Intuition for the "Flip Flop Formula" for the generalized inverse of non-decreasing functions In our lecture, the generalized inverse of a function $F$ is defined as
\begin{equation}
F^{-}(u) := \inf_x \{ F(x) \ge u \}.
\end{equation}
Then we are introduced to the so-called "Flip Flop Formula", i.e. for a non-decreasing function $F$ and its generalized inverse $F^{-}$,
$$x < F^{-}(u) \Leftrightarrow F(x) < u, \forall x, u.$$
Here is the proof: since $F$ is non-decreasing,
\begin{equation}\begin{aligned}
& x < F^{-}(u) = \inf_x \{ F(x) \ge u \} \\
\Leftrightarrow & x \notin \{ F(x) \ge u \} \\
\Leftrightarrow & F(x) < u.
\end{aligned}
\end{equation}
The proof is straightforward, but the formula itself is somewhat abstract. What's the intuition behind it? It would be great if someone could provide something like a graphical illustration.
 A: I have found a counterexample to the formula as stated, but it is valid if the case $x = F^-(u)$ is excluded.
Graphically, $F^-(u_0)$ is the point $x_0$ chosen so that the graph of $F$ to the left of $x_0$ is strictly below $u_0$, and the graph to the right of $x_0$ is non-strictly above $u_0$. That is, for any point $x \neq x_0$, then either $x < x_0$ and $F(x) < u_0$, or $x > x_0$ and $F(x) \geq u_0$.
$F(x_0)$ itself can be either below or above $u_0$: consider the functions $F_1(x) = \lceil x \rceil$ and $F_2(x) = \lfloor x + 1 \rfloor$, which have the same inverse $F^-(u) = \lceil u - 1 \rceil$. Thus, if $u$ is a non-integer, then $x = F^-(u) = \lfloor u \rfloor$ and so $F_1(x) = x < u < F_2(x) = x + 1$.
(I believe the formula is valid as written if the graph of $F$ has no segments with an open left endpoint: that is, if the right-handed limit $\lim_{x' \to x^+} F(x') = x$ for all $x$. This guarantees that $\{x: F(x) \geq u\}$ is topologically closed. Note that $F_1$ above violates this criterion.)
