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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.

Thanks in advance, Lucas

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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 $x_1,x_2 \in int(S).$

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  • $\begingroup$ 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 $\operatorname{int}(S) \subset S \subset \operatorname{cl}(S)$ $\endgroup$
    – Lucas
    Feb 17 at 15:13

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