Lower Semicontinuous function

Question

Let $(M,\delta)$ be Metric space, $f$ be a real function on $M$. Suppose $\{x \in M: f(x) > \alpha\}$ is open set in $M$ forall $\alpha \in \mathbb{R}$. Prove $f$ is a lower semicontinuous function.

Definition: Let $(M,\delta)$ be Metric space. $f$ is a real function on $M$.We say, $f$ is a lower semicontinuous function if:

$\forall$ $\{x_m\}\longrightarrow x \in M$, we have $f(x)\leq \liminf_{m\rightarrow \infty} f(x_m)$

Please help me!

score 2 · Answer 1 · answered Nov 28 '13 at 11:46

2

Hint: Let $\epsilon>0$. Show that $$\liminf_{m \to \infty} f(x_m) \ge f(x)-\epsilon$$ using the fact that $x \in \{y \in M: f(y) > f(x)-\epsilon\}$, which is an open set by assumption.

answered Nov 28 '13 at 11:46

Dustan Levenstein

12,941

My exercise sheet asks one to prove this in the case of a first countable topological space. We do we use first countability? – Danilo Gregorin Afonso Apr 14 '20 at 14:00

score 1 · Answer 2 · answered Nov 28 '13 at 12:00

1

Example of semicontinuous function :

$$ f(x) = \begin{cases} x^2,\ x\neq 0 \\ -1,\ x=0 \end{cases} $$

answered Nov 28 '13 at 12:00

HK Lee

19,964

1

How do you think this example? – Đặng Phước nhật Nov 28 '13 at 12:05
Consider the condition $f(x) < {\rm lim\ inf}\ f(x_m)$ when $x_m\rightarrow x$. for instance $x_m=\frac{1}{m},\ x=0$. – HK Lee Nov 28 '13 at 12:12

Lower Semicontinuous function

2 Answers2