Htykuut

Let $f:R^nrightarrow R$ be a lower semi-continuity function, how to show for any constant $r$ $U={zin R^n : f(z)> r}$ is open ?

Consider a sequence $x_n in mathbb{R}^nbackslash U = E$ and assume $x_nrightarrow x$. Then $f(x) le liminf f(x_n) le r$, so $xin E$ and $E$ 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 $f(y) > c$.