Functions from the Cantor set 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.
 A: Nice question! The answer is "yes".
Notice that, if $x \in \Delta$ and $U \ni x$ is an open interval, then $U \cap \Delta$ is uncountable. So, if there is any such $U$ for which $f$ is monotone decreasing on $U \cap \Delta$, we are done. Therefore, from now on, we assume otherwise: We assume that, for all $x \in \Delta$, and all open sets $U \ni x$, there are $x_0$ and $x_1 \in U \cap \Delta$ with $x_0<x_1$ and $f(x_0) < f(x_1)$. Using this assumption, we will construct an uncounteable (cardinality continuum) subset of $\Delta$ on which $f$ is increasing. 
We now inductively construct a collection of closed intervals $I_{b_1 b_2 \cdots b_n}$ indexed by finite sequences $(b_1, b_2, \ldots, b_n) \in \{ 0, 1\}^n$ with the following properties:


*

*There is a point of $\Delta$ in the interior of each $I_{b_1 b_2 \cdots b_n}$

*We have $I_{b_1 \cdots b_{n-1} b_n} \subset I_{b_1 \cdots b_{n-1}}$.

*For any $x_0 \in I_{b_1 b_2 \cdots b_{n-1} 0} \cap \Delta$ and $x_1 \in I_{b_1 b_2 \cdots b_{n-1} 1} \cap \Delta$, we have $x_0 < x_1$ and $f(x_0) < f(x_1)$.

*The length of $I_{b_1 \cdots b_n}$ is at most $3^{-n}$ (for example).


Given this, we are done. Our set $X$ is 
$$\Delta \cap \bigcap_{n=0}^{\infty} \bigcup_{b_1 b_2 \ldots b_n \in \{0,1 \}} I_{b_1 b_2 \cdots b_n}.$$
As an intersection of closed sets, it is closed. For any infinite bit sequence $b_1 b_2 \cdots$, the intersection $\Delta \cap I_{b_1} \cap I_{b_1 b_2} \cap I_{b_1 b_2 b_3} \cap \cdots$ is a nested intersection of nonempty closed sets so, by compactness of $\Delta$, it is nonempty. By condition 4, this nonempty intersection is a singleton, call it $\{ x_{b_1 b_2 \cdots } \}$. So $X$ is the set of points $\{ x_{b_1 b_2 \cdots } \}$ where $(b_1, b_2, \ldots )$ runs over all continuum many sequences of $0$'s and $1$'s, and is thus uncounteable. If $(b_1, b_2, \ldots)$ and $(b'_1, b'_2,\ldots)$ are two distinct sequences of bits, which first differ in the $n$-th place with $b_n=0$ and $b'_n=1$, then condition 3 says that $x_{b_1 b_2 \cdots } < x_{b'_1 b'_2 \cdots }$ and $f(x_{b_1 b_2 \cdots }) < f(x_{b'_1 b'_2 \cdots })$, so $f$ is monotone on $X$.
So, let's construct the intervals. Take $I_{\emptyset} = [0,1]$. Suppose inductively that we have constructed $I_{b_1 b_2 \cdots b_m}$ for all $m < n$. Since the interior of $I_{b_1 b_2 \cdots b_{n-1}}$ has a point of $\Delta$, by our bolded assumption, there are $x_0$ and $x_1$ in $I_{b_1 b_2 \cdots b_{n-1}}$ with $x_0 < x_1$ and $f(x_0) < f(x_1)$. By continuity, we can find open intervals $U_0 \ni x_0$ and $U_1 \ni x_1$ with $U_0$ to the left of $\tfrac{x_0+x_1}{2}$ and $U_1$ to its right, and with $f|_{U_0} < \tfrac{f(x_0)+f(x_1)}{2}$ and $f|_{U_1} > \tfrac{f(x_0)+f(x_1)}{2}$. Then take $I_{b_1 b_2 \cdots b_{n-1} j}$ to be a closed interval inside $U_j$, with $x_j$ in its interior, and length at most $3^{-n}$. 
A: 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 $y_\alpha.$
