Let be $\phi:X\rightarrow \mathbb{R}$ a lower semi-continuos function then $X = \cup_{n=1}^{\infty}\phi^{-1}(-n,\infty)$.

Why this?

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This is true for any function from $X$ to $\Bbb R$; it has nothing to do with semi-continuity.
$$X=\varphi^{-1}[\Bbb R]=\varphi^{-1}\left[\bigcup_{n\in\Bbb Z^+}(-n,\to)\right]=\bigcup_{n\in\Bbb Z^+}\varphi^{-1}\big[(-n,\to)\big]$$
simply because $\Bbb R=\bigcup\limits_{n\in\Bbb Z^+}(-n,\to)$.
What mean $\to$?, infinite positive or negative? –  user46060 Oct 26 '12 at 15:45
@user46060: I (and others) write $(a,\to)$ instead of $(a,\infty)$, and $(\leftarrow,a)$ instead of $(-\infty,a)$. –  Brian M. Scott Oct 26 '12 at 15:50
Understand, in my case the union is $\cup_{n=1}^{\infty}$ and not $\bigcup_{n\in\Bbb R}$ or these are equivalents? –  user46060 Oct 26 '12 at 16:18
@user46060: Yes, of course. If you read the whole line, the $\Bbb R$ in the last union was an obvious typo. I’ve fixed it now. –  Brian M. Scott Oct 26 '12 at 16:20