Ex4-10

Question:
10. (a) If \displaystyle{f} is deined and continuous on E, show that\{a < f < b\} is relatively open and that\{a \leq f \leq b\} and\{ f = a\} are relatively closed.
(b) Let \displaystyle{f} be a finite function on \mathbb{R^n}. Show that \displaystyle{f} is continuous on \mathbb{R^n} if and only if f^{-1}(G) is open for every open G in \mathbb{R^1}, or if and only if f^{-1}(F)is closed for every closed F in \mathbb{R^1}.
Answer: 
(a) We first prove two Lemma.
Lemma 1: The function f: E \to \R is continuous on E \subseteq \R^n if and only if f^{-1}(G) is relatively open in E for every open set G \subseteq \R.
Proof: Suppose f: E \to \R is continuous on E. Given an open set G \subseteq \R. Let \mathbf{x}_0 \in f^{-1}(G). Then f(\mathbf{x}_0) \in G. Since G is open in \R, there exits \varepsilon > 0 such that (f(\mathbf{x}_0) - \varepsilon, f(\mathbf{x}_0) + \varepsilon) \subseteq G. Since f is continuous at \mathbf{x}_0, there exists \delta > 0 such that if \mathbf{x} \in E and \|\mathbf{x} - \mathbf{x}_0\| < \delta then |f(\mathbf{x}) - f(\mathbf{x}_0)| < \varepsilon. Thus, there is a ball B_\delta(\mathbf{x}_0) such that 
f(E \cap B_\delta(\mathbf{x}_0)) \subseteq (f(\mathbf{x}_0) - \varepsilon, f(\mathbf{x}_0) + \varepsilon) \subseteq G.
Then, we have E \cap B_\delta(\mathbf{x}_0) \subseteq f^{-1}(G). Hence, f^{-1}(G) is relatively open in E. 
Conversely, suppose f^{-1}(G) is relatively open in E for every open set G \subseteq \R. Let \mathbf{x}_0 \in E. Then for every \varepsilon > 0, f^{-1}((f(\mathbf{x}_0)) - \varepsilon, f(\mathbf{x}_0) + \varepsilon)) is relatively open in E. Thus, there exists \delta > 0 such that
B_\delta(\mathbf{x}_0) \cap E \subseteq f^{-1}((f(\mathbf{x}_0)) - \varepsilon, f(\mathbf{x}_0) + \varepsilon)).
Then we have f(B_\delta(\mathbf{x}_0) \cap E) \subseteq (f(\mathbf{x}_0) - \varepsilon, f(\mathbf{x}_0) + \varepsilon). This shows that |f(\mathbf{x}) - f(\mathbf{x}_0)| < \varepsilon whenever \|\mathbf{x} - \mathbf{x}_0\| < \delta and \mathbf{x} \in E. Hence, f is continuous at \mathbf{x}_0.
Lemma 2: The function f: E \to \R is continuous on E \subseteq \R^n if and only if f^{-1}(F) is relatively closed in E for every closed set F \subseteq \R.
Proof: We first note that S \subseteq E is relatively closed in E if and only if E \setminus S is relatively open in E. 
Suppose f: E \to \R is continuous on E. Given F \subseteq \R be a closed set. Then \R \setminus F is open. By Lemma 1, f^{-1}(\R \setminus F) is relatively open in E. Since f^{-1}(\R \setminus F) = E \setminus f^{-1}(F), we have f^{-1}(F) is relatively closed in E.
Conversely, suppose f^{-1}(F) is relatively closed in E for every closed set F \subseteq \R. Then \R \setminus F is open. Thus f^{-1}(\R \setminus F) = E \setminus f^{-1}(F) is relatively open in E for every open set (\R \setminus F) \subseteq \R. Hence, by Lemma 1 f: E \to \R is continuous on E. 
Now, we have
 \{a < f < b\} = \{\mathbf{x} \in E \mid a < f(\mathbf{x}) < b\} \cap E 
= \{\mathbf{x} \in E \mid f(\mathbf{x}) < b\} \cap \{\mathbf{x} \in E \mid f(\mathbf{x}) > a\} 
= f^{-1}((-\infty, b)) \cap f^{-1}((a,\infty))
Note that (a, \infty) and (-\infty, b) are open in \R. Since f is continuous on E, we have f^{-1}((-\infty, b)) and f^{-1}((a,\infty)) are relatively open in E by Lemma 1. Thus, \{a < f < b\} is relatively open.
Similarly, we have \{a \le f \le b\} = f^{-1}((-\infty, b]) \cap f^{-1}([a,\infty)). Note that [a, \infty) and (-\infty, b] are closed in \R. Since f is continuous on E, we have f^{-1}((-\infty, b]) and f^{-1}([a,\infty)) are relatively closed by Lemma 2. Thus, \{a \le f \le b\} is relatively closed in E.
