Let $C$ be a closed and bounded subset of the plane.
A bounding box $B$ for $C$ will by definition be the smallest rectangle with vertical and horizontal sides that contains $C.$
(We allow rectangles with width or height equal to $0.$)
When $C$ is not a singleton then either the height $h$ or the width $w$ (or both) of its bounding box will be non-zero. Define slenderness of $C$ as $\sigma(C)=\frac{h}w$. Then $\sigma(C)$ takes values in the interval $[0,\infty]$ and equals the slope of the the rising diagonal of the bounding box. (We allow the value $\infty=\frac{h}0$ as the slenderness $\sigma(C)$ when $C$ is a vertical line segment of positive height and zero width.)
Question.
Let $C$ be a plane continuum (i.e. a closed, bounded and connected subset of the plane) which is not contained in any straight line.
Must there exist two subcontinua $K_1$ and $K_2$ of $C$ (none of which is a singleton) such that $\sigma(K_1)\not=\sigma(K_2)\ $?
This is a condensed and clarified version of a couple of more verbose questions that I posted on MO, both of which I came up with when trying to answer a Question by Kevin Johnson that is now (May 7, 2023) more than $10$ years unanswered.
Kevin Johnson (https://mathoverflow.net/users/31069/kevin-johnson),
Does every connected set that is not a line segment cross some dyadic square?,
URL (version: 2013-01-31):
https://mathoverflow.net/q/120415
Are there hereditarily square-boxed plane continua?,
URL (version: 2023-05-07):
https://mathoverflow.net/q/446092
The above contains a positive answer of our question here, for the case when $C$ is the graph of a function (even if stated a different way there). (Edit June 1, 2023. A positive answer was added for the case when the plane continuum $C$ is path-connected.)
Find at least one square-boxed subcontinuum,
URL (version: 2023-05-07):
https://mathoverflow.net/q/446317
Result (added June 1, 2023).
Suppose that $C$ is a path-connected
plane continuum and that there is some non-negative $s$ such that $\sigma(K)=s$ for every non-degenerate subcontinuum $K$ of $C.$
Then $C$ must be a straight line segment.
Sketch of proof.
The cases when $s=\infty$ or $s=0$ are easy to deal with and let to the reader. When $0<s<\infty$ then we may stretch or shrink the continuum $C$ and assume without loss of generality that $s=1$, that is the bounding box for every subcontinuum of $C$ is a square. In this case we call $C$ hereditarily square-boxed, hsb for short, and this is the only case that we consider below (and will prove that $C$ must be a line segment with slope $1$ or $-1$).
Take any two different points $D$ and $E$ in $C$. It is well known that path-connected is equivalent to arc-connected (for plane continua), that is there is an arc $\gamma:[0,1]\to C$ with $\gamma(0)=D$ and $\gamma(1)=E$. (An arc has no self-intersections, the map $\gamma$ is a homeomorphism from $[0,1]$ to its image $\gamma([0,1]).$)
Let $B$ be the bounding box for $\gamma([0,1])$. Let $T$ and $Y$ be the top and bottom (closed) edges of $B.$ The arc $\gamma$ must intersect all edges of $B$, and in particular (by reversing direction if necessary) there are $\tau<\nu$ such that $\gamma(\tau)\in T$, $\gamma(\nu)\in Y$, and $\gamma((\tau,\nu))\cap(T\cup Y)=\emptyset.$ Then the bounding box for $\gamma([\tau,\nu])$ must be the same size (same height, a square), and in fact must coincide with the bounding box for the entire $\gamma([0,1]).$
We claim that $\gamma(\tau)$ must be one of the two endpoints of $T$ (that is, either the top left or the top right corner of $B$). Indeed, if not then let $V$ and $W$ be the left and the right (closed) vertical edges of $B$, and let $\mu=\min\{t\in(\tau,\nu):\gamma(t)\in(V\cup W)\}$.
The assumption that $\gamma(\tau)$ is not an endpoint of $T$ (so $\gamma(\tau)\not\in V\cup W$) implies that $\tau<\mu$. Then $\gamma([\mu,\nu])$ must have nonempty intersections with each of $V$ and $W$, but (being compact) must remain some positive distance away from the top edge $T$. So the bounding box for $\gamma([\mu,\nu])$ must be as wide as $B$, but not as tall (so it will not be a square), a contradiction that shows that $\gamma(\tau)$ must be an endpoint of $T$, say without loss of generality the left endpoint, call it $P.$ So, $P=\gamma(\tau)$
is the top left corner point of $B.$ Similarly, $\gamma(\nu)$ must be an endpoint of the bottom edge $Y$, and in fact $Q=\gamma(\nu)$ must be the bottom right corner of $B$, diagonally opposite to $P$ (or else we could again find a subarc whose bounding box is as wide but not as tall as $B$). A similar argument shows that as $t$ goes from $\tau$ to $\nu$, the bounding box for $\gamma([\tau,t])$ must have $P$ as its top left corner point, and in addition its bottom right corner must belong to $\gamma([\tau,t])$. So $\gamma([\tau,\nu])$ must contain every point on the diagonal connecting $P$ to $Q$, and (being an arc) must therefore coincide with this diagonal. Then one could argue that $\tau=0$ for if not, then $\gamma([0,\tau])$ will remain either above or below the diagonal, only intersecting it at $P=\gamma(\tau)$; if say $\gamma([0,\tau))$ remains above the diagonal then it could easily be shown that the width of its bounding box is bigger than its height (even if possibly not as wide as $B$). All in all we picked two different arbitrary points $D$ and $E$, and an arbitrary arc $\gamma$ in $C$ connecting them, and showed that these two points must be the diagonally opposite corners $D=P=\gamma(0)$ and $E=Q=\gamma(1)$ of the bounding box $B$ of $\gamma$, and $\gamma$ must be the diagonal connecting these two points. Say $\ell$ is the line through $D$ and $E$, then it follows that $C\subset\ell.$ Indeed $\ell$ has slope $-1$ (without loss of generality) but if $F$ was a point in $C\setminus\ell$ then we could connect $F$ with one arc in $C$ to $D$ and with another arc in $C$ to $E$, and these two arcs would each have to be the diagonal of the respective bounding box, and will have to have slope $1$ (since $-1$ is already taken by $\ell\ $), and this could not happen since $D$ and $E$ are different points.
This completes the proof (or sketch) that if $C$ is a path-connected hsb continuum in the plane then $C$ must be a line segment with slope $1$ or $-1.$
The question that remains is whether we could remove the condition that the plane continuum $C$ is path-connected.
Thank you and good luck!
(References for a known answer or suggestions for better/accepted terminology would be very much appreciated, along with any answers or comments.)