I would like to prove the equality given in the title, that is $$\inf_{z>y} \inf E_z=\inf \bigcup_{z>y}E_z, \ E_z:=\{x\in \mathbb{R } :f(x)>z\}$$
I don't think any more information is needed but for a reference this is a step in a proof that the generalized inverse of (any) real-valued function is right-continous and the proof is given in Proposition 4.2 in the following dokument (on page 10).
What would be the approach to prove the equality?
Particularly the infimum of the infimum of the left hand side makes me a bit confused. But I suppose it is means that for every $z $ we get a greates lower bound on $E_z $ and we collect all of those greatest lower bound into a set of which we then takes the greatest lower bound?
Most grateful for any help provided!
In the question are not specified the domain and the range of the function $f$. I guess you mean that $f:\Bbb R\to\Bbb R$, whereas in the proof of Proposition 4.2 deal with $f:\bar{\Bbb R}\to\bar{\Bbb R}$. But we don’t need to specify this, we don’t need to know what are $E_z$. It suffice to know that they are non-empty subsets of a linearly ordered space $L$ such that each non-empty open subset $E\subset L$ has $\inf E\in L$. Remark that both $\Bbb R$ and $\bar{\Bbb R}$ satisfy this condition.
Put $i_1=\inf_{z>y} \inf E_z $ and $i_2=\inf \bigcup_{z>y}E_z $. Pick any $z>y$. Since $i_2\le x$ for each $x\in E_z$, $i_2\le\inf E_z$. Since this holds for each $z>y$, $i_2\le \inf_{z>y} \inf E_z=i_1$. Now pick any $x\in \bigcup_{z>y}E_z$. Then $x\in E_z$ for some $z>y$. So $i_1\le\inf E_z\le x$. Since this holds for each $x\in \bigcup_{z>y}E_z$, $i_1\le\inf \bigcup_{z>y}E_z=i_2$.