Well, the definition of upper semicontinuity that I know is (1)
- $F:X\to \mathbb{R}$ is upper semicontinuous at $x_0\in X$ if for all $\varepsilon >0$, there exists a $\delta>0$ such that $x\in X\cap(x_0-\delta,x_0+\delta)\Rightarrow f(x)< f(x_0)+\varepsilon$
- For $x_0\in X$, $\limsup_{x\to x_0}f(x)\leq f(x_0)$
I want to prove $(1)\Leftrightarrow (2)$ This is what I've tried:
$(2)\Rightarrow (1)$: Let $L=\limsup_{x\to x_0}f(x)$. By definition of infimum, given $\varepsilon >0$, there exists $\delta>0$ s.t. $\sup\{f(x):X\cap |x-x_0|<\delta\}<L + \varepsilon$, then, for this $\delta$, we have $X\cap |x-x_0|<\delta\Rightarrow f(x)< L+\varepsilon \leq f(x_0)+\varepsilon$
$(1)\Rightarrow (2)$: Given $\varepsilon>0$, there exists $\delta>0$ s.t. $X\cap |x-x_0|<\delta\Rightarrow f(x)<f(x_0)+\varepsilon \Rightarrow \sup\{f(x):X\cap|x-x_0|<\delta\}\leq f(x_0)+\varepsilon.$ By definition of infimum, $\limsup_{x\to x_0}f(x)\leq \sup\{f(x):X\cap|x-x_0|<\delta\}\leq f(x_0)+\varepsilon$ and $\varepsilon$ was arbitrary. Hence $\limsup_{x\to x_0}f(x)\leq f(x_0)$.
Is this correct?