If space is not continuous, then is calculus wrong?

[Hootenanny]

Without getting too philosophical, I don't accept the premise that mathematics has the ability to be "wrong." It can be used/applied incorrectly, but to suggest that it can be wrong is analogous to assigning blame to a tool for being used improperly, rather than the person who used the tool.

I disagree here. It is possible for mathematics to be inconsistent, in which case, I would consider it to be "wrong".

 

[D H]

Without getting too philosophical, I don't accept the premise that mathematics has the ability to be "wrong." It can be used/applied incorrectly, but to suggest that it can be wrong is analogous to assigning blame to a tool for being used improperly, rather than the person who used the tool.

I disagree here. It is possible for mathematics to be inconsistent, in which case, I would consider it to be "wrong".

What's worse, we don't know if the continuum (the reals) are inconsistent or incomplete. Gödel's incompleteness theorems kinda get in the way.

 

[Studiot]

we don't know if the continuum (the reals) are inconsistent or incomplete

This sounds interesting, you are presumably talking about something other than Cauchy sequences here.

Would you care to elaborate?

 

[D H]

What's worse, we don't know if the continuum (the reals) are inconsistent or incomplete.

This sounds interesting, you are presumably talking about something other than Cauchy sequences here.

Would you care to elaborate?

I already did. You left out the elaboration, which was the very next sentence in the post you quoted: "Gödel's incompleteness theorems kinda get in the way."

 

[pwsnafu]

I already did. You left out the elaboration, which was the very next sentence in the post you quoted: "Gödel's incompleteness theorems kinda get in the way."

There's nothing stopping you moving outside the reals to determine if the reals are inconsistent.

 

[Studiot]

I already did

I know you mentioned Goedel.

However I thought that Cauchy showed the real line to be complete so I would be very interested to see G's theorem used to refute this.

 

[Dembadon]

I disagree here. It is possible for mathematics to be inconsistent, in which case, I would consider it to be "wrong".

I probably lack the mathematical maturity it would take to understand an example, but just in case I'm able to figure it out, could you indulge me and give me an example anyways?

The reason I ask is because, logically, it is possible for an argument to be valid even though the premise(s) is/are inconsistent.

 

[bp_psy]

I know you mentioned Goedel.

However I thought that Cauchy showed the real line to be complete so I would be very interested to see G's theorem used to refute this.

Those are two completely different concepts one deals with the logical foundation of mathematics the other with the existence of gaps in the reals.

 

[HallsofIvy]

Saying that "the reals may be inconsistent" is misleading. What is meant, is not the real line but the axioms defining the real number system. It is a set of axioms, not a set of points or numbers, that is "consistent" or "inconsistent". What Goedel showed was that any set of axioms, large enough to encompass the natural numbers (so this includes any set of axioms large enough to encompass the real numbers, which include the natural numbers) is either inconsistent or incomplete.

"Inconsistent" means that it would be possible to prove both a theorem and its negation. Which would, of course, make a "proof" of anything meaningless.

"Incomplete" means there exist some theorem that can be stated in terms of the axioms but cannot be either proved or disproved in those axioms. (We could, of course, add that theorem as an axiom, but then there would be some other theorem that could neither be proved nor disproved.)

Yes, it it were shown that the axioms for the real numbers (or even just the natural numbers) were inconsistent, that would be a disaster for mathematics. However, that has never been proven and most mathematicians (I would think anyone who thought doing mathematics was worthwhile!) believes that the axioms are consistent and so must be incomplete- which is no big deal. It only means that there will always be more mathematics to discover.

 

[Studiot]

Thank you for that clarification, HOI.

So, if I understand you correctly, the comment about the reals is about axioms, not the real line itself.

In this thread (I think) we are discussing the connection between the real number line itself and spatial reality not the axioms or definitions of either so if you are able to further clarify the relevence that would be grand.

 

[Fredrik]

I probably lack the mathematical maturity it would take to understand an example, but just in case I'm able to figure it out, could you indulge me and give me an example anyways?

The reason I ask is because, logically, it is possible for an argument to be valid even though the premise(s) is/are inconsistent.

I think anyone who has enough mathematical maturity to have heard the words "mathematical maturity" will understand this example:

Definition: A set X is said to be stupid if
(1) X={0,1}.
(2) X={1}.

The mathematics of stupid sets can be said to be "wrong", because the set of axioms that defines the term (the statements (1) and (2)) is inconsistent (i.e. the axioms contradict each other). Let's prove a theorem in the theory of stupid sets:

Theorem: Every stupid set contains the number 5.

