True or false: non degenerate r.v $X$ iff $\mathrm{P}(X
Let $X:\varOmega \to \mathbb{R}$ be a random variable. Random variable $X$ is degenerate if
  for some $c\in \mathbb{R}$ we have $\mathrm{P}(X=c)=1.$ 
True or false? $X$ is a non degenerate random variable iff for some $a\in \mathbb{R}$ we have  $\mathrm{P}(X<a)\in (0,1).$
Attempt.
Converse: true. Suppose that $X$ is a.e. equal to a constant $c$. If $c<a$ then $1=\mathrm{P}(X=c)\leqslant \mathrm{P}(X<a)<1$, contradiction and if  $c\geqslant a$ then $0=\mathrm{P}(X\neq c)\geqslant \mathrm{P}(X<a)>0$, contradiction.
Regarding the other direction I believe the answer is yes (non degenerate: for every $c\in \mathbb{R}$ we have $\mathrm{P}(X=c)<1$), but I haven't been able to reach an $a$, as wanted.
Thanks for the help. 
 A: Hint: Suppose $P(X<a)=0$ or $1$ for all $a$. Let $c=\sup \{a: P(X<a) =0\}$. Show that $P(X<c)=0$ and $P(X <c+\epsilon) =1$ for all $\epsilon >0$. Letting $\epsilon \to 0$ this gives $P(X \leq c)=1$. Hence $P(X=c)=P(X \leq c) -P(X<c)=1-0=1$. 
A: It might be more handsome to prove the equivalence of negations:$$X\text{ is degenerated}\iff P(X<a)\in\{0,1\}\text{ for every }a\in\mathbb R$$

$\implies$ 
Let it be that $P(X=c)=1$. Then evidently $P(X<a)=1$ if $c<a$ and $P(X<a)=0$ otherwise.
$\impliedby$ 
Let it be that $P(X<a)\in\{0,1\}$ for every $a\in\mathbb R$. 
It cannot be that $P(X<a)=0$ for every $a\in\mathbb R$ because this leads to $P(X\in\mathbb R)=0$ which is absurd. So we do have $P(X<a)=1$ for some $a$. 
It cannot be that $P(X<a)=1$ for every $a\in\mathbb R$ because this leads to $P(X\in\varnothing)=1$ which is absurd. So we do have $P(X<b)=0$ for some $b$. 
Then the set $\{a\in\mathbb R\mid P(X<a)=1\}$ is not empty and has a lower bound. Consequently it has an infinum $c$. It can be proved now that $P(X=c)=1$.
Proving that is not difficult.
