(This is a cross-post from here.) Let $A,B\subseteq \mathbb R^d$ be non-empty, such that $B\subseteq \overline A.$ For $S\subseteq\mathbb R^d$ define the relative interior of $S$ by $$\text{ri}(S)=\{s\in S\mid \exists\varepsilon>0:B_\varepsilon(s)\cap\operatorname{aff}(S)\subseteq S\}.$$ Here, $B_\varepsilon(x)$ is the open $d$-dimensional ball with center $x$, $\operatorname{aff}$ denotes the affine hull, $\operatorname{ri}$ denotes the relative interior and $\operatorname{conv}$ denotes the convex hull.

1. Do we have $\operatorname{conv}(A)\cap\operatorname{ri}(\text{conv}(B))\neq\{\emptyset\}?$
2. Or do we even have $A\cap \text{ri}(\text{conv}(B))\neq\{\emptyset\}?$

My intuition is the following (I failed proving it this way): We know that $\text{ri}(\text{conv}(B))$ is non-empty, hence let $x\in\text{ri}(\text{conv}(B)).$ Now $x$ can be approximated by elements in $\text{conv}(\overline A)$, hence also by elements in $\text{conv}(A).$ Choosing a convex combination $a$ close enough to $x$, this should be our solution.

(This is a cross-post from here.) Let $A,B\subseteq \mathbb R^d$ be non-empty, such that $B\subseteq \overline A.$ For $S\subseteq\mathbb R^d$ define the relative interior of $S$ by $$\text{ri}(S)=\{s\in S\mid \exists\varepsilon>0:B_\varepsilon(s)\cap\operatorname{aff}(S)\subseteq S\}.$$ Here, $B_\varepsilon(x)$ is the open $d$-dimensional ball with center $x$, $\operatorname{aff}$ denotes the affine hull, $\operatorname{ri}$ denotes the relative interior and $\operatorname{conv}$ denotes the convex hull.

1. Do we have $\operatorname{conv}(A)\cap\operatorname{ri}(\text{conv}(B))\neq\{\emptyset\}?$
2. Or do we even have $A\cap \text{ri}(\text{conv}(B))\neq\{\emptyset\}?$

My intuition is the following (I failed proving it this way): We know that $\text{ri}(\text{conv}(B))$ is non-empty, hence let $x\in\text{ri}(\text{conv}(B)).$ Now $x$ can be approximated by elements in $\text{conv}(\overline A)$, hence also by elements in $\text{conv}(A).$ Choosing a convex combination $a$ close enough to $x$, this should be our solution.

(This is a cross-post from here.) Let $A,B\subseteq \mathbb R^d$ be non-empty, such that $B\subseteq \overline A.$ For $S\subseteq\mathbb R^d$ define the relative interior of $S$ by $$\text{ri}(S)=\{s\in S\mid \exists\varepsilon>0:B_\varepsilon(s)\cap\operatorname{aff}(S)\subseteq S\}.$$ Here, $B_\varepsilon(x)$ is the open $d$-dimensional ball with center $x$, $\operatorname{aff}$ denotes the affine hull, $\operatorname{ri}$ denotes the relative interior and $\operatorname{conv}$ denotes the convex hull.

1. Do we have $\operatorname{conv}(A)\cap\operatorname{ri}(\text{conv}(B))\neq\{\emptyset\}?$
2. Or do we even have $A\cap \text{ri}(\text{conv}(B))\neq\{\emptyset\}?$

My intuition is the following (I failed proving it this way): We know that $\text{ri}(\text{conv}(B))$ is non-empty, hence let $x\in\text{ri}(\text{conv}(B)).$ Now $x$ can be approximated by elements in $\text{conv}(\overline A)$, hence also by elements in $\text{conv}(A).$ Choosing a convex combination close enough to $x$, this should be our solution $a$.

(This is a cross-post from here.) Let $A,B\subseteq \mathbb R^d$ be non-empty, such that $B\subseteq \overline A.$ For $S\subseteq\mathbb R^d$ define the relative interior of $S$ by $$\text{ri}(S)=\{s\in S\mid \exists\varepsilon>0:B_\varepsilon(s)\cap\operatorname{aff}(S)\subseteq S\}.$$ Here, $B_\varepsilon(x)$ is the open $d$-dimensional ball with center $x$, $\operatorname{aff}$ denotes the affine hull, $\operatorname{ri}$ denotes the relative interior and $\operatorname{conv}$ denotes the convex hull.

1. Do we have $\operatorname{conv}(A)\cap\operatorname{ri}(\text{conv}(B))\neq\{\emptyset\}?$
2. Or do we even have $A\cap \text{ri}(\text{conv}(B))\neq\{\emptyset\}?$

My intuition is the following (I failed proving it this way): We know that $\text{ri}(\text{conv}(B))$ is non-empty, hence let $x\in\text{ri}(\text{conv}(B)).$ Now $x$ can be approximated by elements in $\text{conv}(\overline A)$, hence also by elements in $\text{conv}(A).$ Choosing a convex combination $a$ close enough to $x$, this should be our solution.