Points of continuity on distribution function

Question

From my understanding $F(x) = P(X \le x)$ and $p(x) = {\mathrm d \over \mathrm dx} F(x)$
We denote $F(x-) := \lim_{y \uparrow x} F(y).$

The claim is that $$p(x) = F(x) - F(x-) = P(X \le x) - P(X \lt x),\space~~ \forall \space x \in \mathbb R. $$

Can I get some help understanding why this statement is true? Maybe sketch of a proof? It seems similar to : $$p(x) = {\mathrm d \over \mathrm dx} F(x) = \lim_{y \uparrow x} {F(y) - F(x) \over x-a};$$ however I do not manage to see how this is developed.

This is interesting given the set of continuous points $C_F$ on the cumulative density function is given by $C_f = \{ x \in \mathbb R | F(x-) = F(x+)\}$ for $F(x+)$ defined similarly as $F(x-)$
$= \{ x \in \mathbb R | F(x-) = F(x)\} = \{ x \in \mathbb R | p(x) = 0 \}.$
I do not understand the intuition here. Say we have $U[0,1]$ so $p(x) = 1 \neq 0$ and $F(x) = x$ so is continuous on $[0,1].$

Answer 1

I think in your claim, "$p(x)$" was intended to denote $P(X = x)$, rather than the density.

If $F(x) - F(x-) > 0$ at some point $x$, then $X$ does not have a density, so the claim would not make sense if "$p(x)$" denoted density. Otherwise $F(x) - F(x-) = 0$ for all $x$, in which case the claim also does not make sense if "$p(x)$" denoted the density.