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I wonder about a theorem which can be proven that this theorem is $neither\, provable$ nor $disprovable$ using any kind of mathematical knowledge.


1- Is there any such theorem? or is it proposable?

2- If there exists such a theorem about a fact, how should we think about it? does it mean that this fact is independent of everything?

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You are very confused. First of all, if it can't be proved, it's not a theorem. Second, every well-formed mathematical statement is a theorem in some system (albeit possibly an inconsistent system). Third, this has nothing whatever to do with statistics or probability, so I'm going to remove those tags. – Gerry Myerson Aug 18 '12 at 12:32
I suggest you read up on Gödel's incompleteness theorems, which address this very question. – user22805 Aug 18 '12 at 12:33
@ David Wallace thank you. – Seyhmus Güngören Aug 18 '12 at 12:36
@Gerry thanks. I couldnt find a relevant tag therefore i added them. So it seems bette now. Should I change it from theorem to hypothesis? – Seyhmus Güngören Aug 18 '12 at 12:37
@Seyhmus: statement, sentence, and closed formula are all standard terms for what you’re talking about here — a precise mathematical assertion in a formal system. A theorem of a theory in a formal system is typically taken to mean a statement that can be proved in the theory. – Peter LeFanu Lumsdaine Aug 18 '12 at 17:55
up vote 12 down vote accepted

This is a bit of a semantical issue.

Commonly the word theorem refers to a statement which is provable from a certain theory, for example "Zorn's lemma is a theorem of ZFC" is to say that we can prove Zorn's lemma is true from ZFC.

However sometimes the word "sticks" to a certain assertion (such as Zorn's lemma) and it then becomes a part of the name "Cantor's theorem"; "Hahn-Banach Theorem"; "Tychonoff's theorem"; and so on. For example, the Hahn-Banach theorem is not a theorem of ZF, but it is a theorem of ZFC.

Given a collection of axioms which is "nice" (in the sense that we can easily determine if an arbitrary statement is an axiom); as well strong enough to describe the natural numbers, then if this collection is consistent then there are statements which are true and it cannot prove nor disprove. This is The Incompleteness Theorem mentioned by others.

The Continuum Hypothesis is one example of this. There are models of ZFC in which the continuum hypothesis is true; other models of ZFC in which it is false. Similarly the axiom of choice cannot be proved from the axioms of ZF.

However, we are free to add to ZFC other axioms which are sufficient to prove that the continuum hypothesis is true; or we can add axioms which will prove the continuum hypothesis is false. Of course we cannot add both axioms, as that would generate an inconsistency in the system... and inconsistencies are bad.

For example from the axioms of ZFC+$\lozenge_{\omega_1}$ we can prove the continuum hypothesis is true. On the other hand from ZFC+PFA we can prove that the continuum hypothesis is false.

All in all when we say that $\varphi$ is independent from the theory $T$ it means that $T+\varphi$ is consistent (if $T$ was consistent to begin with, of course). In particular this means that $\varphi$ is not independent of the theory $T+\varphi$. This should answer your second question: no statement is independent from all theories.

share|cite|improve this answer is the name of the technique used on CH and AC – Ciro Santilli 巴拿馬文件 六四事件 法轮功 16 hours ago

To answer if something is proveable, you need to ask if a well-formed formula is provable from some theory (set of axiom). From the other answers, you can find the many example of these from set theory. However, there many examples of statements that you can not prove which most people are very familiar with.

For example, the commutativity axiom for groups is not proveable using the axioms of group theory. You would go about showing this unproveability by producing two models of the group axioms: one which satisfy the axiom of commutativity and one that does not. For example any commutative group (like $\mathbb{Z}$) and non-commutative group $S_3$.

This method of producing models is essentially how you prove even the independence results in set theory.

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Mathematics is only theories based on set of axioms. From them, you can derive truth value of formula in your theory. It is well know that for any theory with arithmetics, there are unprovable formula (see Goedel).

What does it mean ? Just that your unprovable formula can be added to your set and you obtain a new theory. You can add the negation of the formula and you obtain another theory.

Both theory can be seen as valid (as long as the first one is). There is a well known example in geometry, where you can't prove that given a point and a line (disjoint from the point), there exists only one parallel to the line through the point.

If you add this axiom to the other axioms of geometry you obtain the euclidean geometry, but if you don't, you can obtain some hyperbolic geometry. Both are nice geometries with different domains of usefulness. So you can't say something is true independently of everything else, except for god, but it's not the point here ?

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nice)) it was poorly a point. Good catch. – Seyhmus Güngören Aug 18 '12 at 12:52
The question is a good one, because for some axioms (for example, there exist a set in ZF, or there exist 0 in Peano arithmetic), we "feel" they are true because we can't do anything without them. So there is some notion of truth, but it's more metamath (or philosophy) related for now. – Xoff Aug 18 '12 at 12:58
but it means there is no limit for such. There will be always something new? – Seyhmus Güngören Aug 18 '12 at 13:08
yes, we are only limited by our imagination. But it's a huge limitation. And we like beautiful and small. Hard to find such new things... – Xoff Aug 18 '12 at 13:11
what's does god have to do with it?! – wim Oct 8 '12 at 1:18

Yes, there are many such “theorems”. These include well-known conjectures such as Continuum Hypothesis, and also many more obscure ones, like Whitehead problem. One might also include axiom of choice among those (well, assuming ZF is not inconsistent).

You might want to take a look at the list at Wikipedia.

That said, your question has nothing to do with statistics or probability theory, really.

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Exactly I couldnt even find a Tag for it then I decided to take attraction of some probabilistic guys. I am aware of continuum hypothesis but it is non provable in the ZF as long as i remember. Does it mean that no other mathematical knowledge can prove or disprove it? – Seyhmus Güngören Aug 18 '12 at 12:35
@SeyhmusGüngören: What do you mean exactly by “mathematical knowledge”??? – tomasz Aug 18 '12 at 12:35
I mean there are ways to use the mathematic. It might happen that one uses set theory and prove that it is not provable with the methods in set theory. But mathematic is not restricted to set theory. It shouldnt block anybody to come up with another solution using some different theory? or am I mistaken. Here: it says: "Impossibility of proof and disproof in ZFC" but in ZFC. I thought it was not general enough. – Seyhmus Güngören Aug 18 '12 at 12:42
@SeyhmusGüngören: You have to take some assumptions to prove a theorem. Sure, you can take different axioms and using those, prove any theorem along with its negation (except perhaps the $x\neq x$ kind), like Gerry Myerson remarked in a comment to your question. But using the commonly accepted axioms, you can't prove those statements, and, as Gödel's theorems show, for any sensible and reasonably comprehensive axiom system, there will be undecidable statements. – tomasz Aug 18 '12 at 12:45

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