QQuestionMathematics
QuestionMathematics
Exercise 4.6.5. [Used in Exercise 7.4.1.] Let $(a, b) \subseteq \mathbb{R}$ be a non-degenerate open bounded interval, and let $f:(a, b) \rightarrow \mathbb{R}$ be a function. Suppose that $f$ is continuous, strictly increasing and bounded. Let $F:[a, b] \rightarrow \mathbb{R}$ be defined by
F(x)= \begin{cases}\operatorname{glb} f((a, b)), & \text { if } x=a \\ f(x), & \text { if } a<x<b \\ \operatorname{lub} f((a, b)), & \text { if } x=b\end{cases}
(1) Prove that $F$ is continuous.
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Answer
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Step 1::
We will prove that $F$ is continuous at $x=a$, $x \in (a,b)$, and $x=b$ separately.
Step 2:.1: Continuity at $x=a$
This shows that $F$ is continuous at $x=a$.
Let $m = \operatorname{glb} f((a,b))$. |f(x) - f(c)| = |f(x) - m| < \epsilon \quad \text{ whenever } \quad |x-c| < \delta_1 \tag{1} m \leq f(x) < m + \epsilon \quad \text{ whenever } \quad a < x < a + \delta_2 \tag{2} |F(x) - F(a)| = |F(x) - m| < \epsilon \quad \text{ whenever } \quad |x-a| < \delta
Step 3:.2: Continuity in $(a,b)$
Since $f$ is continuous on $(a,b)$, it follows that $F(x) = f(x)$ is also continuous on $(a,b)$.
Step 4:.3: Continuity at $x=b$
This shows that $F$ is continuous at $x=b$.
The proof is similar to Step 1.1. |f(x) - f(d)| = |f(x) - M| < \epsilon \quad \text{ whenever } \quad |x-d| < \delta_1 \tag{3} M - \epsilon < f(x) \leq M \quad \text{ whenever } \quad b - \delta_2 < x < b \tag{4} |F(x) - F(b)| = |F(x) - M| < \epsilon \quad \text{ whenever } \quad |x-b| < \delta
Step 5::
Since $F$ is continuous at $x=a$, in $(a,b)$, and at $x=b$, we conclude that $F$ is continuous on the closed interval $[a,b]$.
Final Answer
The function $F$ is continuous on the closed interval $[a,b]$.
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