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I came up with this question while trying to solve the following MO one: Does every connected set that is not a line segment cross some dyadic square?

Suppose $C$ is a plane continuum (i.e. a compact, connected set in the plane), contained in the square $S=[-1,1]^2$, but missing the closed vertical line segment $I = \{0\}\times[0,1]$. Suppose $C$ contains the points $P(0,-1)$, $Q(-1,1)$, and $R(1,1)$. Two examples of such continua are shown in the pictures, in green, and the excluded line segment $I$ is shown in red. Note that in general we do not require that $C$ be path-connected, e.g. it might contain copies of the $\sin\frac1x\,$ curve, or of the pseudo-arc (the only path-connected components of which are points), or something else: The only requirements are that $C$ is a continuum in $S$, misses $I$, and contains $P(0,-1)$, $Q(-1,1)$, and $R(1,1)$.

       

Question. Does $C$ necessarily contain a subcontinuum $K$ that contains $P(0,-1)$, intersects both the 1st and the 2nd Quadrants, and such that $K$ "reaches further up" in either the 1st Quadrant or in the 2nd Quadrant: That is $y_1\not=y_2$ where $y_1 = \max\{y: (x,y)\in K, x>0\}$ and $y_2 = \max\{y: (x,y)\in K, x<0\}$?

Comment. It is not difficult to show that for every $z\in[0,1]$ there is a subcontinuum $K_z$ of $C$ with $P(0,-1)\in K_z$ and such that $z = \max\{y: (x,y)\in K_z\}$. To do this, for each $n>0$ take a path from $P(0,-1)$ in the $\frac1n$-neigborhood of $C$ and reaching up to the line $y=z$ (but no further up), then a subsequence of these paths would converge to some $K_z$ that works. We may ask (the contrapositive?): Does there exist a continuum $C$ (contained in $S$, missing $I$, and containing $P(0,-1)$, $Q(-1,1)$, and $R(1,1)$) such that no matter how we pick $z\in[0,1]$ and a subcontinuum $K_z$ containing $P(0,-1)$ and reaching up the line $y=z$ (but no further up), we would always have that $K_z$ intersects the line $y=z\,$ both in the 1st and in the 2nd Quadrant?

Comment. I thought I would work with the specific points $P(0,-1)$, $Q(-1,1)$, and $R(1,1)$, but the condition for $C$ may be weakened to only that $C$ intersects the bottom edge of the square $S$, misses $I$, and intersects the top edge of $S$ in both the 1st and the 2nd Quadrants, and then ask if a subcontinuum $K$ necessarily exist that intersects the bottom edge of $S$, and reaches further up either into the 1st Quadrant or into the 2nd Quadrant.

Thank you :)

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    $\begingroup$ Nice question. I suspect that the answer is no, i.e. a continuum may exist that violates your requirement. I would try to construct it as a nested intersection of continua $K_n$, where every subcontinuum of $K_n$ that contains both $P$ and a point in either of the two quadrants at height $q_n$ must also contain a point at the same height in the other quadrant. Here $q_n$ is a sequence that is dense in the interval $[0,1]$. The construction in each step would be based on a thickened version of the $\sin(1/x)$-continuum. I cannot claim to have thought out the details, so this may not work ... $\endgroup$ – Lasse Rempe-Gillen Oct 26 '14 at 22:21
  • $\begingroup$ Does the following simpler question have a positive answer? >> Let $C\subseteq [0,1]^2$ be a continuum with $(0,0), (1,1) \in C$. Is there a continuum $K\subseteq C$ such that $\textrm{sup}\{y\in [0,1]: (\exists x\in [0,1]): (x,y)\in K\} < 1$? << $\endgroup$ – Dominic van der Zypen Nov 10 '14 at 16:07
  • $\begingroup$ @dominiczypen yes, quite similar to my comments about $K_z$ $\endgroup$ – Mirko Nov 10 '14 at 16:57
  • $\begingroup$ Right - I'm sorry. In that context, I think it would be rather surprising if the answer to your question were "no". $\endgroup$ – Dominic van der Zypen Nov 12 '14 at 7:08
  • $\begingroup$ @LasseRempe-Gillen I think you are right, and (contrary to my hopes/expectations :) there is a continuum (with constraints as described in the question) every subcontinuum of which reaches just as high in the 1st Quadrant as it does in the 2nd. I may post an answer later: My construction is very specific, based on the union of certain vertical intervals (with gradually increasing height) with base-points at the "end"-points of the Cantor set, together with some $\sin(\frac1x)$ type curves inserted in between. Surprising such a continuum may contain many arcs, but at hindsight it is also easy. $\endgroup$ – Mirko Jan 5 '16 at 15:57
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I will answer my own question. The answer turned out the opposite to what I expected or hoped for, to my surprise, and I guess to Dominic van der Zypen's surprise. So, Lasse Rempe-Gillen's intuition worked better, he did guess the correct answer (see comments above, following the question), although he suggested a recursive construction without verifying the details, whereas what I came up with today is a pretty specific construction based on the Cantor set (and some $\sin\frac1x$ type curves). One could make a picture, which I do below since this is perhaps the easiest way to explain it.

