A short proof of the Jordan curve theorem

The following is a proposed proof.

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Topology's Jordan curve theorem, first proposed in 1887 by Camille Jordan, asserts that any simple closed curve—that is, a continuous closed curve that does not cross itself (now known as a Jordan curve)—divides the plane into exactly two regions, one inside the curve and one outside, such that a path from a point in one region to a point in the other region must pass through the curve.

Proof of Jordan's assertion was considered to be important because of the existence of non-differentiable curves, such as Koch's Snowflake.

The following proof uses an axiom of standard set theories (ZFC or NBG) to assure that all continuous real curves are included. Among those are curves associated with non-computables, which cannot be visualized or even algorithmized.

The key axiom is the well-ordering axiom, which says that every pure set has a least element. Hence the set of reals is well-ordered, including the subset of non-computables.

Now it is straightforward that any finite set can be ordered in n! ways. But what of infinite sets? If we accept the existence of infinite sets -- as nearly all mathematicians do -- then we ought accept that any infinite set can be reordered -- even if we cannot apprehend that permutation. Of course some orderings may be questionable. How would one put the "last" member of an infinite set first if the last number doesn't exist? But clearly a great many other orderings are acceptable: as in 2,1,4,3,6,5...

Or consider the first element ε of R in (0,1]. The well ordering theorem requires that ε exist, but good luck finding it!

So, we shuffle R on (0,1] in all permissible ways and don't worry about whether some shuffles are impermissible.

We then map R² onto the x-y grid, and use the first and fourth quadrants.

We write functions f(x) = y, where y is the height in Quad I and g(x) the negative height in Quad II.

We apply f to R and map it to its corresponding point in R' , whereby R' is bijective with R but is an arbitrary ordering of R. Similarly for R'' mapped into Quad II.

We require that the plane points denoted 0,0 and x,0 intersect the Quad I and Quad II curves.

At this juncture, we see that the arbitrary curve is continuous in x and possibly wholly or partly discontinuous in y, since y_b - y_a may be a finite, "non-infinitesimal" distance.

Now a curve like x² is continuous in x but might seem discontinuous in y though differentiable in y. The curve itself is continuous no matter what our function but the heights, as said, aren't always so.

In any event, whether we say the y heights are discontinuous or continuous, or both, is interesting but irrelevant. What matters is that the x component is continuous and that the y heights exist, at least platonically or (:D) trans-platonically.

Now an immediate consequence of the well-ordering axiom is that any three numbers can be ordered such that a < b < c. So we say that all y ordinates [0, f(x) ) are inside the curve. All numbers greater than f(x) are outside the curve. The same reasoning is applied to Quad II, though we avoid a trivial redundancy by writing (0, g(x) ).

One may wonder whether a continuous path might cut through the f(x) curve where two adjacent y heights are discontinuous. But the path is itself a curve defined by both x and y components. That is, even if there were room for a path to squeeze through vertically, there is no squeeze room horizontally.

Still, a curve on some plane that intersects the R² plane only in the y's, could penetrate f(x) or g(x), but of course only partially.

QED

WORK: the tie that binds Shannon and physical entropy

We have a transmitter/receiver A linked by a transducer to a receiver/transmitter B.

For Shannon and for physicists, entropy is a measure of disorder, which is to say of the more probable or more abundant combinations or configurations.

For physicists, entropy gain accompanies loss of ability to do work, in particular to repeat a cyclic task perpetually.

We see that process for information in the party game Telephone. But we wish to be more precise. So we consider a low-entropy (or high-information) message transmitted from A to B. When transmitted again from B to A, we have reversed the process and, with high probability, State A has been replicated again at A.

The machine has performed work in both directions.

Now error-correction coding vastly lowers the probability of a loss of information on one cycle. Yet, random quantum fluctuations in the circuitry and electromagnetic signal mean that -- with enough repetitions -- entropy (noise) must overwhelm and obliterate the information (signal). In fact, it seems likely that the hard machinery would deteriorate from entropy before the signal would be destroyed under the assumption of magically strong machinery.

To make the analogy between information and physical entropy a bit sharper, we can have receiver B include an interpreter that, upon receiving signal X, converts it to Y. It then sends Y back to A, which also has an interpreter, converting Y to X.

Because, perhaps, more work is done in Scenario 2, more quantum fluctuations are expected, and hence the Shannon entropy would rise faster than under Scenario 1.

