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Given $f: \mathbb{R} \to \mathbb{R}$ and \begin{align*} f(x) = \begin{cases} x^2 & \text{if}\; x \not= 0 \\ -1 &\text{if}\; x = 0 \end{cases} \end{align*} Please help me prove it follows this definition. $f$ is lower semi-continuous iff $f^{-1} ((-\infty,M]):=\{ x \in \mathbb{R}: f(x) \leq M \}$ is closed. I want to show $f^{-1} ((-\infty,0])$ is closed, then f is lower semi continuous.

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    $\begingroup$ And what do you get for $f^{-1} ((-\infty,0])$? $\endgroup$
    – Martin R
    Aug 26, 2019 at 11:24

2 Answers 2

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$f^{-1} ((-\infty,0])=\{x \in \mathbb R: f(x) \le 0\}=\{0\}.$

Can you proceed ?

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  • $\begingroup$ why? I don't understand $\endgroup$ Aug 26, 2019 at 11:30
  • $\begingroup$ @PpooKkyPaeko What don't you understand ? Why is $f^{-1}((-\infty,0]) = \{0\}$, or how to proceed from here? $\endgroup$ Aug 26, 2019 at 11:43
  • $\begingroup$ How to proceed? $\endgroup$ Aug 26, 2019 at 11:46
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Hint

The function $f(x)$ maps $[-x_0,x_0]$ to $\{-1\}\cup (0,x_0^2]\subseteq(-\infty,M]$ with $M=x_0^2$. What if either $x_0$ is $0$ or $M<-1$?

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