Is a point a Jordan Curve?

Question

My days of listening to mathematical lectures are a few years past (I pretty much only need applied math like actually solving integrations, but no more theoretical math since years) and yet I have recently found a puzzle that intrigued me: It is actually an attempt of a disguised mathematical problem. The puzzle, because it used normal english in sloppy wording, it clearly allowed to cheat the actual math problem by using a single point as the solution as having only one point available does not allow to place several distinct points on the loop to create a polygon woth equal sides.

Now, after reading up on the intended mathematical problem - the inscribed square problem - the demanded for the shape of the loop was, that it has to be a Jordan-Loop. Also, nothin in the problem indicates, that the square needs to have a side length $ a\neq 0$ but the wording in the puzzle was, as I mentioned sloppy.

This lead to an odd thought that I couldn't find in my old notes or the definition text books:

Is a point a valid Jordan-Loop, even if an edge case?

For myself I worked out that it might be sitting on the edge and that there might be an argument that a point might be a Jordan Loop, based on that it fits the following parameters:

1. It is a curve (of length 0) from a starting point (itself) to an end point (itself). As they are identical it is a closed loop.
2. The curve doesn't (seem to) intersect itself to me [but we never covered how to proove that besides just look at it in the class] (see also 3)
3. There is a continuous function that can achieve this curve, derived from the unit circle: $\phi(t)=(\cos(t)\sin(t))r ,t\in[0,2\pi],r=0 $
4. The unit circle is one of the standard examples of a Jordan Curve.

On the other hand, my mind boggled at one thing, yelling at me that something was amiss there:

• Doesn't a curve of length 0 violate something? Or was it just a demand of Jordan-curves to have length>0?!

Something in this made my mind grind to a halt.

So, is a Point a valid Jordan-Loop or do my assumptions violate some commandment I have forgotten? (If yes, please tell me how.)

Answer 1

A curve is by definition a continuous function $\gamma: [0,1]\to\mathbb R^2$. Being a Jordan curve adds a few additional conditions.

In order for a single point to be a curve, your function $[0,1]\to\mathbb R^2$ has to be a constant function. Since, for example, $\gamma(\frac13)=\gamma(\frac23)$, it does not satisfy the "no self-intersections" part of the requirements for a Jordan curve. The only deviation from injectivity that is allowed (and required) is $\gamma(0)=\gamma(1)$.

Answer 2

The curve does "intersect itself" since it meets the origin many times.

Edit in response to comment. I think what I wrote is a proof. This comment from @RicardoBuring spells it out a bit more by clarifying an implicit definition:

The technical meaning of "the curve doesn't intersect itself" is the condition that the function is injective on the interval with one endpoint removed

Answer 3

You have to start with a definition. Once you have a definition of a Jordan curve, then you can ask if a singleton point meets that definition. Some Googling gives a couple of fairly reasonable definitions:

Definition: A Jordan curve is a plane curve which is topologically equivalent to (a homeomorphic image of) the unit circle, i.e., it is simple (i.e., no self-intersections) and closed (i.e., the curve encloses an area; it has no endpoints). [source]

and

Definition: A Jordan curve or a simple closed curve in the plane $\mathbb{R}^2$ is the image $C$ of an injective continuous map of a circle into the plane, $\varphi: S^1 \to \mathbb{R}^2$. [source]

These two definitions are equivalent (though this statement may not be entirely obvious and, if we are being really rigorous, requires proof), so we may use either to conclude that a singleton point is not a Jordan curve.

In the case of the first definition, a singleton point is not homeomorphic to a circle (this is an exercise in point-set topology; if you don't know any topology, skip to the second definition). In the case of the second definition, we can note that if $\varphi : \mathbb{S}^1 \to \mathbb{R}^2$ is defined by $\varphi(x) = p$ (where $p$ is our distinguished point in the plane), then we must have $\varphi(x) = p = \varphi(y)$ for any $x,y\in S^1$. But $S^1$ contains more than one point, so injectivity fails.

Again, the key point is that you have to use the definitions. Whenever you want to know "Is an $X$ a $Y$?" or "Is $X$ an example of $Y$?", start by carefully understanding the definition of $Y$. Once you are sure that you understand that definition, then decide if $X$ satisfies that definition. Don't rely on heuristics and intuition.