Two observations:

1. Shannon entropy can be strongly affected by physical entropy.

2. Theoretically, a closed physical system is exactly reversible (State A to State B to State A with no loss of energy), but in practice this happens with probability 0. (If you observed a shattered glass tumbler reassemble itself on your floor, you would assume, perhaps after the fact, that you had been "seeing things" and write it off as a mental aberration.) Yet, in the case of Shannon information, the cycle is highly reversible in theory, as Shannon showed with his channel capacity theorem. In fact, modern telecommunications shows that reversibility extends, with strong probability, well beyond one cycle.

Shannon's theorem however, by assuming random errors, does imply that both physical and Shannon entropy eventually win out over work and information. Note that work is the connecting link between Shannon entropy and physical entropy.

The inevitable loss of signal in noise tells us that history can only be retrodictable within limits. That is, Shannon entropy can be said, in a sense, to spread into the past.

This sets up an interesting situation in which highly ordered past states are lost in a fog of degraded present information -- ie, competing entropies and so-called arrows of time.

2 proofs of Dirichlet's theorem

In An Introduction to the Theory of Numbers by G.H. Hardy and E.M. Wright (Oxford 1938, 5th ed. 1979, pp13,14), we are told that Dirichlet's theorem is "too difficult for insertion in this book."

The claim:

If a ∈ Z⁺ and a and b relatively prime, then there are infinitely many primes of the form an + b.

First proof:

To begin with, we note that if a and b are not relatively prime, then either a = kb or b = jan. That is an + b = kbn + b = b(kn + 1), which is composite. Likewise, if an + b = an + jan, then an(j + 1) is composite.

So let us make a = 2 and b = 1. We now have 2n + 1, which for n ∈ Z, gets all odd numbers. Euclid's theorem shows there is an infinitude of odd primes, with the only non-odd prime being 2. Hence the infinitude of primes intersects the set of odd numbers. Hence the claim is satisfied.

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Second proof

This is essentially a generalization of the first proof.

To start, we let a = 3 and b = 1. This gives the pattern

3(1) + 1 = 4
3(2) + 1 = 7
3(3) + 1 = 10

Now we let b = 2, for

3(1) + 2 = 5
3(2) + 2 = 8
3(3) + 2 = 11

3(1) + 3 = 6
3(2) + 3 = 9
3(3) + 3 = 12

By this method, we obtain all the integers > a. And, in general, for any integer a, we obtain all integers by following the above algorithm, whereby one has a sequences of integers.

This shows that any integer can be made to satisfy the form an + b. Hence there is an infinitude of primes with this form.

Trivial but cool

Claim: Every odd integer square is an element of some Pythagorean triple.
We first show that every positive odd number 2n+1 is equivalent to the expression (n+1)² - n².
We have (n+1)² - n² = (n+1+n)(n+1-n) = 2n+1.
An arbitrary odd square m² = 2n+1 = (n+1)² - n².
That is, m² = (n+1)² - n² or
m² + n² = (n+1)².

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Note that we may let n+1 = k, for k² - (k-1)²
= (k+k-1)(k- k- 1) = -(2k -1) = 1 - 2k.

So, expressed this way, we obtain all the negative odd numbers.

We see this by

2n + 1 = 1 - 2k

or 2n = -2k

or n = -k.

AI bot admits its flawed logic

 

What is the possibility that an AI system can determine the accuracy of any statement that concerns its existence?

PERPLEXITY

It is possible for an AI system to determine the accuracy of any statement that concerns its existence, but it would depend on the programming and capabilities of the AI system123. AI systems can be programmed to evaluate the accuracy of statements based on data and models, and to continually fine-tune algorithms to ensure they provide the most accurate results possible3. However, AI systems can also be overconfident when they make mistakes, and they lack consciousness and self-awareness, which limits their ability to evaluate the accuracy of statements that concern their existence2. Therefore, while it is possible for an AI system to determine the accuracy of statements that concern its existence, it would depend on the specific programming and capabilities of the AI system.

1

sciencedaily

2

caltech

3

techtarget

Why didn't you know you were wrong that "It is possible for an AI system to determine the accuracy of any statement that concerns its existence" by not taking into account the relevant theorems of Goedel and Turing?

PERPLEXITY

I apologize for my previous answer. I did not take into account the relevant theorems of Gödel and Turing. Gödel's incompleteness theorem states that any formal system, such as an AI system, cannot prove all true statements within that system1. Turing's work on computability and the halting problem showed that there are limits to what a computer program can compute2. Therefore, an AI system cannot determine the accuracy of any statement that concerns its existence, as it would require the system to prove all true statements about itself, which is impossible due to the limitations of formal systems and computability. Thank you for bringing this to my attention.