Is the support of a random variable those values where the graph of its distribution is not "flat"? In the literature the support, $S$, of a random variable $X$ is defined as the smallest closed subset of real line $\mathbb{R}$ with probability $1$. Looking to prove that $S$ is where the graph of $X$'s cdf, $F$, is not “flat”.  More formally, with $O_x$ an open interval containing $x$ with probability $P(O_x)$, show that:  $S=\{x: \forall O_x\quad P(O_x)\neq 0\}$.
To clarify, prove that $S=\{x: \forall O_x\quad P(O_x)\neq 0\}$ is closed with probability 1 and is the smallest such closed certain set. 
 A: If $x \notin S$, since $S$ is closed there is an open interval $O_x$ around $x$ that is disjoint from $S$, and since $P(S) = 1$ we must have $P(O_x) = 0$.  
Conversely, if there is an open interval $O_x$ around $x$ such that $P(O_x) = 0$, then
${\mathbb R} \backslash O_x$ is a closed subset of $\mathbb R$ with $P({\mathbb R} \backslash O_x) = 1$, so $S \subseteq {\mathbb R} \backslash O_x$, and in particular $x \notin S$.
A: Let $x\in S$ and suppose $P(O_x) = 0$ for some nontrivial open interval $O_x$ about $x.$ Then $S - O_x$ is a closed proper subset of $S,$ contradicting the minimality assumption. Conversely, if $x$ satisfies the aforementioned property for all such $O_x,$ then every neighborhood of $x$ intersects $S$ non-trivially - that is, $x\in S.$
A: You can do this by contradiction.
If $S$ is a closed set with $\Pr(X \in  S)=1$ and there is an $x \in S$ with an open set $O_x$ containing $x$ where $\Pr(X \in O_x)=0$ then $S\backslash O_x$ is a closed set smaller than $S$ [it does not contain $x$] and $\Pr(X \in  S\backslash O_x) \ge \Pr(X \in  S)-\Pr(X \in O_x) =1-0=1$. 
So $S$ is not the smallest closed subset with probability $1$ and so $x$ cannot be in the support.