Proof: Let X be an arbitrary stupid set. We will prove that $5\in X$ by deriving a contradiction from the assumption that $5\notin X$. So suppose that $5\notin X$. Since X is stupid, $\{1\}=X=\{0,1\}$, but this contradicts $\{1\}\neq\{0,1\}$.

Since we didn't have to use the assumption $5\notin X$, the method we used to prove this can be used to prove any statement. Let P be an arbitrary statement.

Theorem: P

Proof: Suppose that P is false. Let X be an arbitrary stupid set. Then $\{1\}=X=\{0,1\}$. Since this contradicts $\{1\}\neq\{0,1\}$, P must be true.

So in the theory of stupid sets (and in all other inconsistent theories), every statement is true.

 

[D H]

I probably lack the mathematical maturity it would take to understand an example, but just in case I'm able to figure it out, could you indulge me and give me an example anyways?

The reason I ask is because, logically, it is possible for an argument to be valid even though the premise(s) is/are inconsistent.

So in the theory of stupid sets (and in all other inconsistent theories), every statement is true.

As is every statement's contradiction.

Dembadon, there are some mathematicians who intentionally play with inconsistent mathematics (e.g., http://plato.stanford.edu/entries/mathematics-inconsistent/, http://books.google.com/books?id=pipfBn8mCTwC). The only way forward with such systems is to throw out reductio ad absurdum, or proof by contradiction. This is an extremely powerful tool that most mathematicians are loath to surrender. That is why Halls in an earlier post said that most mathematicians deal with Gödel's incompleteness theorems by believing "that the axioms [of the reals] are consistent and so must be incomplete".

 

[Dembadon]

Thank you for the example, Fredrik, and the explanation, D H.

 

[lugita15]

What's worse, we don't know if the continuum (the reals) are inconsistent or incomplete. Gödel's incompleteness theorems kinda get in the way.

Actually the first order theory of real closed fields is complete and provably consistent, as shown by Tarski, because it is too weak for Godel's Theorem to apply. You can only talk about the natural numbers, and thus make the system susceptible to Godel, if you use the second order theory of real numbers.

 

[Studiot]

Actually the first order theory of real closed fields is complete and provably consistent, as shown by Tarski, because it is too weak for Godel's Theorem to apply. You can only talk about the natural numbers, and thus make the system susceptible to Godel, if you use the second order theory of real numbers.

Please explain further.

This philosophy of maths / very pure maths is beyond my normal field, but interesting.

 

[lugita15]

Please explain further.

This philosophy of maths / very pure maths is beyond my normal field, but interesting.

OK, a first order theory is a theory that only allows quantification over individuals, not quantification over sets. So in the first order theory of real numbers, you can say "for all real numbers x, ...", but you can't say "for all sets F of real numbers, ..." (for that you'd need the second-order theory).

Now the standard first-order theory of natural numbers is called Peano Arithmetic (PA), and Godel's theorem applies to any theory which can prove all the facts about natural numbers that PA can prove. But it turns out that in the theory of real closed fields (RCF), which is the standard first-order theory of real numbers, we can't even talk about natural numbers! The axioms of RCF are all the axioms for fields, plus one additional axiom which is stated http://en.wikipedia.org/wiki/Real_closed_field#Definitions". You may be thinking, if the field axioms define both the multiplicative identity (1) and addition, that would be enough to talk about natural numbers; you could just say that natural numbers are 1, 1+1, 1+1+1, and so on. The problem is, what does "and so on" mean? It's not precise enough for a formal theory.

Here's a proper definition of natural numbers. A set F of real numbers is called hereditary if it is closed under the successor operation; in other words, $x+1 \in F$ whenever $x \in F$. Then we can say that a natural number is a real number which belongs to all hereditary sets containing 1. If you stare at that long enough, you'll find that it works: 1 is a natural number because it belongs to all sets containing 1, so in particular it belongs to all hereditary sets containing 1. 2 is a natural number because it is 1+1, so any hereditary set that contains 1 automatically contains 2. Etc. Since this definition used the phrase "ALL hereditary sets", any statement about natural numbers is necessarily second-order in the theory of real numbers.

Because of all this, Godel's theorem doesn't apply to RCL, so Tarski was able to show that RCL is complete, consistent, and even decidable (meaning you can write an algorithm which decides whether any given first-order statement about real numbers is true or false, something you can never dream of doing with the natural numbers).

 

[SteveL27]

Because of all this, Godel's theorem doesn't apply to RCL, so Tarski was able to show that RCL is complete, consistent, and even decidable (meaning you can write an algorithm which decides whether any given first-order statement about real numbers is true or false, something you can never dream of doing with the natural numbers).

