Let $S$ be a closed convex set & $x$ be an extreme pt of $S$ then $S-\{x\}$ is Let $S$ be a closed convex set and $x$ be an extreme point of $S$, then $S-\{x\}$ is


*

*Convex

*Not Convex

*May or may not be convex


I am thinking that the convexity doesn't fail even if we remove the extreme points. It's just my intuition from finite dimensional spaces. What about the answer? Is there anything different if the space is infinite dimensional? Please help me.
Thank you.
 A: If $a$ is an extreme point of $S$, then $a=\alpha x_1+(1-\alpha)x_2$ for some $x_1,x_2\in S$ implies $x_1=x_2=a$. (I.e., $a$ cannot be written as convex combination of points from $S\setminus\{a\}$.)
This means that $S\setminus\{a\}$ is convex. If you take any two points $x_1,x_2\in S\setminus\{a\}$ and make a convex combination $x=\alpha x_1+(1-\alpha)x_2$, then $x\in S$ (since the set $S$ is convex) and $x\ne a$ (because of the condition in the definition of extreme point).

In fact, these conditions are equivalent. (Which is not too difficult to show.)
Here is a quote from Conway's book A Course in Functional Analysis, p.142

Proposition 7.3. If $K$ is as convex subset of a vector space $X$ and $a\in K$, then the following statements are equivalent:
  
  
*
  
*$a$ is an extreme point of $K$
  
*If $x_1,x_2\in X$ and $a=\frac12(x_1+x_2)$, then either $x_1\notin K$ or $x_2\notin K$ or $x_1=x_2=a$.
  
*If $x_1,x_2\in X$, $0<t<1$, and $a=tx_1+(1-t)x_2$, then either $x_1\notin K$ or $x_2\notin K$ or $x_1=x_2=a$.
  
*If $x_1,\dots,x_n\in K$ and $a$ is in the convex hull of $\{x_1,\dots,x_n\}$, then $a=x_k$ for some $k$.
  
*$K\setminus\{a\}$ is a convex set.
  

Proof of this proposition is omitted in this book. (It is left as an exercise.)
