TITLE: Functions from the Cantor set
QUESTION [11 upvotes]: Consider the Cantor set $\Delta\subset [0,1]$, and let $f\colon \Delta\to [0,\infty)$ be a continuous injection. Must $f$ be monotone on some uncountable closed subset of $\Delta$? Note that that Van Der Waerden's construction is not applicable here.

REPLY [3 votes]: This was proved, not just for continuous but for Borel-measurable functions, by F. M. Filipczak, Sur les fonctions continues relativement monotones, Fund. Math. 58 (1966), 75–87. (See Andreas Blass, A partition theorem for perfect sets, Proc. Amer. Math. Soc. 82 (1981), 271–277, for a generalization.) It is not true for arbitrary functions:
Theorem. For any set $Y\subseteq\mathbb R$ of condinality $\mathfrak c,$ there is an injection $f:\mathbb R\to Y$ which is not monotone on any set of cardinality $\mathfrak c.$
Proof. Let $\mathbb R=\{x_\alpha:\alpha\lt\mathfrak c\},\ x_\alpha\ne x_\beta$ for $\alpha\ne\beta.$
Let $\{g_\beta:\beta\lt\mathfrak c\}$ be the set of all strictly monotone countable functions $g\subseteq\mathbb R\times Y,$ and let $D_\beta$ be the domain of $g_\beta.$
For $\alpha\lt\mathfrak c,$ let $Y_\alpha$ be the set of all $y\in Y$ such that, for some $\beta\lt\alpha,$ at least one of the following holds:
$$y=\sup\{g_\beta(x):x\in D_\beta,\ x\lt x_\alpha\};$$
$$y=\inf\{g_\beta(x):x\in D_\beta,\ x\lt x_\alpha\};$$
$$y=\sup\{g_\beta(x):x\in D_\beta,\ x\gt x_\alpha\};$$
$$y=\inf\{g_\beta(x):x\in D_\beta,\ x\gt x_\alpha\}.$$
Thus $|Y_\alpha|\le|\alpha|\cdot4\lt\mathfrak c.$
We define $y_\alpha$ for $\alpha\lt\mathfrak c$ by transfinite induction: assuming that $y_\beta$ has already been defined for all $\beta\lt\alpha,$ we choose $y_\alpha\in Y\setminus\left(Y_\alpha\cup\{x_\beta:\beta\lt\alpha\}\right).$
Now $f=\{(x_\alpha,y_\alpha):\alpha\lt\mathfrak c\}$ is an injection from $\mathbb R$ to $Y.$
Assume for a contradiction that $f|X$ is monotone for some $X\subseteq\mathbb R$ with $|X|=\mathfrak c.$ Choose a countable dense subset $D$ of $X.$ Then there is an ordinal $\beta\lt\mathfrak c$ such that $f|D=f_\beta$ amd $D=D_\beta.$
Let $X_0=\{x\in X:x\text{ is a limit point of }X\text{ and }f|X\text{ is continuous at }x\}.$ Then $|X_0|=\mathfrak c,$ since $|X|=\mathfrak c$ and $X\setminus X_0$ is countable.
Choose $\alpha$ so that $\beta\lt\alpha\lt\mathfrak c$ and $x_\alpha\in X_0.$ Then $f|X$ is continuous at $x_\alpha,$ and $x_\alpha$ is a limit point of $X$ and therefore also of $D_\beta.$ It follows that $f(x_\alpha)\in Y_\alpha,$ but this contradicts the way we chose $y_\alpha.$