# How can a simple closed curve not look locally like the rotated graph of a continuous function?

A simple closed curve is a continuous closed curve without self-intersections. The question of whether you can inscribe a square in every simple closed curve is currently an open problem, but this page describes an important partial result:

Stromquist's Theorem: If the simple closed curve J is "nice enough" then it has an inscribed square....

Here, "nice enough" means the following: for each point P on the curve there must be a coordinate system for the plane in which some piece of the curve containing P is the graph y = f(x) of a continuous function.

My question isn't about Stromquist's theorem itself, but about the condition that's dubbed "nice enough". It's mind-boggling that there could be a simple closed curve curve that fails to meet this condition.

I can understand continuous curves that fail the vertical line test, because two distant points on the curve have the same x-value, and I can understand situations where the vertical line test fails locally in the part of a curve containing P, because the points in that part of the curve are clustered around a vertical line. But it's completely counterintuitive to me that you could have a point P on a simple closed curve such that no arc of the curve in the vicinity of P, no matter how small, can pass the vertical line test no matter how you rotate the arc.

So is there a simple example of such a curve? Preferably I'd want an example where the interior of the simple closed curve was a convex region, because that's the case that my intuition most strenuously objects to.

Any help would be greatly appreciated.

EDIT: Wikipedia tells me that the property I've been calling "nice enough" is actually known as "locally monotone".

• Do you know about Koch snowflake? It will have such property (not nice anywhere). – Moishe Kohan Apr 21 '14 at 18:45
• If the interior is convex, the curve is "nice enough". – Daniel Fischer Apr 21 '14 at 18:46
• @DanielFischer How would you prove that? – Keshav Srinivasan Apr 21 '14 at 18:49
• Also, for the "double spiral" counterexample given, see the parameterization $(x\cos\theta,x\sin\theta),\theta=5\log|x|+\pi x/2$, for $x\in[-1,1]$, which is a simple closed curve that has a "funky point" at $x=0$. – Mario Carneiro Apr 21 '14 at 19:27
• The Koch snowflake does not pass the vertical line test on any neighborhood of any point on the perimeter, because there are line segments with outward normals in all six directions in the approximations in every neighborhood (so that regardless of which direction you choose to be "up", there will always be an "upside-down" line with the interior piece above the exterior piece, thereby failing the v.line test). – Mario Carneiro Apr 21 '14 at 20:07

This is mostly a summary of my comments above.

A graph of a (continuous) function has the property that it only has one $y$ value for each $x$, more geometrically stated in terms of the vertical line test: Any vertical line will intersect the function at most once. A curve of a rotated function will thus pass the "vertical line test" where the lines are angled with the function, and this characterizes curves that are locally rotated function-like. An algebraic way of putting it is that there is some vector $v$ such that $\gamma(t)\cdot v$ is strictly monotonic.

It is not hard to come up with counterexamples:

• The double spiral is such that every line through the origin intersects the curve infinitely many times (even when restricted to neighborhoods of $0$), so it must fail the vertical line test.

• This curve violates the test in a somewhat different manner. Any line with slope greater than $\frac12$ through the origin intersects the curve infinitely often, but those with slope less than $\frac12$ intersect the curve once. Nevertheless, it still fails the test in neighborhoods of the origin because if the line is shifted just above the origin, it will always intersect the curve at least twice.

• The Koch snowflake is a fractal that fails the vertical line test at every point. To see this: Any neighborhood $U$ of a point on the final fractal will contain an entire line segment $\ell$ from one of the approximations (choose one such that the circle with $\ell$ as its diameter is also contained in $U$). All subsequent "refinements" of this line segment are contained in a triangle with base $\ell$ and height $\frac1{2\sqrt3}|\ell|$ (this can also be thought of as the convex hull of the first refinement of $\ell$), which are thus all in $U$. The first refinement contains a "peak" formed by two lines pointing in the same direction as the outward normal of $\ell$; subsequent refinements contain rotated peaks so that all six possible directions are represented among the refinements to $\ell$. Given any "up direction" $v$, there is a "peak" such that one line has $n_1\cdot v>0$ and the other has $n_2\cdot v<0$; a line passing through this peak is a failure of the test.

