# semi-continuous function

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.

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

$$f^{-1} ((-\infty,0])=\{x \in \mathbb R: f(x) \le 0\}=\{0\}.$$
• @PpooKkyPaeko What don't you understand ? Why is $f^{-1}((-\infty,0]) = \{0\}$, or how to proceed from here? Aug 26, 2019 at 11:43
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 $$M<-1$$?