Let $R$ be the "right half" of the usual middle third Cantor set, that is the intersection of the latter with $[\frac23,1]$. For each $r\in R$ the continuum $C$ that I will construct will contain the lower semicircle centered at the origin, of radius $r$. If $C_r$ denotes this (closed) semicircle, then the first picture below shows $C_1$, $C_{\frac23}$, $C_{\frac79}$, $C_{\frac89}$. (Note $\frac23=\frac69<\frac79<\frac89<\frac99=1$.) Each $C_r$ will be extended with a couple of vertical line segments at its endpoints, going up in the 1st and in the 2nd Quadrant. $C_{\frac23}$ will be extended with vertical line segments of length $1$, thus reaching the horizontal line $y=1$ (and intersecting it at points $(\pm\frac23,1)$). $C_{\frac79}$ and $C_{\frac89}$ will each be extended with vertical line segments of length $\frac12$. (I will comment on a formula that gives the exact length of the vertical line segments, depending on $r$, but I feel I should do that later.) $C_1$ will be the only one that is not extended (or formally extended with vertical line segments of length $0$). What was described so far is shown on the first picture below.

Neptune Trident, step 1

Corresponding to each middle-third $(a,b)$ that was removed in the construction of (the "right-half" $R$) of the Cantor set, we will add a "double $\sin\frac1x$ type" arc that "connects" the semicircles $C_a$ and $C_b$, including the attached vertical line segments. For example, for the middle-third $(\frac79,\frac89)$, this arc will alternate from points at height $\frac12$ in the 1st Quadrant to points at height $\frac12$ in the 2nd Quadrant, remaining all the time "between" $C_\frac79$ and $C_\frac89$ (with the attached vertical segments). One "end" of this arc will "approach" $C_\frac79$ (including its attached vertical segments), and the other "end" will "approach" $C_\frac89$ (with its attached vertical segments). This is illustrated in the next picture.

Neptune Trident, step 2

The main idea here is that if $K$ is any subcontinuum of $C$ (where $C$ is the continuum we are about to construct from all the $C_r$, attached vertical line segments, and connecting "double $\sin\frac1x$ type" arcs), and if $K$ contains the bottom point $P(0,-1)$ and at least one point on $C_\frac79$, then $K$ must also contain the connecting arc between $C_\frac79$ and $C_\frac89$. Since the "ends" of that arc "approach" $C_\frac79$ and $C_\frac89$, it follows that $K$ must contain both $C_\frac79$ and $C_\frac89$ (including respective vertical line segments). Such a subcontinuum $K$ must intersect the horizontal line $y=\frac12$ in both the 1st and the 2nd Quadrants (at points $(\pm\frac79,\frac12)$, $(\pm\frac89,\frac12)$, plus a bunch of points on the arc).

As a minor variation of this argument, $K$ might be a subcontinuum (of $C$) that does not contain any point of $C_\frac79$ (or of the respective vertical line segments), but does contain at least one point of the connecting arc between $C_\frac79$ and $C_\frac89$ (and also contains the bottom point $P(0,-1)$). In this case $K$ must contain just "one end" of the connecting arc: Namely, that end which approaches $C_\frac89$. It follows that in this case $K$ must contain $C_\frac89$ with its vertical line segments. $K$ will be contained in the closed half-plane below the horizontal line $y=\frac12$, and will meet this line at $(\pm\frac89,\frac12)$, plus a bunch of points on the arc (in both the 1st and in the 2nd Quadrant).