Thank you for that explanation. Curious about one point: You are saying that first order statements are decidable. Are second-order statements decidable too? Or not? For example the least upper bound property, which quantifies over sets (for all nonempty sets bounded above etc.) is a second order property. Does this mean there are undecidable second-order statements about real closed fields?

 

[D H]

Here's a proper definition of natural numbers. A set F of real numbers is called hereditary if it is closed under the successor operation; in other words, $x+1 \in F$ whenever $x \in F$. Then we can say that a natural number is a real number which belongs to all hereditary sets containing 1.

Hmmm. The set of all hereditary set that contain 1 looks to me to be more like $\mathbb Z$ rather than $\mathbb N$. Isn't $\mathbb N$ (or a set polymorphic to it) the hereditary set that contains 1 but contains no element whose successor is 1?

Regarding the reals, isn't the concept of a supremum second order?

 

[lugita15]

Thank you for that explanation. Curious about one point: You are saying that first order statements are decidable. Are second-order statements decidable too? Or not? For example the least upper bound property, which quantifies over sets (for all nonempty sets bounded above etc.) is a second order property. Does this mean there are undecidable second-order statements about real closed fields?

Yes, any second-order theory of real numbers have undecidable statements. The axioms of the standard second-order theory of real numbers are the axioms for fields along with the least-upper-bound axiom, which states that every nonempty set which has an upper bound has a least upper bound (as you mentioned, this is a second-order statement). Then you can define the natural numbers using a second-order statement about real numbers, and in fact you can prove (it's a pretty simple exercise) all 5 of Peano's axioms, including the full second-order principle of mathematical induction, which begins with "for all of natural numbers ...".

So the second-order theory of real numbers contains within it all of second-order Peano arithmetic, which means that Godel's theorem definitely applies.

 

[lugita15]

Hmmm. The set of all hereditary set that contain 1 looks to me to be more like $\mathbb Z$ rather than $\mathbb N$. Isn't $\mathbb N$ (or a set polymorphic to it) the hereditary set that contains 1 but contains no element whose successor is 1?

Regarding the reals, isn't the concept of a supremum second order?

$\mathbb Z$ is definitely not the intersection of all the hereditary sets containing 1; $\mathbb N$ is a hereditary set containing 1, but it doesn't contain the rest of the integers. And the interval $\left[1,\infty\right)$ is a hereditary set containing 1, so $\mathbb N$ is not the only hereditary set containing 1.

Yes, the definition of supremum is second-order, which is why we can't use the least-upper-bound axiom for the first-order theory. Instead, we have to use a more complicated axiom which yields the same consequences as LUB, at least for first-order statements. For instance, we can take the intermediate value theorem for all polynomials as our axiom. See http://en.wikipedia.org/wiki/Real_closed_field#Definitions" for other equivalent axioms.

(The fact that you're not actually using LUB comes with a cost: even the algebraic numbers satisfy the first-order theory. This means that somewhere deep in the proofs that pi and e are transcendental, we are implicitly using a second-order consequence of the least upper-bound property. I wonder what that is ...)

 

[D H]

$\mathbb Z$ is definitely not the intersection of all the hereditary sets containing 1; $\mathbb N$ is a hereditary set containing 1, but it doesn't contain the rest of the integers. And the interval $\left[1,\infty\right)$ is a hereditary set containing 1, so $\mathbb N$ is not the only hereditary set containing 1.

Got it. I just didn't follow your directions the first time around. I didn't stare at it long enough.


Anyhow, we are getting quite far afield from the OP. Then again, the original post represented a misconception that was dealt with in the first few posts of this thread. Whether space is discrete or continuous has nothing to do with the validity of calculus. Or the reals for that matter.

 

[Studiot]

Lugita, thank you for these clear and handy explanations.

 

[Studiot]

Here is another way to look at the question
(This was inspired by another recent thread)

If we roll a die n times the probability of a specific sequence is

$${P_n} = {\left( {\frac{1}{6}} \right)^n}$$

Now let n tend to infinity

$${\left[ {{P_n}} \right]_{n \to \infty }} = \left( {\frac{1}{6}} \right)_{n \to \infty }^n = 0$$

That is the probability of any particular sequence becomes vanishingly small.

Yet we assert that if we add all of these up we get a finite total.

$$\sum\limits_{n = 1}^\infty {\left( {{P_n}} \right)} = 1$$

Which is essentially the same process as the probabilisitc calculation/view in quantum mechanics.