# Convex set also has convex closure (alternative proof)

While studying convex analysis, i tried to write a proof for corollary 2 which i do not know if it is indeed correct.

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 $$\lambda \in(0,1)$$. I understood the proof regarding this one.

Corollary 1:

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

Answered here: Convex set also has convex interior (Corollary)

Corollary 2:

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

My attempt for corollary 2(alternative proof(?)):

For cl $$S$$ to be convex, one has to let $$x_{1}$$, $$x_{2}$$ $$\in$$ cl $$S$$ and show that $$\lambda x_{1}+(1-\lambda) x_{2} \in cl(S)$$.

Let $$x_{1}$$, $$x_{2}$$ $$\in$$ $$int(S)$$ $$\implies$$ $$\lambda x_{1}+(1-\lambda) x_{2} \in$$ int $$S$$ by corollary 1.

Furthermore, since $$\operatorname{int}(S) \subset S \subset \operatorname{cl}(S)$$ we conclude that $$x_{1}, x_{2} \in cl(S)$$ and that $$\lambda x_{1}+(1-\lambda) x_{2} \in$$ cl $$S$$. Therefore, $$cl(S)$$ is convex.

Original proof:

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)

Can someone help me? I dont know if my proof is correct.

Your proof is not correct. You take $$x_1,x_2 \in cl(S)$$ and later you asume that $$x_1,x_2 \in int(S).$$
You can not asume that $$x_1,x_2 \in int(S).$$
• I don't understand. What i did was letting $x_{1}, x_{2} \in \operatorname{int}(S)$ and then come to the conclusion that they belong to $cl(S)$ through the inclusion $\operatorname{int}(S) \subset S \subset \operatorname{cl}(S)$ Feb 17 at 15:13