Chapter A3, Problem 24E

To determine

To prove: The given statement and reconstruct the proof from the given diagram.

Answer to Problem 24E

Here, $g\circ f$ is continuous at c.

Explanation of Solution

Given:

  $f$ is continuous at $x=c$ and $g$ is continuous at $f\left(c\right)$ then $g\circ f$ is continuous at c.

  

Calculation:

From the continuity of $g$ , for any $\epsilon >0$ , there is $\delta >0$ such that $0<|f\left(x\right)-f\left(c\right)|<\delta$

  $\begin{array}{l}0<|f\left(x\right)-f\left(c\right)|<\delta \\ \Rightarrow |g\left(f\left(x\right)\right)-g\left(f\left(c\right)\right)|<\epsilon \end{array}$

The second inequality also holds when $f\left(x\right)=f\left(c\right)$ .

So,

  $\begin{array}{l}|f\left(x\right)-f\left(c\right)|<\delta \\ \Rightarrow |g\left(f\left(x\right)\right)-g\left(f\left(c\right)\right)|<\epsilon \end{array}$

But from the continuity of $f$ , there is a $\gamma >0$ such that,

  $\begin{array}{l}0<|x-c|<\gamma \\ \Rightarrow |f\left(x\right)-f\left(c\right)|<\delta \end{array}$

So, for any $\epsilon >0$ , there is a $\gamma >0$ such that,

  $\begin{array}{l}0<|x-c|<\gamma \\ \Rightarrow |g\left(f\left(x\right)\right)-g\left(f\left(c\right)\right)|<\epsilon \end{array}$

This means that $g\left(f\left(x\right)\right)=g\circ f$ is continuous at $c$ .

Conclusion:

Therefore, $g\circ f$ is continuous at c.