Prove that the composition of continuous functions is continuous using the $\epsilon-\delta$ definition of continuity.

Question

Definition of Continuity: A function $f:A \rightarrow \mathbb{R}$ is continuous at a point $c \in A$ if, for all $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $|x - c| < \delta$ (and $x \in A$) it follows that $|f(x) - f(c)| < \epsilon$.

Theorem to be Proved: Given $f:A \rightarrow \mathbb{R}$ and $g:B \rightarrow \mathbb{R}$, assume that the range $f(A) = \{f(x):x \in A\}$ is contained in the domain $B$ so that the composition $g \circ f(x) = g(f(x))$ is defined on $A$. If $f$ is continuous at $c \in A$, and if $g$ is continuous at $f(c) \in B$, then $g \circ f$ is continuous at $c$.

My Proof Attempt: Let $\epsilon > 0$.

Then there exists a $\delta_1 > 0$ such that $|x-c|<\delta_1$ implies $|f(x) - f(c)| < \epsilon$ (for $x \in A$).

There also exists a $\delta_2 > 0$ such that $|b-f(c)|<\delta_2$ implies $|g(b) - g(f(c))| < \epsilon$ (for $b \in B$).

Let $\delta = min\{\epsilon, \delta_1, \delta_2\}$. Then, for all $b \in B$ contained in the $\epsilon$-neighborhood of $f(c)$ which are in the range $f(A)$, we have $$|b-f(c)| = |f(x)-f(c)| < \delta \leq \delta_2$$ $$\Rightarrow |g(f(x))-g(f(c))| < \epsilon. $$ So, given an arbitrary $\epsilon > 0$, we found a $\delta > 0$ such that $|x-c| < \delta$ implies $|g(f(x)) - g(f(c))| < \epsilon$. Thus, the composition $g \circ f$ is continuous at $c$.

Answer 1

No. Your last paragraph just doesn't make any sense.

You say $|b-f(c)| = |f(x)-f(c)|$. Why? $f(x) \ne b$ (although $f(x)$ is close to $b$) so why would $|f(x)-f(x)|=|b-f(c)|$. And you say $|f(x)-f(c)|< \delta$. Why? You do have $|x-c|< \delta_1$so that means $|f(x)-f(c)|< \epsilon$ but that doesn't have anything to do with $|f(x)-f(x)|$ being more or less than delta.

Here's a hint. Labels are just labels. And as $f$ and $g$ being continuous means for any $\epsilon> 0$ there is a $\gamma$ so that $|y-c|< \gamma$ implies $|g(y)-g(c)| < \epsilon$. And for any $\zeta > 0$ there is a $\delta$ so that $|x-b|< \delta \implies |f(x)-f(b)| < \zeta$.

Does that look weird? Why? All I did was change the labels.

Can you take it from there?

Hint:

Use $\epsilon$ to define $\gamma$ as above. Set $\zeta = \gamma$. Use $\zeta$ to find $\delta$ as above. Then if $|x-b|< \delta$... what happens(!)