinations?

Question

I'm trying to follow this proof for the one in the title above.

It is here, from page 3 for the only if portion.

What I don't get is why they introduce this:

$$ y = (1-\lambda_m)[\sum_{i=1}^m \frac{\lambda_i}{1-\lambda_m}y_i] + \lambda_m y_m$$

since with each sum, the outside $(1-\lambda_m)$ simply cancels out the denominator. Isn't it the same as writing out:

$ y = \sum_{i=1}^m \lambda_iy_i + \lambda_m y_m$ ?

Answer 1

By definition a set $A$ is convex if $(1-t)x+ty \in A$ whenever $0\leq t \leq 1$ and $x,y \in A$. So the idea in this proof is to write $y$ as a convex combination of just two elements of $A$. We use induction hypothesis to see that $\sum\limits_{i=1}^m \frac {\lambda_i} {1-\lambda_m}y_i$ belongs to $A$. Call this $z$. Then we use the equaltio $y = (1-\lambda_m)z+\lambda_m y_m$ to conclude that $y \in A$.