$G_{1}$ and $G_{2}$ are CDFs of some non-degenerate distributions specified on $\mathbb{R}$ and $\widetilde{G}_{1}$, $\widetilde{G}_{2}$ are corresponding generalized inverses ($\widetilde{f}(u)=\inf \lbrace t: f(t) \geq u \rbrace$). Prove the existance of $u_{1}<u_{2}$, such that: $$ \widetilde{G}_{1}(u_{1}) < \widetilde{G}_{1}(u_{2}) \\ \widetilde{G}_{2}(u_{1}) < \widetilde{G}_{2}(u_{2}) $$ and $ u_{1},u_{2} \in \mathbb{T}(\widetilde{G}_{1}) \cap \mathbb{T}(\widetilde{G}_{2}) $, where $ \mathbb{T}(f) = \lbrace x \in \mathbb{R}$ such that $f$ is continuous in x and $f(x)$ is finite $\rbrace $
I know that: $\widetilde{G}_{1},\widetilde{G}_{2}$ are non-decreasing and set of discontinuity points of any monotonic function might be at least countable. Thus, it's easy to show that $ u_{1},u_{2} \in \mathbb{T}(\widetilde{G}_{1}) \cap \mathbb{T}(\widetilde{G}_{2}) $, but I don't know how to prove the two remaining inequalities are satisfied. Thanks for any help.
$G$ is a monotonic increasing(non decreasing) function. So we know$$ a\le b \Leftrightarrow G(a) \le G(b)$$(Strict inequality if continuous inceasing over the period containing $a,b$).What if we put $a=\widetilde {G}(u_1)$ and $b$ similarly by choosing any $u_1\lt u_2$ in $\mathbb{T}(\widetilde{G}_{1}) \cap \mathbb{T}(\widetilde{G}_{2})$ ? Consider the fact that if $u\in \mathbb{T}(\widetilde{G}_{1}) \cap \mathbb{T}(\widetilde{G}_{2})$ then at $u$ ,$G$ doesn't have any flat region (i.e. is one one) ,so we can write $u=G(\widetilde{G} (u))$