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The continuity law for composition states, informally, that:

$$\text{IF } f \text{ is continuous at } a \ \text{ AND }\ g \text{ is continuous at } f(a) \text{ THEN } g(f(a)) \text{ is continuous at } a$$

Or, using the limit definition of continuity:

$$ \lim_{x \to a} f(x) = f(a) \wedge \lim_{x \to f(a)} g(x) = g(f(a)) \Longrightarrow \lim_{x \to a} g(f(x)) = g(f(a))$$

Or, using the formal epsilon-delta definitions:

$$ \forall \epsilon > 0, \exists \delta_1 > 0, \forall x \in \mathbb{R}, |x - a| < \delta_1 \Longrightarrow |f(x) - f(a)| < \epsilon \tag{1}\label{1}$$

$$ \forall \epsilon > 0, \exists \delta_2 > 0, \forall x \in \mathbb{R}, |x - f(a)| < \delta_2 \Longrightarrow |g(x) - g(f(a))| < \epsilon \tag{2}\label{2}$$

$$ \forall \epsilon > 0, \exists \delta > 0, \forall x \in \mathbb{R}, |x - a| < \delta \Longrightarrow |g(f(x)) - g(f(a))| < \epsilon \tag{3}\label{3}$$

$$ \eqref{1} \wedge \eqref{2} \Longrightarrow \eqref{3}$$


Proof:

We assume \eqref{1} and \eqref{2}. We want to show \eqref{3}.

Let $\epsilon > 0$.

We will use the same $\epsilon$ in \eqref{2}, giving us:

$$ \exists \delta_2 > 0, \forall x \in \mathbb{R}, |x - f(a)| < \delta_2 \Longrightarrow |g(x) - g(f(a))| < \epsilon \tag{2a}\label{2a} $$

Let $\delta_2 > 0$ such that the following is true:

$$ \forall x \in \mathbb{R}, |x - f(a)| < \delta_2 \Longrightarrow |g(x) - g(f(a))| < \epsilon \tag{2b}\label{2b}$$

We will substitute $\delta_2$ for $\epsilon$ in $\eqref{1}$ to obtain:

$$\exists \delta_1 > 0, \forall x \in \mathbb{R}, |x - a| < \delta_1 \Longrightarrow |f(x) - f(a)| < \delta_2\tag{1a}\label{1a} $$

Let $\delta_1 > 0$ such that the following is true:

$$\forall x \in \mathbb{R}, |x - a| < \delta_1 \Longrightarrow |f(x) - f(a)| < \delta_2\tag{1b}\label{1b}$$

Let $x \in \mathbb{R}$.

We will use the same $x$ in \eqref{1b} to obtain:

$$ |x - a| < \delta_1 \Longrightarrow |f(x) - f(a)| < \delta_2\tag{1c}\label{1c}$$

Let $\delta = \delta_1$.

We assume $ |x - a| < \delta = \delta_1$.

We want to show:

$$ |g(f(x)) - g(f(a))| < \epsilon $$

By our assumption and \eqref{1c}, we know:

$$|f(x) - f(a)| < \delta_2 \tag{1d}\label{1d}$$

We substitute $f(x)$ for $x$ in \eqref{2b} to obtain:

$$|f(x) - f(a)| < \delta_2 \Longrightarrow |g(f(x)) - g(f(a))| < \epsilon \tag{2c}\label{2c}$$

Finally, we substitute \eqref{1d} in \eqref{2c} to obtain our desired conclusion:

$$|g(f(x)) - g(f(a))| < \epsilon$$

$\blacksquare$


And here is where I'm stuck. All I need to show is $0 < |f(x) - f(a)| $, and I will be able to complete the proof. But I don't know how to show that. Any ideas?


Edit: Proof has been updated and completed! If there's anything that can be improved, please let me know.

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    $\begingroup$ I think you mean $f$ to be continuous at $a$, not at $g$. And you mean $g(f(x))$, not $g(f(a))$. $\endgroup$ – Arturo Magidin May 27 '19 at 4:32
  • $\begingroup$ @ArturoMagidin Thanks for the correction! $\endgroup$ – Calculemus May 27 '19 at 4:34
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Your definition of continuity is wrong here - it is just $|x-a|<\delta$, and not $0<|x-a|<\delta$. For continuity, the case $x=a$ is valid too.

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  • $\begingroup$ Oh wow, I completely forgot about that! Thanks! $\endgroup$ – Calculemus May 27 '19 at 5:06

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