prove that $abs(f)$ is continuous at p. Is the converse true? let E be a subset of the reals, and suppose $f:E \to$ the reals, is continuous at p. 
this is a two part question:
First prove that $|f|$ is continuous at p. Is the converse true? 
second set g(x) = $\sqrt{|f(x)|}$, x is an element of E. prove that g is continuous at p.
for the first part i can visually understand how part a works but i dont know how to algebraically approach it. 
for the second part I'm pretty sure i understand how to do it if i can prove the first part. 
 A: You have $|x| \le |x-y|+|y|$, so $|x| -|y|\le |x-y|$. Reversing the roles of $x,y$ we get $||x| -|y||\le |x-y|$. This shows that the function $x \mapsto |x|$ is continuous (in fact Lipschitz continuous with rank 1).
The converse is not true as you lose sign information. The function $f(x) = \begin{cases} (-1)^{\lfloor \frac{1}{x} \rfloor}, & x \neq 0 \\ 1, & \text{otherwise} \end{cases}$ has $|f(x)| = 1$ for all $x$, so $x \mapsto |f(x)|$ is continuous, but $f$ is not.
Note that for $x,y \ge 0$, we have $(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) = x-y$.
If $\max(x,y) >0$, we can write $\sqrt{x}-\sqrt{y} \le \frac{x-y}{\sqrt{x}+\sqrt{y}}$. Reversing the roles of $x,y$ we get $|\sqrt{x}-\sqrt{y}| \le \frac{|x-y|}{\sqrt{x}+\sqrt{y}}$. We also have $|\sqrt{x}-\sqrt{y}| \le  \sqrt{x}+\sqrt{y} $, so we have $|\sqrt{x}-\sqrt{y}| \le \min(\sqrt{x}+\sqrt{y}, \frac{|x-y|}{\sqrt{x}+\sqrt{y}})$.
For $a\ge0$, we have $\min(x,\frac{a}{x}) \le \sqrt{a}$, hence we have $|\sqrt{x}-\sqrt{y}| \le \sqrt{|x-y|}$. This is valid for all $x,y \ge 0$. Continuity of $x \mapsto \sqrt{x}$ on $[0, \infty)$ follows from this. Continuity of $x \mapsto \sqrt{|x|}$ follows as well.
It follows that $x \mapsto \sqrt{|f(x)|}$ is continuous.
