Convex closure (need for proof explanation)

Question

I got stuck while reading a proof regarding convex sets (Corollary 2). I was supposed to use the first statement and Corollary 1 below, if necessary.

First statement(theorem):

Let $S$ be a convex set in $R^{n}$ with a nonempty interior. Let $x_{1} \in \operatorname{cl} S$ and $x_{2} \in$ int S. Then $\lambda x_{1}+(1-\lambda) x_{2} \in$ int $S$ for each $\lambda \in(0,1)$.

Corollary 1:

Let $S$ be a convex set. Then int $S$ is convex.

Corollary 2:

Let $S$ be a convex set with a nonempty interior. Then cl $S$ is convex.

-> Proof by the Author:

Let $x_{1}, x_{2} \in \operatorname{cl} S .$ Pick $z \in$ int $S$ (by assumption, int $\left.S \neq \varnothing\right) .$ By the theorem, $\lambda x_{2}+(1-\lambda) z \in$ int $S$ for each $\lambda \in(0,1) .$ Now fix $\mu \in(0,1) .$ By the theorem, $\mu \mathbf{x}_{1}+(1-\mu)\left[\lambda \mathbf{x}_{2}+(1-\lambda) \mathbf{z}\right] \in$ int $S \subset S$ for each $\lambda \in(0,1) .$ If we take the limit as $\lambda$ approaches $1,$ it follows that $\mu \mathrm{x}_{1}+(1-\mu) \mathrm{x}_{2} \in cl(S),$ and the proof is complete.

From the book: Nonlinear Programming (Theory and Algorithms)

Where am i stuck?

I just cannot understand the part where the author says:

"If we take the limit as $\lambda$ approaches $1,$ it follows that $\mu \mathrm{x}_{1}+(1-\mu) \mathrm{x}_{2} \in cl(S)..."$

I can see that the limit equals $\mu \mathrm{x}_{1}+(1-\mu) \mathrm{x}_{2}$. However, two things are really bothering me:

1. I thought that $\lambda$ was a fixed number in $(0, 1)$ and because of that i should not be able to just "consider $\lambda$ as it approaches 1". Why can he do that?

2. I cannot see why $\mu \mathrm{x}_{1}+(1-\mu) \mathrm{x}_{2} \in cl(S)$. It has something to do with the set inclusions $int(S) \subset S \subset cl(S)$? Or does it have something to do with the fact that $S$ is convex? I also considered the possibility that the limit points are also in $S$ since $cl(S)$ is closed.

Can someone help me? I am really struggling.

Thanks in advance, Lucas

Answer 1

(#1) The statement is true for every choice of $\lambda \in (0,1)$. That means it's true for $\lambda=0.9, \lambda=0.99, \lambda=0.999$, etc., and you can choose these $\lambda$ such that their limit is $1$.

(#2) You have a sequence of points $\mu \mathbf{x}_{1}+(1-\mu)\left[\lambda \mathbf{x}_{2}+(1-\lambda) \mathbf{z}\right] \in S$. Taking the limit as $\lambda \to 1$, you can see the points approach $\mu \mathbf{x}_{1}+(1-\mu)\mathbf{x_2}$. In other words, this is a convergent sequence of points contained in $S$; now apply your definition of closure.