If $f_n(z)$ converges uniformly to $f(z)$ on a compact set $A\subset\mathbb{c}$ can it be proven $|f_n(z)|$ also converges uniformly to $|f(z)|$? For $f:\mathbb{R} \rightarrow \mathbb{R}$ I can “visualize“ this as follows:
If $f_n(x)$ is confined to an epsilon tube around $f(x)$ and if the section of tube is entirely on positive, we have trivially $|f_n(x)|= f_n(x)$ and $|f(x)| =f(x)$. 
On the other hand if the tube is entirely on negative side, we have $|f_n(x)|= -f_n(x)$ and $|f(x)| =-f(x)$ and the tube is reflected from negative side of the axis to positive side. 
When we have the tube partially on the positive and negative side, the negative part of the tube is reflected on positive side, and a bottom flattened tube is formed where bottom coincides with $0$.
We have $|f_n(x)|$ inside an epsilon tube around $|f(x)|$.
(See three pictures attached).
I am having difficulty expressing this in mathematical language for real functions (forget about generalizing it to other topological spaces).
Is it provable? Is there a theorem or proof of this in a book which I can use as a reference in bibliography?
Thank you in advance.



 A: Sure. Just note the inequality
$$
||f_n(x)|-|f(x)||\le|f_n(x)-f(x)|.
$$
If the right hand side can be made small uniformly in $x$ for large $n$, the same happens to the left hand side, with the same $n$ for the given $\epsilon$.
The inequality $||a|-|b||\le |a-b|$ for complex numbers follows from the triangle inequality. The given inequality is equivalent to
$$
-|a-b|\le|a|-|b|\le|a-b|
$$
The right inequality is just the usual triangle inequality
$$
|a|=|a-b+b|\le|a-b|+|b|.
$$
The left inequality is again the triangle inequality
$$
|b|=|b-a+a|\le|b-a|+|a|=|a-b|+|a|.
$$
A: To prove that $ | |a| - |b| | \le |a-b|$ we just use the triangle inequality twice and the definition of absolute value for real numbers:
First$$ |a| = |(a-b) + b| \le |a-b| + |b|$$
so$$ |a| - |b| \le |a-b|$$
Similarly
$$|b| = |(b-a) + a| \le |b-a| +|a|$$
so $$|b| - |a| \le |b-a|$$
which we can rewrite as 
$$-(|a| - |b|) \le |-(a-b)| = |a-b|$$
We know that $| |a| - |b| |$ is equal to either $(|a|-|b|)$ or $-(|a|-|b|)$, but we've just proved that both of them are $\le |a-b|$, so 
$$ | |a| - |b| | \le |a-b|$$
