2
$\begingroup$

Let $A$ convex set

$(1)$ If $y\in\bar{A}$ and $x\in\text{int}(A)$ then $\lambda x+(1-\lambda)y\in\text{int}(A)$, $\forall\,0<\lambda<1$

$(2)$ $\bar{A}$ and $\text{int}(A)$ are convex sets.

My teacher said (without showing how) that $(1)$ and $(2)$ can be used to show that

If $\text{int}(A)\neq \phi$ then $\bar{A}=\text{closure}(\text{int}(A))$ and $\text{int}(A)=\text{int}(\bar{A})$

But I am not sure why this is true. Does anyone know how this is true?

$\endgroup$
0
$\begingroup$

$\bar{A}=\text{closure}(\text{int}(A))$ by (1) you can show that y is a limit of points in the interior as $ \lambda $ goes to 1. That is true only if the interior is non empty. For the secend equlity think of $\bar{A} / \partial \bar{A}$ what are the set members(think of (1) once more)?

| cite | improve this answer | |
$\endgroup$

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.