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


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.