Unmasking the completeness of $\mathbb{R}$ So the classical proof that a regulated (= having one-sided limits everywhere) function $f$: $[a, b]\longrightarrow\mathbb{R}$ is the uniform limit of step functions makes heavy use of completeness of $\mathbb{R}$ in the form that open covers of $[a, b]$ have finite subcovers.
Building upon the observation that in the same style we can show that piecewise uniformly continuous functions (or equivalently, regulated + piecewise continuous) are uniform limits of step functions, I am wondering if the above proof can be split into a completeness and non-completeness part, so the question is:
Is any of the following (formally weaker) statements
(1) regulated functions are uniform limits of piecewise uniformly continuous functions
(2) regulated functions are uniform limits of regulated piecewise continuous functions
NOT an instance of the completeness of $\mathbb{R}$ and hence provable without it?
*Definition: Piecewise sth. is taken to mean that there is a partition $a=a_0<a_1<\ldots <a_n=b$ such that the restriction of $f$ to each $(a_{k-1}, a_k)$ has that property and $f$ does whatever it likes at the $a_k$.
 A: No.
Let's drop some requirements on the definition of "regulated". Say that $f:X\to M$, where $X$ is totally ordered and $M$ is metric, is semi-regulated if it has one-sided limits everywhere (so we don't get completeness bundled with the definition).
Now your statements are of the form: if $f$ is semi-regulated then there exists a sequence $f_n$ in a class $C$ of functions such that $f_n\to f$ uniformly. The converse can also be stated that if $f_n$ a sequence in $C$ and $f_n\to f$ uniformly then $f$ is semi-regulated.
Now the first statement will imply similar statement if you replace $C$ with a superset and the second if you replace $C$ with a subset.
Now let's look at the requirements for the first statement. It's relatively easy to prove it if $C$ are piecewise constant functions and $X$ is compact. On the other hand even if $C$ are piecewise continuous functions there's a counter example that shows that $X$ can't be non-compact:
Let $X=\mathbb Q \cap [0,1]$, and consider the function $\xi_E$ where $E$ is truncation of the decimal expansion of $\sqrt2$:
$$f(x)=\begin{cases}
1 & x\mbox{ is a truncation of the decimal expansion of }\sqrt2\\
0 & \mbox{otherwise}
\end{cases}$$
$f$ is semi-regulated since if $f(a) = 0$ you have an neighbourhood of $a$ such that $f(x)=0$ there and if $f(a) = 1$ there is a neigbourhood of $a$ such that $f(x)=0$ except at $a$.
Now assume theres a sequence $f_n$ that converges to $f$ uniformly, this means that eventually $f_n(x)>1/2$ iff $x\in E$. Now $f_n$ being piecewise continuous means that it's restriction by removing a finite number of points is continuous. This in turn means that the inverse image under the restriction of $(1/2,\infty)$ will be $E$ minus a finite number of points, but since $E$ is not open the restriction is not continuous as assumed.
The second (reverse) statement on the other hand holds even if $X$ is not compact as it only relies on the preservation of (single-sided) limits under uniform convergence. If $f_n$ is only a Cauchy-sequence we require $M$ to be complete in order to guarantee that it's limit takes values only in $M$.
