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

Why this?

1 Answers 1

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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 $\Bbb R=\bigcup\limits_{n\in\Bbb Z^+}(-n,\to)$.

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    What mean $\to$?, infinite positive or negative?2012-10-26
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    @user46060: I (and others) write $(a,\to)$ instead of $(a,\infty)$, and $(\leftarrow,a)$ instead of $(-\infty,a)$.2012-10-26
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    Understand, in my case the union is $\cup_{n=1}^{\infty}$ and not $\bigcup_{n\in\Bbb R}$ or these are equivalents?2012-10-26
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    @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.2012-10-26