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?


$\bar{A}=\text{closure}(\text{int}(A))$ 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)?

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