Drexel dragonThe Math ForumDonate to the Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Topic: easiest proof of Jordan Curve Theorem as a corollary of Moebius
theorem #1982 Correcting Math

Replies: 7   Last Post: Aug 11, 2014 3:19 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]

Posts: 13,144
Registered: 3/31/08
proof of Jordan Curve Theorem from Moebius theorem #1985 Correcting Math
Posted: Aug 11, 2014 1:25 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

Jordan Curve Theorem: Now Wikipedia describes this theorem as saying, draw a closed curve in the plane and that closed curve divides the plane into two regions of all the points inside the closed curve and all the points exterior to the closed curve. So that if we take any point inside and any point exterior to the closed curve and followed a path containing those two points we end up crossing the closed curve.

Moebius Theorem: basically says that 4 mutual adjacencies in the plane is the maximum there can be.

Alright, here are 4 mutual adjacencies where the region B is a closed loop that is landlocked so no 5th adjacency can reach it.


So, how to prove the Jordan Curve theorem given the Moebius theorem? Well we construct a 5th region of P that encloses all of the M,O,J,B regions:


Now we construct a path out of B from an interior point of B to an interior point of P, the region that encloses all the other regions. If such a path is constructible that does not cross the borderline of B or the closed loop of B, means that such a path would be a 5th mutual adjacency of the region B, so that the Moebius Theorem is no longer true because such a construction gives a 5th mutual adjacency.

Now here is the proof again of the Moebius Theorem.

proof of Moebius theorem-- 4 mutual adjacencies is the maximum:

Proginoskes wrote 17 Feb 2006:
> The usual proof of Mobius's proof is through the use of Euler's
> formula
> for polyhedra:
> v - e + f = 2
> In the "map" with 5 mutual adjacencies, f = 5, e = C(5,2) = 10, and v <= 2/3 * 10. (Thus v <= 6.) Substituting into this equation, we get
> 2 = v - e + f <= 6 - 10 + 5 = 1.

Now maybe we can wrestle and wrangle the Euler formula to give a direct proof of Jordan Curve theorem, although I doubt it, because unless we use a generalized statement-- Moebius, we get into the bane of mathematics proof-- case studies. When your proof is case studies, more than 3 cases, means you have no proof. Because a case study proof has to prove at the end why the proof requires umpteen number of cases. Not because you say it needs these cases to study, but why is a special number of cases needed that is integral to the statement you want to prove.

You see, when we prove FLT from Beal, we need no case study at all. And when we prove Jordan Curve from Moebius, we need no case studies at all for one is bound up inside the other as a corollary. Too many old timers of Old Math believe in case studies as proofs, but the only use those case studies are, is to give the person a better understanding of the statement, and they do not serve as a proof of the statement.


Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2015. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.