I need to prove the following theorem:
Given $f(x)$ with domain $Df$ and $g(x)$ with domain $Dg$:
With $f$ and $g$ above, let $a$ be in the domain of $f(g(x))$. Then $f(g(x))$ is continuous at $a$ if $g$ is continuous at $a$ and $f$ is continuous at $g(a)$.
I am at a loss of how to approach this problem.
Any help is appreciated.
Proof:
The definition of continuity is $$ \forall \epsilon > 0 : \exists \delta > 0 : 0 < |x − x_0| < \delta \implies |f(x)−f(x_0)| < \epsilon $$
showing that the preimage of all opens is open is enough to show that the function is continuous
let $U$ open in the image of the domain of $fog$, show that $(fog)^{-1}(U)$ is open.
$(fog)^{-1}(U)=g^{-1}(f^{-1}(U))$ and since $f$ is continuous so $f^{-1}(U)$ is open and since g is continuous then $g^{-1}(f^{-1}(U))$ is open so $f$o$g$ is continuous
Let $g:D_g\to \mathbb R$
and
$f: D_f \to \mathbb R$
$a \in D_g.$
if g is continuous at a
$\lim_{x\to a}g(x)=g(a)$
and
f continuous at g(a)
$\lim_{x \to g(a)}f(x)=f(g(a))$
and
$\color{red}{g(D_g) C D_f}$
( included)
then
$lim_{x \to a}f(g(x))=f(g(a)).$
which means $fog$ is continuous at $a$.
