I have a doubt about upper semicontinuous functions. I have two definitions and I need to show that they are equivalente.
$f$ is upper semi-continuous at $x_{0}$ if for every $\varepsilon>0$, exists $\delta>0$ such that $f(x)<f(x_0)+\varepsilon$ if $x \in (x_0-\delta, x_0+\delta)$.
For the particular case of a metric space, I need to prove that $f$ is upper semicontinuous in $x_0$ if and only if
$$ \limsup _{x\to x_{0}}f(x)\leq f(x_{0}).$$
Thanks for any help,
Cleto
I realised that it was just a calculus exercise. I will indicate the answer. Maybe it will be usefull for someone. Thanks for the comments.
Suppose that for all $\epsilon >0$, there exists $\delta >0$ such that $x \in (x_0 - \delta, x_0 + \delta ) \cap X$ implies $f(x) \leq f(x_0) + \frac{\epsilon}{2}.$
Let $c$ and $(x_n) \in X \backslash \{x_0\}$ such that $\displaystyle{\lim_{n\rightarrow \infty} x_n = x_0}$ and $\displaystyle{\lim_{n\rightarrow \infty} f(x_n)} = c$. Let $n_0 \in \mathbb{N}$ such that $n>n_0 \Rightarrow \left|x_n - x_0\right| < \delta \ \textrm{e} \ \left|f(x_n) -c\right| < \frac{\epsilon}{2}.$ If $n>n_0$, as $f$ is upper semicontinuous at $x_0$, we have $ c -\frac{\epsilon}{2} < f(x_n) < f(x_0) +\frac{\epsilon}{2}$ In this way, we have $c < f(x_0) + \epsilon. $ As $\varepsilon$ is arbitrary, we conclude that $f(x_0)$ is an aderence value and, as $c$ is arbitrary $ \limsup_{x\rightarrow x_0} f(x) \leq f(x_0), $ as we want to show.
On the other hand, if $ \limsup_{x\rightarrow x_0} f(x) \leq f(x_0), $, the biggest aderence avalue of $f$ at $x_0$ is shorter than $f(x_0)$. This way, with the exception of a finite number of values, $(f(x_n))$ with $(x_n) \rightarrow x_0$ is shorter than $f(x_0)$. Let us consider only subsequence of $(f(x_n))$ bunded by $f(x_0)$. Suppose that $f$ is not upper semicontinuous at $x_0$, and that exists $\epsilon>0$ such that, for all $\delta >0$, there exists $x \in X$, $\left|x-x_0\right|< \delta$, such that $f(x_0) + \epsilon \leq f(x)$. We can define a sequence such that $\left|x_n-x_0\right| < \frac{1}{n}$ satisfies $f(x_0) + \epsilon \leq f(x_n).$ Then, the sequence $(f(x_n))$ is lower bounded and has a convergent subsequence $(f(x_{n_k}))$. Let $c = \displaystyle{\lim_{n_k \rightarrow \infty} f(x_{n_k})}$. By our construction, $f(x_0) \leq c$. As $c$ is an adherence valueof $(f(x_n))$, we conclude that $f(x_0) \leq c \leq \limsup_{x\rightarrow x_0} f(x), $ a contradiction.