Upper semicontinuous function I want to ask a theorem.
"A function f is upper semicontinuous relative to a set E if and only if {$x\in E|f \ge a$} is relative closed for all finite a."
I know "{$x\in E|f \ge a$} is relative closed to E" is equivalent to "{$x\in E|f \lt a$} is relative open to E"
But how can  I prove {$x\in E|f \lt a$} is relative open to E directly.
I have no idea,please give me some suggestion,thank you!!!
 A: Suppose $f$ is upper semicontinuous on $E$.  Then for each $x \in E$ and each $\epsilon > 0$, there exists $\delta>0$ such that $|x - y| < \delta$ implies $f(y) \in (-\infty, f(x) + \epsilon)$.
In other words, for all $y$ near $x$, the image of $y$, $f(y)$, cannot be too far "above" $f(x)$.  If it is above $f(x)$, then it is close to $f(x)$.  The only way it can be "far" from $f(x)$ is to be below $f(x)$.
Alright, so we want to show for each $a \in \Bbb R$, $\{ x \in E \mid f(x) \geq a \}$ is closed in $E$, i.e., $\{x \in E \mid f(x) < a \}$ is open.
Let $z \in \{x \in E \mid f(x) < a \}$.  Then $f(z) < a$.  We can express $a$ as $f(z) + \epsilon$ for some $\epsilon >0$ (actually, for $\epsilon = a - f(z) >0$), so we know by the upper semicontinuity of $f$ at $z$ that there is $\delta > 0$ such that $|z - y| < \delta$ implies $f(y) \in (-\infty, f(z) + \epsilon)$.  Since $f(z) + \epsilon = a$, that means for some neighborhood around $z$, all the points in that neighborhood have image in $(-\infty, a)$, i.e., have image less than $a$.
So, if you are familiar with open ball notation, for each $y \in B(z,\delta)$, $f(y) < a$.  That means $B(z, \delta) \cap E$, which is relatively open in $E$, is an $E$-neighborhood contained in $\{x \in E \mid f(x) < a \}$, which proves $\{x \in E \mid f(x) < a \}$ is relatively open in $E$.
