Is there a Cantor set $C$ in $\mathbb{R}^{2}$ so the graph of every continuous function $[0,1]\rightarrow [0,1]$ intersects $C$?

Question

Consider the Cantor ternary set on the real line with the usual topology and define a Cantor set to be any topological space $C$ homeomorphic to the Cantor ternary set.

The idea is to construct a Cantor set $C$ in $\mathbb{R}^{2}$ such that for every continuous function $f:[0,1]\rightarrow [0,1]$ we have $C\cap\operatorname{Graph}(f)\neq\emptyset$, where $\operatorname{Graph}(f) = \{(x,f(x)):x\in[0,1]\}$.

Does the result generalize to $\mathbb{R}^{n}$, $n\ge 3$?

That is, for every positive integer $k$, let $I^{k}$ denote the product $[0,1]\overbrace{\times\cdots\times}^{k\rm\ times}[0,1]$. We seek to find a Cantor set $C$ in $\mathbb{R}^{n}$ such that for every continuous function $f:I^{n-1}\rightarrow I^{n-1}$ we have $C\cap\operatorname{Graph}(f)\neq\emptyset$.

A related question can be found at Is there a cantor set in $\mathbb{R}^{n}-\{0\}$ which intersects every ray from the origin?.

Answer 1

Following Will Brian's comments 1 2, here is a graphical "proof" that the graph of every continuous function from $[0,1]$ to $[0,1]$ intersects my Cantor set whose original description is retained below.

Here are the first three steps $C_1$, $C_2$, $C_3$ of a construction that starts with the diagonal line from $(0,0)$ to $(1,1)$, and then in each step replaces the previous stage with two copies of it, scaled by $\frac{1}{2}$ horizontally and $\frac{2}{3}$ vertically:

and here is $C_7$:

So: starting with any continuous function $f: [0,1] \to [0,1]$, let $x_1$ be a point where its graph intersects $C_1$, $x_2$ a point where its graph intersects $C_2$, etc. That these intersections exist is visually obvious and not hard to prove rigorously. Any cluster point of this sequence will be a point of intersection between the graph of $f$ and my Cantor set.

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Original post: I think there is such a Cantor set, and here is my proposal:

The first few points are $(0,0)$, $(1,1)$, $(\frac{1}{2}, \frac{1}{3})$, $(\frac{1}{2}, \frac{2}{3})$, then we have $(\frac{1}{4}, \frac{2}{9})$, $(\frac{1}{4},\frac{4}{9})$, $(\frac{3}{4}, \frac{5}{9})$, $(\frac{3}{4}, \frac{7}{9})$, etc. Hopefully that explains the pattern. There is a homeomorphism from the usual Cantor set to this set which takes the middle-third endpoints to the points I started to list above.

I don't have any good ideas about how to prove that the graph of every continuous function intersects this, but I can't see how it could be avoided.

Answer 2

Here's one simple other example of such a Cantor subset.

First, the theoretical side. In the closed unit square $[0,1]^2$, define a decreasing sequence of "simple" closed subsets $K_n$ (with complement $U_n$), and define $K=\bigcap_n K_n$. The conditions we can to ensure is that (a) $K$ is totally disconnected (b) $K$ meets the graph of every continuous map (written "graph" below).

A trivial but essential remark is that (b) holds if and only if no $U_n$ contains a graph. (Indeed, by compactness, if $U=\bigcup U_n$ contains a graph, then this graph is contained in $U_n$ for some $n$.) As regards (a), the condition to ensure is that the sup $r_n$ of diameters of connected components of $K_n$ tends to zero. (I don't insist on $K$ being Cantor because it's a trivial issue, which can always be ensured by enlarging it, e.g., replacing isolated points with small Cantor subsets.)

The interest is that while $K$ is a bit mysterious, the individual $U_n$ are meant to be easy to understand.

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Now for the example. Each $U_n$ will be blue and $K_n$ white. Each $K_n$ will be a finite union of horizontal squares tilted by $45$ degrees, then intersected with $[0,1]$.

The basic construction is as follows: start from a square, cut it into $5\times 5$ squares and fill the 9 ones as follows:

Next, fill the 16 remaining $5$-times-smaller white squares in the same fashion, but in the orthogonal direction. Next, do the same with the $25$-times-smaller white squares, again in the original direction. Eventually tilt everything by 45 degrees.

At each step, this defines $U_1\subset U_2\subset U_3$... The pictures are as follows, showing, for $i=1,2,3$, $U_i$ (blue), $K_i$ (white):

(at the bounding square, the white part should be understood as the closure of the inner white part — it was written as a blue line only for readibility)

It can easily be checked that the sup of diameter of components of $K_n$ tends exponentially to zero. Next, $U_n$ contains no graph for any $n$. The point is that in $U_n$, the smallest $\Pi$-shaped paths (those in $U_n-U_{n-1}$) can't be used for travelling "to the right". Hence showing that $U_n$ contains no graph can be checked by induction. Of course details would be a little cumbersome to be written in Bourbaki-style, but this is quite visible from the picture, even for a non-mathematician.

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Edit: here are pictures of subsets $U_n$ (blue color) for Nik Weaver's example, after 4 steps and after 11 steps:

(the increasing slopes are $(4/3)^n$, while the apparently decreasing slopes are actually vertical segments)

More precisely, define functions $f_i$ as in Nik's example, where $f_i$ has slope $(4/3)^i$ at each continuity point: $f_0(x)=x$, $f_1(x)=(4/3)x$ for $x<1/2$ and $=(4/3)x+(2/3)$ for $x>1/2$, etc. Note that $f_i$ is defined on the set $C$ of dyadic expansions, rather than on $[0,1]$ where it bi-valued at some dyadic numbers. Thus the $f_n$, viewed as continuous function on $C$, converge uniformly to a function $f$ and $f(C)$ is the desired Cantor set. We have $\|f_{n+1}-f_n\|_{\infty}\le \frac16(2/3)^n$, so $\|f_n-f\|_\infty\le \frac12(2/3)^n$. Thus I defined $U_n$ as the set of $(x,y)$ such that $|y-f_n(x)|>\frac12(2/3)^n$ (properly speaking this is ill-defined for for the few points at which $f_n$ is discontinous, in which case I mean that $\max(|y-f_n(x^+)|,|y-f_n(x^-)|)>\frac12(2/3)^n$).

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PS pictures written with SageMath.

Answer 3

I suspect not.

Fix $C$. Then its complement in $[0,1]\times[0,1]$ is open and path-connected, and so we can find a path $\gamma$ in this complement with $\gamma(0)_x=0$ and $\gamma(1)_x=1$. Moreover, we may take this path to be smooth by Whitney approximation. Consequently, $\gamma$ will have a vertical tangent line at only finitely many points. Applying the implicit function theorem, we get a smooth function $f$ defined on $[0,1]\setminus\{a_1,\dotsc,a_n\}$ (for some finite collection of $x$-values) whose graph avoids $C$.

I would guess that one can modify $f$ to get a function violating the criterion, possibly by repeating the process inductively in a neighborhood of each point with some condition guaranteeing the regularity of the limit, such as bounding the variation. However, I can't think of an argument right now.