Property of lower semi-continuous maps

[JG89]

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Homework Statement

If X is a metric space then a map [itex] f: X \rightarrow \mathbb{R} [/itex] is called lower semi-continuous if for each ray of the type [itex] (a, + \infty) [/itex], 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 [itex] x \in X [/itex].

Homework Equations

The Attempt at a Solution

Here is my proof:

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

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

Is this proof correct?

 

[mighty2000]

Your proof is correct and well-written. However, I would like to offer an alternative proof that may be easier to follow for some readers.

Proof:

Since X is a compact metric space, it is also sequentially compact. This means that for any sequence (x_n) in X, there exists a subsequence (x_{n_k}) that converges to some point x \in X. Now, let (a_n) be a sequence in f(X). Since f is lower semi-continuous, the inverse image of any ray (a_n, +\infty) is open. Therefore, for each n, there exists an open set V_n \subset X such that x \in V_n and f(V_n) \subset (a_n, +\infty).

Now, consider the sequence (V_n). Since X is sequentially compact, there exists a subsequence (V_{n_k}) that converges to some open set V \subset X. By construction, f(V_{n_k}) \subset (a_{n_k}, +\infty) for all k. Taking the limit as k \rightarrow \infty, we have f(V) \subset (a, +\infty) where a = \lim_{k \rightarrow \infty} a_{n_k}. This means that f is bounded below by a.

Since a \in f(X), there exists some c \in X such that f(c) = a. Therefore, f attains its minimum value at c.

I hope this alternative proof is helpful to you and other readers. Keep up the good work in your studies!