Continuing with the construction, we now add $C_\frac{25}{27}$ and $C_\frac{26}{27}$ along with vertical line segments of length $\frac14$ (reaching up to horizontal line $y=\frac14$). Note that $\frac89=\frac{24}{27}<\frac{25}{27}<\frac{26}{27}<\frac{27}{27}=1$, and $C_\frac{25}{27}$ and $C_\frac{26}{27}$ correspond to the endpoints of the removed middle-third $(\frac{25}{27},\frac{26}{27})$. We also add a "double $\sin\frac1x$ type arc" connecting $C_\frac{25}{27}$ and $C_\frac{26}{27}$ (and attached vertical line segments), alternating between points at height $\frac14$ in the 1st and in the 2nd Quadrants. (Picture enclosed further down.)

Also, add $C_\frac{19}{27}$ and $C_\frac{20}{27}$ along with vertical line segments of length $\frac34$ (reaching up to horizontal line $y=\frac34$). Note that $\frac23=\frac{18}{27}<\frac{19}{27}<\frac{20}{27}<\frac{21}{27}=\frac78$, and $C_\frac{19}{27}$ and $C_\frac{20}{27}$ correspond to the endpoints of the removed middle-third $(\frac{19}{27},\frac{20}{27})$. We also add a "double $\sin\frac1x$ type arc" connecting $C_\frac{19}{27}$ and $C_\frac{20}{27}$ (and attached vertical line segments), alternating between points at height $\frac34$ in the 1st and in the 2nd Quadrants.

$C_\frac{25}{27}$ and $C_\frac{26}{27}$ (with vertical line segments and connecting arc), and $C_\frac{19}{27}$ and $C_\frac{20}{27}$ (with vertical line segments and connecting arc) are illustrated at the next picture.

Neptune Trident, step 3

The continuum $C$ will consist of all lower semicircles $C_r$ for $r\in R$, together with respective vertical line segments, and with connecting arcs, as indicated above. (The precise length of the vertical line segments will be described, as was promised, further down. Hopefully the above description is clear enough, for now.)

If $K$ is any subcontinuum of $C$ that contains the bottom point $P(0,-1)$ and at least one point of the connecting arc between $C_\frac{25}{27}$ and $C_\frac{26}{27}$, then $K$ must contain at least "one end" of this connecting arc: At least the end which approaches $C_\frac{26}{27}$. It follows that in this case $K$ must contain $C_\frac{26}{27}$ with its vertical line segments (reaching height $\frac14$). Such a $K$ will meet the horizontal line $y=\frac14$ at least at $(\pm\frac{26}{27},\frac14)$ (plus a bunch of points on the arc) in both the 1st and in the 2nd Quadrant.

Similarly, if $K$ is any subcontinuum of $C$ that contains the bottom point $P(0,-1)$ and at least one point of the connecting arc between $C_\frac{19}{27}$ and $C_\frac{20}{27}$, then $K$ must contain at least "one end" of this connecting arc: At least the end which approaches $C_\frac{20}{27}$. It follows that in this case $K$ must contain $C_\frac{20}{27}$ with its vertical line segments (reaching height $\frac34$). Such a $K$ will meet the horizontal line $y=\frac34$ at least at $(\pm\frac{20}{27},\frac34)$ (plus a bunch of points on the arc) in both the 1st and in the 2nd Quadrant.

Re the exact length of the vertical line segments, the main idea here is a well-known continuous map from the product space $2^{\mathbb N}$ (where $2=\{0,1\}$ with the discrete topology, and $\mathbb N=\{1,2,3,\dots\}$) onto the unit interval $[0,1]$. To make things precise, if $s=\langle s_1,s_2,s_3,\dots\rangle\in2^{\mathbb N}$ then one defines $\displaystyle\varphi(s)=\sum_{n\in\mathbb N}\frac{2s_n}{3^n}$. Then $\varphi$ is a homeomorphism between $2^{\mathbb N}$ and the usual middle-third Cantor set. For example $\varphi\langle 0,1,1,\dots\rangle=0+\frac29+\frac2{27}+\cdots=\frac13$ and $\varphi\langle 1,0,0,\dots\rangle=\frac23+0+0+\cdots=\frac23$.
One also defines a continuous $\psi$ from $2^{\mathbb N}$ onto $[0,1]$ by $\displaystyle\psi(s)=\sum_{n\in\mathbb N}\frac{s_n}{2^n}$.
Note that $\psi\langle 0,1,1,\dots\rangle=0+\frac14+\frac18+\cdots=\frac12=\frac12+0+0+\cdots=\varphi\langle 1,0,0,\dots\rangle$.
That is, $(\psi\circ\varphi^{-1})(\frac13)=\frac12=(\psi\circ\varphi^{-1})(\frac23)$. The map $\psi\circ\varphi^{-1}$ is a continuous map from the usual middle-third Cantor set onto the unit interval $[0,1]$ such that, for each removed middle-third, both its end-points are sent to the same point in $[0,1]$. (As illustrated above, both endpoints of $(\frac13,\frac23)$ are sent to $\frac12$.)

