Exercise 4.3.3

Question

\begin{enumerate}[label=(\alph*)] 
  \item Supply a proof for Theorem 4.3.9 (Composition of continuous functions) using the $\epsilon-\delta$ characterization of continuity.
  \item Give another proof of this theorem using the sequential characterization of continuity (from Theorem 4.3.2 (iii)).
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item Let $f$ is continuous at $c$ and $g$ be continuous at $f(c)$. We will show $g \circ f$ is continuous at $c$. Let $\epsilon > 0$ be arbitrary, we want $|g(f(x)) - g(f(c))| < \epsilon$ for $|x-c| < \delta$. Pick $\alpha > 0$ so that $|y - f(c)| < \alpha$ implies $|g(y) - g(f(c))| < \epsilon$ (possible since $g$ is continuous at $f(c)$) and pick $\delta > 0$ so that $|x-c|<\delta$ implies $|f(x)-f(c)|<\alpha$. Putting all of this together we have $$|x-c| < \delta \implies |f(x) - f(c)| < \alpha \implies |g(f(x)) - g(f(c))| < \epsilon$$ \item Let $(x_n) o c$, we know $f(x_n)$ is a sequence converging to $f(c)$ since $f$ is continuous at $c$, and since $g$ is continuous at $f(c)$ any sequence $(y_n) o f(c)$ has $g(y_n) o g(f(c))$. Letting $y_n = f(x_n)$ gives $g(f(x_n)) o g(f(c))$ as desired. \end{enumerate}

Exercise 4.3.3

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Exercise 4.3.3

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