# Property of lower semi-continuous maps

1. Feb 15, 2012

### JG89

1. The problem statement, all variables and given/known data

If X is a metric space then a map $f: X \rightarrow \mathbb{R}$ is called lower semi-continuous if for each ray of the type $(a, + \infty)$, the inverse image of the ray under f is also open. If X is a compact metric space, prove that f is bounded below, and that f actually obtains its minimum value for some $x \in X$.

2. Relevant equations

3. The attempt at a solution

Here is my proof:

Let $U_a = (a, + \infty)$. Let $U = \{U_a : a \in f(X)\}$. Let $V_a = f^{-1}(U_a)$ and let $V = \{V_a : a \in f(X) \}$. Then V is an open cover of X and so it reduces to a finite subcovering $\{ V_{a_1}, V_{a_2}, ... , V_{a_n}$. Let $b = min \{a_1, a_2, ... , a_n$. Then $f(V_b) = (b, + \infty) \supset (a_i, + \infty)$ for $a_i \neq b$.

Now we show that f is bounded below by b. Choose any $x \in X$. $x \in V_{a_i}$ for some $i \in \{1, 2, ... , n \}$. Thus $f(x) \in (a_i, + \infty) \subset (b, + \infty)$. Thus f is bounded below. Furthermore, since $b \in U_b \subset f(X)$, there is a c in X such that $f(c) = b$. Thus f also attains its minimum value of b at the point c of X.

Is this proof correct?