2

Suppose $(X,d)$ is a metric space and a real-valued function $f$. Denote the oscillation of $f$ restricted to a closed subset $F \subseteq X$ at a point $x \in F$ by $$oscf|_{F}(x)=\inf_{U \ni x \space open} \sup \{ d(f(x_1), f(x_2)):x_1,x_2\in U \cap F \}$$

Claim: $oscf|_{F}$ is upper semicontinuous.

Definition: $oscf|_{F}:X\rightarrow \mathbb{R}$ is upper semicontinuous if for all real number $c$, the set $\{ x \in X: oscf|_{F}(x) < c \}$ is open in $X$.

My attempt: Let a real number $c$ be given and $oscf|_{F}(y) <c.$ By definition of infimum, there exists an open neighourhood $U$ of $y$ such that $\sup \{ d(f(x_1), f(x_2):x_1,x_2 \in U\cap F \} <c$. If $z \in U\cap F$, then $U$ is an open neighbourhood of $z$. Therefore $oscf|_{F}(z) <c.$ Hence, $U \cap F \subseteq \{ x \in F: oscf|_{F}(x) < c\}$ and $U \cap F$ is open in $F$. So the set $\{ x \in F: oscf|_{F}(x) < c\}$ is open in $F$ and $oscf|_{F}$ is upper semicontinuous.

Is my proof correct?

Martin Sleziak
  • 53,687
Idonknow
  • 15,643

1 Answers1

3

The structure of your proof is correct. You just need to be more precise about a few details and in showing that $osc\,f|_F(z) < c.$

Suppose $y \in \{x \in X: osc \, f|_F(x) < c \},$ with $osc \, f|_F(y) = a < c.$ Choosing $b$ such that $a < b < c$, there exists an open neighborhood $U$ of $y$ such that $\sup\{d(f(x_1),f(x_2)): x_1, \, x_2 \in U \cap F \} \leqslant b$.

Consider any $z \in U \cap F$. Since $U$ is open there exists an open neighborhood $V$ of $z$ such that $V \subset U$.

If $z_1, z_2 \in V \cap F \subset U \cap F$, then $d(f(z_1),f(z_2)) \leqslant \sup\{d(f(x_1),f(x_2)): x_1, \, x_2 \in U \cap F \} \leqslant b$ and, hence,

$$\sup \{d(f(z_1),f(z_2)): z_1,\, z_2 \in V\cap F \} \leqslant b, $$

which implies

$$osc \, f|_F(z) = \inf_{V, \, z \in V} \,\sup\{d(f(z_1),f(z_2)): z_1,\, z_2 \in V\cap F \}\leqslant b < c.$$

Thus, $U \subset \{x \in X: osc \, f|_F(x) < c \} $ and $\{x \in X: osc \, f|_F(x) < c \} $ is open in $X$.

RRL
  • 90,707