The composition of continuous functions on $\mathbb{R}$ is continuous

Question

I really need help in some theorems of real analysis. I don't even know where to start them. If someone know any of the following theorems, then please either give the proof or direct me to any good cites where I can find them because I couldn't find them anywhere.

Theorem: Let A, B be subsets of R, let f: -> R be continuous on A, and let g: B -> R be continuous on B. If f(A) is a subset if B, then the composite function g o f: A -> R is continuous on A.

Theorem: Let $A\subseteq\mathbb{R}$, let $f:A\to\mathbb{R}$, and let $|f|$ be defined by $|f|(x) := |f(x)|$ for all $x\in A$.

1. If $f$ is continuous at a point $c\in A$, then $|f|$ is continuous at $c$.
2. If $f$ is continuous on $A$, then $|f|$ is continuous on $A$.

and

Theorem: Let $A\subseteq\mathbb{R}$, let $f: A\to \mathbb{R}$, and let $f(x) \ge 0$ for all $x\in A$. We let $\sqrt{f}$ be defined for $x\in A$ by $(\sqrt{f})(x) = \sqrt{f(x)}$.

1. If $f$ is continuous at a point $c\in A$, then $|f|$ is continuous at $c$.
2. If $f$ is continuous on $A$, then $|f|$ is continuous on $A$.

Answer 1

The two theorems that you cite are special cases of the following. If for some reason you require proofs in the special cases, you can always replace $g$ with either $|\cdot|$ or $\sqrt{\cdot}$. Note that the proof basically consists of carefully writing down the definitions, then following your nose.

Theorem: Let $A\subseteq \mathbb{R}$, let $g: A\to \mathbb{R}$ and $f : f(A) \to \mathbb{R}$.

1. If $f$ is continuous at $c\in A$ and $g$ is continuous at $f(c) \in f(A)$, then $g\circ f$ is continuous at $c$.
2. If $f$ is continuous on $A$ and $g$ is continuous on $f(A)$, then $g\circ f$ is continuous on $A$.

Proof (1): Let $\varepsilon > 0$. By the hypothesis that $g$ is continuous at $f(c)$, there exists some $\delta' > 0$ such that if $y\in f(A)$ and $|y - f(c)| < \delta'$ then $|g(y) - g(f(c))| < \varepsilon$. By the hypothesis that $f$ is continuous at $c$, there exists some $\delta > 0$ such that if $x\in A$ and $|x-c| < \delta$ then $|f(x) - f(x)| < \delta'$. But then, chasing the above inequalities around, we have $$ |x-c| < \delta \implies |f(x) - f(c)| < \delta' \implies |g(f(x)) - g(f(c))| = | (g\circ f)(x) - (g\circ f)(c) | < \varepsilon.$$ Therefore the function $g\circ f : A \to \mathbb{R}$ is continuous at $c$.

Proof (2): It is necessary to show that for $c\in A$, we have that $g\circ f$ is continuous at $c$. So let $c\in A$. Since $f$ is continuous at $c$ and $g$ is continuous at $f(c)$ by hypothesis, it follows from part (1) that $g\circ f$ is continuous at $c$, which is the desired result.