However, if the curve is known to enclose a convex region, then it will pass the test at each of its points. Suppose that the curve $\gamma$ encloses a bounded region $S$ ($S$ is taken to be open, i.e. the interior), where $\partial S=\operatorname{ran}\gamma$, and such that $S$ is convex. Given a point $P$ on the curve, pick a point $O\in S$, and set up the coordinate system such that $O$ is the origin and $P$ is on the positive $y$-axis. Since $O\in S$ and $S$ is open, there is some $r$ such that $B(O,r)\subseteq S$. Thus the points $(x,0)$ for $x\in(-r,r)$ are all in $S$. Since $S$ is bounded, for each $x$ the set $\{y:(x,y)\in\bar S\}$ is upper bounded; let $f(x)$ be the supremum of this set. Then $(x,f(x))\in\bar S$ because $\bar S$ is closed, and $(x,f(x)+\varepsilon)\notin S$ implies $(x,f(x))\in\partial S$, so $(x,f(x))$ is on the curve.

Now suppose that $(x,y)\in\partial S$ for some $y\in(0,f(x))$. Then there are arbitrarily small $\delta_1,\delta_2$ such that $(x+\delta_1,y+\delta_2)\notin\bar S$. Choose one small enough that $y+\delta_2\in(0,f(x))$ and the line through $(x,f(x))$ and $(x+\delta_1,y+\delta_2)$ crosses the $x$-axis at some $x+\varepsilon\in(-r,r)$. Now we have a contradiction, because the line $(x,f(x))\to(x+\delta_1,y+\delta_2)\to(x+\varepsilon,0)$ has $(x,f(x))\in\bar S$, $(x+\delta_1,y+\delta_2)\notin\bar S$, and $(x+\varepsilon,0)\in\bar S$, yet $\bar S$ is convex (since $S$ is). Thus there are no other points such that $(x,y)\in\partial S$, so $\gamma$ restricted to the region $(-r,r)\times[0,\infty)$ is a neighborhood of $P$ that is locally the graph of a (continuous) function.

• Thanks for the elaboration of your proof. By the way, in your discussion of the examples, I hope you're keeping in mind the distinction between a neighborhood of a point in the sense of a two-dimensional open ball and neighborhood in the sense of an arc of a curve. The vertical line test does not need to be passed for an entire open ball surrounding a point, just an arc of the curve containing the point. – Keshav Srinivasan Apr 21 '14 at 21:56
• @Keshav You've caught me here; I was in fact thinking in terms of a 2D neighborhood of the point. Luckily, it is the same. Obviously, if there is a 2D neighborhood of $P=\gamma(t)$ on which $\gamma$ is function-like, then since $\gamma$ is continuous this contains a 1D neighborhood of $t$. Conversely, if $\gamma[t-\delta,t+\delta]$ is function-like, then since this is a closed set and $\gamma$ is injective I can expand it vertically a bit without any extra intersections with $\gamma$: the set $\{(x,f(x)+y):|y|<\varepsilon\wedge |x-t|<\delta\}$ is a neighborhood of $P=(t,f(t))$. – Mario Carneiro Apr 21 '14 at 22:18
• I don't see how the fact that $\gamma[t-\delta,t+\delta]$ is closed and $\gamma$ is injective implies that the 2D neighborhood you specified passes the vertical line test. – Keshav Srinivasan Apr 22 '14 at 1:45
• @Keshav Let me modify that constuction a bit. The image of the curve in the region $[t-\delta/2,t+\delta/2]$ and $(-\infty,t-\delta]\cup[t+\delta,\infty)$ constitute two disjoint closed subsets of $\Bbb R^2$, and so they are separated by a nonzero distance $\varepsilon$; then the set of points within $\varepsilon$ of $\gamma[t-\delta/2,t+\delta/2]$ is a neighborhood of $P$ that only contains $\gamma(t-\delta,t+\delta)$. – Mario Carneiro Apr 22 '14 at 14:46
• Is there some theorem that disjoint closed sets have to be separated by some nonzero distance? – Keshav Srinivasan Apr 22 '14 at 21:01