Let $(X,\|\cdot\|)$ be a Banach space and $U_{X}$ its closed unit ball. According to Himmelber, Fixed points of compact multifunctions, a subset $B \subset X$ is said to be almost convex if given $\varepsilon >0$ if given $\{x_{1},\ldots x_{n}\}\subset B$ there are $\{z_{1},\ldots, z_{n}\}\subset B$ such that $\|x_{i}-z_{i}\|\leq \varepsilon$ and $\mathrm{co}(z_{1},\ldots, z_{n})\subset B$ ("co" denotes the convex hull). Of course, a convex set is almost convex.
It is not very hard to show (for instance, in the Euclidean plane) that in general, the nonempty intersection of almost convex sets is not an almost convex set. I am looking for an example of a bounded and almost convex (but not convex) subset of $X$ (infinite dimensional), say $B$, such that if $C$ is any family of almost convex subsets de $B$ with nonempty intersection, then $\cap_{A\in C}A\neq \emptyset$ is an almost set. I have tried with the following example.
Example: Let $X$ be the Banach space, endowed its usual supremum norm, of real continuous functions defined in $[0,1]$, and $P[0,1]$ the Berstein polynomials of the functions belonging to $U_{X}$, recall $$ P[0,1]:=\big\{ \sum_{i=0}^{n} \binom{n}{i} x\big( \frac{i}{n} \big) t^{i}(1-t)^{n-i}: x\in U_{X}, n\in\mathbb{N} \big\} , $$ which is a dense (and convex) subset of $U_{X}$. It Also, consider the sets $E[0,1]:=\{\exp(-nt):n\in\mathbb{N}\} $ and $B:=P[0,1]\cup E[0,1]$. Then, it is easy to check that $B$ is almost convex but not $B$ convex.
Now, let $C\subset B$ be any family of almost convex sets such that $A^{*}:=\cap_{A\in C}A\neq \emptyset$. We consider 3 cases:
Case 1: There is $P_{0}[0,1]\subset P[0,1]$ such that $P_{0}[0,1]\subset A$, $P_{0}[0,1]$ convex and dense in $A$ for each $A\in C$. Then, $P_{0}[0,1]\subset A^{*}$ and therefore $A^{*}$ is almost convex.
Case 2: $ A_{0}\subset E[0,1] \setminus P[0,1]$ for some $A_{0}\in C$. Then, it is easy to check that $A_{0}$ is almost convex if, and only if, $A_{0}=\{x_{0}\}$ for some $x_{0}\in E[0,1]$. As we are assuming that $A^{*}\neq\emptyset$, we have that $A=\{x_{0}\}$ for each $A\in C$ and then, trivially, $A^{*}$ is almost convex.
Case 3: There are $P_{0}\subset P[0,1]$ and $E_{0}[0,1]\subset E[0,1]$ such that $A_{0} = P_{0}[0,1]\cup E_{0}[0,1]$ for some $A_{0}\in C$.
I don't know how to prove (if it is true) that in Case 3 $A^{*}$ is almost convex. We can "relax" the assumptions by taking $S[0,1]:=\sin(2\pi t)$ (or any other non-polynomial function)
¿Some body can help? Somebody know an example of such set $B$?
Many thanks in advance for your comments.
There is no set $B$ for which you are looking for. Indeed, suppose for a contradiction that $B$ is a required set. Then $B$ is noncollinear, that is, contained in no straight line, because otherwise $B$ is convex. Let $\{x_1,x_2,x_3\}\subset B$ be any noncollinear set. It is easy to show that there exists $\varepsilon>0$ such that for each $\{z_{1},z_{2}, z_{3}\}\subset X$ with $\|x_{i}-z_{i}\|\leq \varepsilon$ for each $i\in\{1,2,3\}$, the set $\{z_1,z_2,z_3\}$ is noncollinear too. Suppose now that $\mathrm{co}(z_{1},z_{2}, z_{3})\subset B$. Let $z'_3$ be the midpoint of the segment $[z_1,z_2]$. Put $A_1=\mathrm{co}(z_{1},z_{3}, z'_{3})\setminus (z_3,z'_3)$ and $A_2=\mathrm{co}(z_{2},z_{3}, z'_{3})\setminus (z_3,z'_3)$. Then $A_1$ and $A_2$ are almost convex subsets of $B$, but $A_1\cap A_2=\{z_3,z'_3\}$ is not almost convex, a contradiction.