Composition of two continuous functions is continuous

Question

Let $f, g$ be functions

• $f$ is continuous at a,
• $\operatorname{f}(a) = b \in \operatorname{Dom}(f)$,
• $g$ is continuous at b.

Then $g\circ f$ is continuous at $a$.

Proof: Let $\varepsilon>0$,

$$\exists\delta'>0, \text{ s.t. } \lvert g(y)-g(b)\rvert<\varepsilon\qquad\forall y\in\operatorname{Dom}(g) \text{ s.t. } \lvert\ y-b\rvert < \delta'$$

$$\exists\delta>0, \text{ s.t. } \lvert f(x)-f(a)\rvert<\delta'\qquad\forall x\in\operatorname{Dom}(f) \text{ s.t. } \lvert\ x-a\rvert < \delta$$

$$\implies \forall x\in\operatorname{Dom}(g\circ f) \text{ s.t. } \lvert x-a \rvert < \delta, \text{ we have:}$$

$$\lvert(g\circ f)(x) - (g\circ f)(x)\rvert = \lvert g(f(x)) - g(f(a)) \rvert < \varepsilon, \qquad \text{ since } \lvert f(x)-f(a)\rvert<\delta'$$

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Why does $\lvert f(x)-f(a)\rvert<\delta'\implies \lvert g(f(x)) - g(f(a)) \rvert < \varepsilon$?

Answer 1

Because $\delta'$ was chosen so that whenever an element $y$ of the domain of $g$ is $\delta'$-close to $b$, i.e.

$$|y-b|<\delta' $$ then the corresponding images under $g$ are automatically $\epsilon$-close, i.e.

$$|g(y)-g(b)|<\epsilon $$ because of the continuity of $g$.

Here, the two domain elements of $g$ are $\underbrace{f(x)}_{y}$ and $\underbrace{f(a)}_{b}$, so the guaranteed result is that

$$ |g(\;\underbrace{f(x)}_{y}\;)-g(\;\underbrace{f(a)}_{b}\;)|<\epsilon $$