3

This could be something truly stupid. But let me describe it: Let's take a proof by contradiction, we suppose $¬ A$, and when an absurdity comes from this, we deduce $¬¬A$ must be true.

  • In this case, we verified what happens when something is false and found an absurdity, and hence it must be true, but we actually didn't verify if it being true would imply some problem.

Now take - for example - Russell's paradox. We verify if it is true and an absurdity happens and then, we verify if it is false and then another absurdity happens.

  • Notice that in this case, we verified what happens if it is truth or false.

So in general what guarantee do we have that $¬A$ causes an absurdity, $¬¬A$ is true and $¬¬A$ does not cause another absurd just as in Russell's paradox?

$$\tiny \text{Anybody with a heart upvotes love.}$$

Red Banana
  • 23,956
  • 20
  • 91
  • 192
  • So far as I understand it, maths is more about proving things in a certain environment (not quite sure what term to use here) and slowly but surely constructing theorems about that environment. When we use proof by contradiction, we deduce that a statement must either be true in our environment (in this case the type of logic we are using) or that it is not applicable in our environment. – user400188 Jan 20 '17 at 03:49
  • I think it is an axiom. The guarantee seems to be that all of the mathematics are axiomatized from a logic that support $A=\lnot\lnot A$. – Masacroso Jan 20 '17 at 03:49
  • Yes, it is implicitly assumes that the axioms we are using do not have contradictions like this. It is impossible to know if for sure (except for fairly limited axiom systems) but we haven't found such a logical contradiction yet in the current systems, which are designed to avoid the problems of Russell's paradox. If the axioms are inconsistent, however, then they are problematic for a variety of reasons. – Thomas Andrews Jan 20 '17 at 03:49
  • But why did we have question both yes and no for Russell's paradox and we don't do the same for proofs by contradiction? – Red Banana Jan 20 '17 at 04:01

2 Answers2

1

You did not understand Russell's paradox correctly. Here it is formally (using natural deduction): $ \def\imp{\Rightarrow} \def\eq{\Leftrightarrow} $

If $\exists S\ \forall x\ ( x \in S \eq \neg x \in x )$:

  Let $R$ be such that $\forall x\ ( x \in R \eq \neg x \in x )$.   [by $\exists$-elim]

  $R \in R \eq \neg R \in R$.   [by $\forall$-elim]

  $R \in R \imp \neg R \in R$.   [by $\eq$-elim]

  $\neg R \in R \imp R \in R$.   [by $\eq$-elim]

  If $R \in R$:

    $\neg R \in R$.   [by $\imp$-elim]

    Contradiction.

  $\neg R \in R$.   [by $\imp$-intro and $\neg$-intro]

  $R \in R$.   [by $\imp$-elim]

  Contradiction.

$\neg \exists S\ \forall x\ ( x \in S \eq \neg x \in x )$   [by $\imp$-intro and $\neg$-intro]

In particular, we do not have a sentence that is proven true and false at the same time, except under the false assumption of the existence of a Russell collection. The above proof is valid in ZF set theory, and also in many other systems, as long as they support the rules used. It thus implies that such systems cannot have unrestricted specification of collections.

Community
  • 1
user21820
  • 57,693
  • 9
  • 98
  • 256
  • Yes, but you have translated it to a question of whether it exists or not and the answer is no. Isn't it different of asking if the set of all the sets which do not contain themselves contains itself (in which the answer is nor yes nor no)? Or should all the questions alike be translated in this way? – Red Banana Jan 20 '17 at 05:07
  • 1
    @OppaHilbertStyle: The answer is that we cannot even construct the Russell collection in ZF set theory, so we are unable to say "The Russell set does not exist.". Rather, we can say "There is no set that behaves like the Russell predicate.". The Russell predicate is defined as $R$ such that $\forall x\ ( R(x) \eq \neg x \in x )$. Does this answer your question? If not I can further explain. The key is that there is a difference between introducing a new predicate symbol defined by some first-order formula (we always can), and claiming the existence of a set that reflects the predicate. – user21820 Jan 20 '17 at 15:07
  • 1
    @OppaHilbertStyle: See http://math.stackexchange.com/a/1864310/21820 for a precise explanation of when and how we can introduce new predicate-symbols and function-symbols in any first-order formal system. This is how we can talk about the Russell collection indirectly in ZF set theory, but we can never directly construct it as an object, and the proof above shows that no object with such behaviour exists. – user21820 Jan 20 '17 at 15:09
0

This is my understanding:

When we work with proofs by contradiction, We take as an axiom that every statement is either true or false, but not both.

Thus, If $\neg A$ is false, then (by another axiom) $A$ is true.

However, when it comes to Russell's paradox, you encounter a statement that is both true and false. Thus, the above "setting" is invalid (since some of the axioms are not true) and we cannot apply the "proof by contradiction" logic.

Pawel
  • 4,871
  • Your last paragraph is correct regarding naive set theory. That is precisely why we had to discard it. However, see my answer for what remains valid in ZF set theory, which is the mainstream foundation for modern mathematics. – user21820 Jan 20 '17 at 04:35