Moreover, note that \{f = a\} = f^{-1}(\{a\}). Since \{a\} is closed in \R, we have \{f = a\} is relatively closed in E.
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(b) 
(Statement on Open Sets)
Assume that x \in f^{-1}(G), then f(x) \in G .
Firstly, we prove that if \displaystyle{f} is continuous on \mathbb{R^n} then f^{-1}(G) is open for every open G in \mathbb{R^1}.
Because G is an open set, there exist a \varepsilon >0 , 
(f(x)-\varepsilon,f(x)+\varepsilon) \subset G 
And follow definition of continuous function. 
For each \varepsilon >0 , there exist a \delta>0 such that,
|f(x_1)-f(x_0)| < \varepsilon, whenever |x_1-x_0|<\delta .
|f(x_1)-f(x)| < \varepsilon, then f(x_1) \subset G for |x_1-x|<\delta .
x_1 \in f^{-1}(G) for |x_1-x|<\delta ,
(x-\delta,x+\delta) \subset f^{-1}(G),
Then we can get that f^{-1}(G) is open set.
Secondly , we prove that if f^{-1}(G) is open for every open G in \mathbb{R^1} then \displaystyle{f} is continuous on \mathbb{R^n}.
Assume \varepsilon >0,x \in \mathbb{R^1}.
 
G=(f(x)-\varepsilon,f(x)+\varepsilon) is an open set in\mathbb{R^1} and f^{-1}(G) also an open set.
Hence, there exist \delta<0 such that,
(x-\delta,x+\delta) \subset f^{-1}(G), And x_1 \in (x-\delta,x+\delta) ,f(x_1) \in G ,
We can get that,
|f(x_1)-f(x)| < \varepsilon, whenever |x_1-x|<\delta .
Therefore, \displaystyle{f} is continuous on \mathbb{R^n}.
(Statement on Closed Sets)
Using the fact that a set is open if and only if its complement is closed, the statement on the closed sets immediately follows from that of open sets. Nevertheless, we can still give a direct proof as follows.
Suppose f  is continuous on \R^n. Let F \subseteq \R be a closed set. Suppose \{\mathbf{x}_k\} is a sequence in f^{-1}(F) that converges to some \mathbf{x}_0 \in \R^n. Since f is continuous at \mathbf{x}_0, the sequence \{f(\mathbf{x}_k)\} is a sequence in F that converges to the point f(\mathbf{x}_0) \in \R. But since F is closed in \R, we have f(\mathbf{x}_0) \in F. Therefore, \mathbf{x}_0 \in f^{-1}(F). This proves that f^{-1}(F) is closed in \R^n. 
Conversely, suppose that f is not continuous at some \mathbf{x}_0 \in \R^n. We want to show that there is a closed set F \subseteq \R such that f^{-1}(F) is not closed in \R^n. There exists \varepsilon > 0 such that for any \delta > 0, there is a point \mathbf{x} \in B_\delta(\mathbf{x}_0) such that|f(\mathbf{x}) - f(\mathbf{x}_0)| \ge \varepsilon.
For k \in \Z^+, let \mathbf{x}_k \in B_{\frac{1}{k}}(\mathbf{x}_0) be such that|f(\mathbf{x}_k) - f(\mathbf{x}_0)| \ge \varepsilon.
Since \|\mathbf{x}_k - \mathbf{x}_0\| < \dfrac{1}{k} for all k \in \Z^+, the sequence \{\mathbf{x}_k\} in \R^n converges to \mathbf{x}_0. LetF = \{a \in \R \mid |a - f(\mathbf{x}_0)| \ge \varepsilon\}. 
Then F = \left(B_\varepsilon(f(\mathbf{x}_0)\right)^c, so that F is closed in \R and f(\mathbf{x}_k) \in F for all k \in \Z^+. But f(\mathbf{x}_0) \notin F. Therefore, the set f^{-1}(F) contains the sequence \{\mathbf{x}_k\} but does not contain its limit point \mathbf{x}_0. This implies that f^{-1}(F) is not closed.
Proof on the Statements of Closed Sets using Sets Complement
First suppose that f: \R^n \to \R is continuous. Let F \subseteq \R be a closed set. Then \R \setminus F is open. Thus by the result on open sets we have f^{-1}(\R \setminus F) is open in \R^n. Since f^{-1}(\R \setminus F) = \R^n \setminus f^{-1}(F), we find that f^{-1}(F) is closed in \R^n. 
Conversely, suppose that f^{-1}(F) is closed in \R^n for every closed subset F \subseteq \R. Let G \subseteq \R be an open set. Then \R \setminus G is closed. Thus \R^n \setminus f^{-1}(G) = f^{-1}(\R \setminus G) is closed in \R^n. It follows that f^{-1}(G) is open for every open set G \subseteq \R, which then implies that f: \R^n \to \R is continuous.