This question arose from my thinking about order-preserving isomorphisms on the Real numbers. These functions must be injections (“one-to-one”)
$$a\ne b \implies f(a)\ne f(b)$$
and preserve order
$$a\le b \implies f(a)\le f(b)$$
In other words, I knew that these functions must be strictly increasing, perhaps with a finite number of points with zero slope. For if there were any intervals $(a,b)$ on which the function had zero or decreasing slope, it would no longer be an injection or preserve order: two different inputs would have the same output.
But then I realized that even an infinite, albeit countable, number of points would suffice, for example, a function that is increasing everywhere except at every integer; something like $f(x)=x+sin(x)$. This function’s graph looks like a smoothed-out staircase, and its derivative $f^{\prime}(x)=1+cos(x)$ is tangent to the $x$-axis at those points.
So then taking a step further, I wondered if perhaps an uncountable number of points would suffice, provided that these points weren’t “together in a straight line”, or put more precisely, they didn’t make an interval.
My first thought was an uncountable set of irrational numbers within an interval, say, $[1,2]$. This function would be increasing at all rational points, but have zero slope at all irrational points. (There are uncountably many irrational numbers within $[1,2]$.) If such a function exists, and is continuous and differentiable, its derivative would be something like the Dirichlet function:
$$
g^{\prime}(x) =
\begin{cases}
1 &: &x < 1\\
1 &: &x \in [1,2]\cap\mathbb{Q}\\
0 &: &x \in [1,2]-\mathbb{Q}\\
1 &: &x > 2\\
\end{cases}
$$
I'm not sure if that’s possible. If it isn’t, then the points in my uncountable set must all be isolated, that is, there must exist an $\epsilon$-neighborhood around each point that doesn’t contain any other points in the set.
Is it possible to select uncountably many isolated points?
Best Answer
For $x\in\mathbb R$ and $\epsilon\gt0$ let $B(x,\epsilon)$ denote the $\epsilon$-neighborhood $(x-\epsilon,x+\epsilon).$ A set $X\subseteq\mathbb R$ is discrete if, for each point $x\in X,$ there is a number $\epsilon\gt0$ such that $B(x,\epsilon)\cap X=\{x\}.$
Every discrete subset of $\mathbb R$ is countable.
Proof. Assume for a contradiction that there is an uncountable discrete set $X\subseteq\mathbb R.$ For $n\in\mathbb N$ let $X_n=\{x\in X:B(x,\frac1n)\cap X=\{x\}\}.$ Then $X=\bigcup_{n\in\mathbb N}X_n,$ so $X_n$ is uncountable for some $n.$
Choose $n\in\mathbb N$ so that $X_n$ is uncountable. Then the sets $B(x,\frac1{2n}),\ x\in X_n$ constitute an uncountable family of pairwise disjoint open intervals, which is impossible.
However, the fact that discrete subsets of $\mathbb R$ are countable seems unrelated to your question about slopes of order-isomorphisms of $\mathbb R'$
There is a strictly increasing continuously differentiable bijection $f:\mathbb R\to\mathbb R$ such that $\{x:f'(x)=0\}$ is uncountable and even has positive Lebesgue measure.
Proof. Let $C$ be a compact nowhere dense subset of $\mathbb R$ with positive Lebesgue measure, e.g., a so-called fat Cantor set. Define $$g(x)=d(x,C)=\min\{|x-y|:y\in C\},$$ the distance from $x$ to $C,$ and define $$f(x)=\int_0^xg(t)dt.$$ Since $g$ is continuous and nonnegative, $f$ is continuously differentiable and nondecreasing. In fact, since $C$ is nowhere dense, $\{x:f'(x)\gt0\}$ is everywhere dense, whence $f$ is strictly increasing. Since $f(-\infty)=-\infty$ and $f(+\infty)=+\infty,\ $ $f$ is a bijection. Finally, the set $\{x:f'(x)=0\}=\{x:g(x)=0\}=C$ has positive Lebesgue measure.