The length of the vertical line segments attached at each $C_r$ is, essentially, given by the map $\psi\circ\varphi^{-1}$, with some adjustment. Instead of sending $\frac13$ and $\frac23$ to $\frac12$, we would like to send $\frac79$ and $\frac89$ to $\frac12$. Also, larger $r\in R$ are sent to smaller height, like $\frac{25}{27}$ and $\frac{26}{27}$ are sent to $\frac14$, while $\frac{19}{27}$ to $\frac{20}{27}$ are sent to $\frac34$. Formally, what I think works, is to send $r\in R$ to $(\psi\circ\varphi^{-1})(3(1-r))$. For example when $r=\frac79$ we have $(\psi\circ\varphi^{-1})(3(1-\frac79))=(\psi\circ\varphi^{-1})(3\cdot\frac29)=(\psi\circ\varphi^{-1})(\frac23)=\frac12$. When $r=\frac89$ we have $(\psi\circ\varphi^{-1})(3(1-\frac89))=(\psi\circ\varphi^{-1})(3\cdot\frac19)=(\psi\circ\varphi^{-1})(\frac13)=\frac12$. When $r=\frac23$ we have $(\psi\circ\varphi^{-1})(3(1-\frac23))=(\psi\circ\varphi^{-1})(3\cdot\frac13)=(\psi\circ\varphi^{-1})(1)=\frac12+\frac14+\frac18+\cdots=1$. When $r=1$ we have $(\psi\circ\varphi^{-1})(3(1-1))=(\psi\circ\varphi^{-1})(0)=0$. All these values agree with the heights of the vertical line segments from our initial less formal description.

(As a side remark, in my question I required that the continuum $C$ must contain the points $Q(-1,1)$ and $R(1,1)$. The continuum $C$ that I described in this answer does not contain these points, but to rectify this, we only need to add the horizontal line segments from the point $(\frac23,1)$ to the point $(1,1)$, and from the point $(-1,1)$ to the point $(-\frac23,1)$. I prefer instead to work with the more relaxed version as stated towards the end of the question, requiring $C$ to intersect the horizontal line $y=1$ in both the 1st and the 2nd Quadrants, but not necessarily at $Q(-1,1)$ and $R(1,1)$.)

Note that for every $z\in(0,1)$ the continuum $C$ constructed in the present answer will contain a subcontinuum $K_z$ such that $P(0,-1)\in K_z$, and $K_z$ intersects the horizontal line $y=z$ and is contained in the closed half-plane below this line. Specifically, $K_z$ will consist of all $C_r$ with $r\in R$ (the right-half of the middle-third Cantor set) such that $(\psi\circ\varphi^{-1})(3(1-r))\le z$, along with respective vertical line segments, and necessary connecting arcs. (The set of these $r$ will be of the form $[r_0,1]\cap R$ for a suitable $r_0$, namely $r_0$ would be the solution of the equation $(\psi\circ\varphi^{-1})(3(1-r))=z$ for $r$ in terms of $z$. For most $z\in(0,1)$ this solution will be unique. For countably many values of $z$, there will be two solutions, so in this case $K_z$ will not be uniquely determined. More precisely, if $z=\frac p{2^k}\in(0,1)$ for some integers $p,k$ (we may assume that $p$ is odd) then there will be a middle-third $(r_0,r_1)$ such that $(\psi\circ\varphi^{-1})(3(1-r_1))=(\psi\circ\varphi^{-1})(3(1-r_0))=z$. One could come up with a precise formula, but perhaps this is unnecessary.)

(Credit goes to Ivan Johansen for his Graph https://www.padowan.dk/ , which I used to make the pictures.)

P.S. I am tempted to call the continuum $C$ described above the Neptune Trident. This might be a little imprecise since the middle dent is missing (but the more precise "bident" or "polydent" just don't sound right). At hindsight the construction wasn't overly difficult and perhaps $C$ does not need a special name, but it doesn't hurt to propose one.

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