Convex set for a set of points in 2d plane There are set of five points $A(0,0) ,B(1,1) ,C(2,0) ,D(2,2).E(0,2) F(1.5,1.5)$
$S=\{A,B,C,D,E,F\}$

Please tell me whether my understanding is correct or not!


*

*The points $A,C,D,E$ forms a convex set for $S$.(i.e the convex set of $S$ contains $A,C,D,E$ and not all points of $S$).

*If asked whether Set $S$ is a convex set or not the answer is no.

*The convex set of a set $S$ need not contain all the points in $S$.
 A: The points $A,C,D,E$ do not form a convex set. This in the sense that set $\{A,C,D,E\}$ is not a convex set. You could highly say that the square marked by corners $A,C,D,E$ is convex. This square is the smallest convex set that contains these points. Secondly, it also contains the points $B$ and $F$ so that it also is the smallest convex set that contains set $S$. It is also named the convex hull of set $S$. I think that you call it the convex set of set $S$. Set $S$ is not convex. A set is convex iff it coincides with its convex hull.
A: Note, that the convex set has to be convex, and $\{A,C,D,E\}$ is not, e.g. it does not contain the line from $A$ to $E$.
The convex hull is the whole set, not only its boundary.
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Be aware, that algorithms that calculate the convex hull actually only output its boundary. This is because it is impractical to generate an infinite output. Hence, a finite representation is required and the boundary is exactly such a thing (i.e. it is finite if the input set itself was finite).
To conclude, your convex set is the whole rectangle with $A,C,D,E$ as corners.
I hope this helps $\ddot\smile$
