Below is the definition of a lower semicontinuous function:
We say that $\; f\;$ is lower semi-continuous at $\; x_{0} \;$ if for every $\; \epsilon \gt 0\;$ there exists a neighborhood $\; \mathcal U \;$ of $\;x_{0} \;$ such that $\; f(x)\ge f(x_{0})-\epsilon \;$ for all $ \; x \in \mathcal U\;$ when $\;f(x_{0})<+\infty \;$ , and $\;f(x)\;$ tends to $\; +\infty \;$ as $\;x\;$ tends towards $\; x_{0}\;$ when $\; f(x_{0})=+\infty\; $ . Equivalently, in the case of a metric space, this can be expressed as:
$\; \liminf _{x\to x_{0}}f(x)\ge f(x_{0})\;$ .
I'd like to know why the equivalence in the definition holds. I thought that will be a connection between the $\; \epsilon$-definition of infimum of a bounded set and the definition above but I'm missing the details.
I would appreciate if somebody could help me through this.
Thanks in advance!!!
In the case of metric space $(X,d)$ "there exists a neighborhood $\mathcal U$" is replaced by "there exists $\delta>0$" and neighborhood $B(x_0,\delta)$ is considered ($x\in B(x_0,\delta) \iff d(x_0,x)<\delta$). First definition (containing neighborhood $\mathcal U$) can be now reformulated as:
In simple words (1.-st case): for every $\epsilon>0$ all $x$-es thet are sufficiently close to $x_0$ have value $f(x)\ge f(x_0)-\epsilon$. So there is no sequence $x_n\rightarrow x_0$ such that $f(x_n)\rightarrow a<f(x_0)$. Therefore $\liminf_{x\rightarrow x_0}f(x)\ge f(x_0)$. 2.-nd case is similar.