Statement: Polytopes are convex, thus: if $x, y \in P $ then $\lambda x + (1-\lambda)y \in P$ for $0 \leq \lambda \leq 1$
I understand that somehow it should mean that if we draw a line from $x$ to $y$ then it should stay inside a polytope and it won't cross the borders.
But I do not understand the formal statement. Could someone explain? Thank you
EDIT: I guess I don't understand how $\lambda x + (1-\lambda)y \in P$ defines the line segment...
On
The formal statement formalize the segment between the point $x$ and $y$. As you can see in the extreme value of $\lambda$ there are point $x$ ($\lambda = 1$) and $y$ ($\lambda = 0$). And for other value of the $\lambda$ the formal statement show a linear combination of $x$ and $y$. As $\lambda$ is between $0$ and $1$, the determined point is located between $x$ and $y$.
On
$P$ is the polytope.
$\forall x,y \in P$ means for any $x,y$ (any two points) that are in $P$.
$\lambda x + (1-\lambda) y, 0 \leq \lambda \leq 1$ is the line segement between $x$ and $y$.
Hence the statement is just what you stated, for any two points that you pick, form the line segment and the line segment should be part of the polytope.
Edit:
$\lambda x + (1-\lambda)y = y + \lambda (x-y)$,
when $\lambda = 0$, we are at $y$.
when $\lambda >0$, we are moving along the direction of $x-y$.
when $\lambda = 1$, we reach the $x$ location.
Hence it is a line segment connecting $x$ and $y$.
On
$\lambda x + (1-\lambda) y$ expresses a straight line segment, since it is the sum of a linear transformation of $x$ and a linear transformation of $y$.
And its endpoints are $x$ (at $\lambda = 1$) and $y$ (at $\lambda = 0$).
So "the line from $x$ to $y$ is fully in $P$" is precisely equivalent to the formal statement you were having trouble with.
On
the line segment connecting two points $x,y \in R^n$ is the set
$$\{x+t(y-x) ~ | \quad t \in [0,1] \}$$ in other words: the line starting from $x$ moving toward $y$ which is same as moving along the direction $y-x$, and stop on $y$. and note that the later set can be rewritten equivalently as follow
$$\{tx+(1-t)y ~ | \quad t \in [0,1]\}$$
Assume $x,y \in P$. Look at $f(\lambda) = \lambda x + (1-\lambda)y$.
Note that $f(0) = y$ and $f(1) = x$, and $f(1/2) = (x+y)/2$. As you try out different values of $\lambda \in (0,1)$ you will see that $f(\lambda)$ is indeed a parameterization of a line segment between $x$ and $y$.
Then, as you said, your statement says that if you find any pair of points in $P$, the line segment between them is also in $P$, which is a property of a convex set.
EDIT
To build some intuition, let $x = (1,0)$ and $y = (0,1)$. Note that $$ f(\lambda) = \lambda (1,0) + (1-\lambda)(0,1) = (\lambda, 1- \lambda). $$ Notice that as $\lambda$ varies from $0$ to $1$, you begin at $(0,1)=y$ and trace out the part of the line $y=1-x$ until you end up at $(1,0)=x$. The same happens between any two pairs of points in any space, feel free to experiment with other values to get some intuition how it works.