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:
- 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.
- 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)
- 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 $
- 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.)
Best Answer
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 $\gamma(0)=\gamma(1)$.