# upper/lower semi-continuous from topological definition via Example

Let me first state the definition of semi-continuous function.

Let $$X$$ be a space and $$f : X \rightarrow \mathbb{R}$$ a real-valued function on $$X$$. Then $$f$$ is upper semi-continuous if $$f^{-1} (-\infty, a)$$ is open for each $$a$$ in $$\mathbb{R}$$; $$f$$ is lower semi-continous if $$f^{-1}(a, \infty)$$ is open for each $$a$$ in $$\mathbb{R}$$.

and I know that

A function $$f$$ is continuous iff it is both upper and lower semi-continuous.

First I want to prove \begin{align} f(x) = \begin{cases} \frac{1}{x} \quad & x<0, \\ 0 \quad & x=0, \\ -\frac{1}{x} & x>0. \end{cases} \end{align} is upper semi-continuous. [From analysis, since $$\lim_{x\rightarrow 0} f(x) = -\infty < f(0)=0$$, I know this function $$f$$ is upper semi-continuous]

From the topological definition I have \begin{align} f^{-1} (-\infty, a) = \begin{cases} (-\frac{1}{|a|}, 0) \cup (0, \frac{1}{|a|}) \quad & a<0 \\ (-\infty, 0) \cup (0, \infty) \quad & a=0 \\ (-\infty, 0) \cup \{0\} \cup (0, \infty) = \mathbb{R} & a \geq 0 \end{cases} \end{align} so any case the inverse image is open so $$f$$ is upper semi- continuous.

But

\begin{align} f^{-1} ( a, \infty) = \begin{cases} \phi \quad & a \geq 0 \\ (-\infty, -\frac{1}{|a|}) \cup \{0\} \cup (\frac{1}{|a|}, \infty) & a <0 \end{cases} \end{align} this seems also open for arbitrary $$a\in \mathbb{R}$$...

However this cannot be true since $$f$$ is not continuous. [$$f(0) \neq \lim_{x\rightarrow 0} f(x)$$.] What's wrong with my approach?

$$(-\infty, -\frac{1}{|a|}) \cup \{0\} \cup (\frac{1}{|a|}, \infty)$$ is not open.
• I see your point. since $\{0\}$ is a closed set in $\mathbb{R}$, $(-\infty, -\frac{1}{|a|}) \cup \{0\} \cup (\frac{1}{|a|}, \infty)$ is not open. Oct 15, 2019 at 15:47