Jtdylktuy

Does anyone knows how to prove this:

Let $f: (X, d) rightarrow mathbb{R}$ be an upper semi-continious function.

Prove that $f$ is u.s.c. if and only if $ { x | f(x) geq z } $ is closed $forall z in mathbb{R}$

Hope someone knows

Let $A_z := {x , |, f(x) ge z}$, for all $z in Bbb R$. Suppose $A_z$ is closed for every $z$. Then given $x_0 in X$ and $epsilon > 0$, the set $U_{x_0,epsilon} := X setminus A_{f(x_0) + epsilon}$ is an open neighborhood of $x_0$ such that $f(x) < f(x_0) + epsilon$ for all $x in U_{x_0,epsilon}$. Hence $f$ is upper semi-continuous.

Conversely, suppose $f$ is upper semi-continuous. Fix $zin Bbb R$. Given $x_0 in X setminus A_z$, set $epsilon = z - f(x_0)$. Then $epsilon > 0$. So by upper semi-continuity of $f$, there exists an open neighborhood $U$ of $x_0$ such that $f(x) < f(x_0) + epsilon = z$ for all $xin U$. Hence $x_0 in U subset X setminus A_z$. It follows that $X setminus A_z$ is open, so $A_z$ is closed.

We have
begin{align*}
{ f le s } text{ closed for all } s in mathbb{R}
iff & { f > -s } text{ open for all } s in mathbb{R} \
iff & { f > s } text{ open for all } s in mathbb{R}
end{align*}
Now a subset $A subseteq X$ is called open in $X$, when
begin{equation} tag{$star$} label{eq:offen}
forall a in A exists delta > 0: B_{delta}(a) subset A.
end{equation}
Because $f$ is lower semi continuous on $X$, we have
begin{equation*}
forall c < f(x) exists delta > 0: f(y) > c forall y in B_{delta}(x).
end{equation*}
(follows from the standard $varepsilon-delta$-definition with $varepsilon := f(x) - c$.)

Now let $A := { f > s }$.
From definition eqref{eq:offen} the proposition follows, since for all $a in A { c in mathbb{R}: f(x) > c}$ exists a $delta > 0$, so that for all $y$ with $| x - y | < delta$, we have $y in A$ and therefore $f(y) > c$.