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Yablo's paradox arises from considering the following infinite set of sentences:

$$(S_1): \mbox{for all }k > 1, S_k\mbox{ is false} \\ (S_2): \mbox{for all }k > 2, S_k\mbox{ is false} \\ (S_3): \mbox{for all }k > 3, S_k\mbox{ is false} \\ \vdots \\ (S_i): \mbox{for all }k > i, S_k\mbox{ is false} \\ \vdots $$ [from Yablo's paradox]

If $S_1$ is true, then $S_2$ and every $i>2$ $S_i$ are false. then there is an $i>2$, $S_i$ is true(from $S_2$), then contradicted. If $S_1$ is false, in the same manner, can lead to contradicted too.

The paradox is disturbing me, do anyone here post any comments?

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closed as not a real question by vadim123, Amzoti, Micah, Zev Chonoles, rschwieb Jun 13 '13 at 16:39

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Since when do paradoxes have to be self-referential? $\{x: x \notin x\}$. – nullUser Jun 13 '13 at 3:43
@nullUser: The Russell paradox is self-reference in hidden way, please rephrase completely it, you will find out. – logician Jun 13 '13 at 3:51
There is an obvious contradiction... It's like writing x=1 only and x=2 only... Clearly that's not possible... Same thing here. The problem comes from the fact that statement 1 refers to statement 3 and statement 2 refers to statement 3... And because of that regardless of what value of statement 1 is give. Statement 3 ends up with both values... – frogeyedpeas Jun 13 '13 at 5:26

Here is an alternative formulation:

There does not exist a set $S\subseteq{\Bbb N}$ such that for each $n\in\Bbb N$, $n\in S$ iff $\forall k>n\, k\notin S$.

First, we prove $S$ is nonempty. If $1\in S$, we are done; otherwise, $\exists k>1$ such that $k\in S$.

Given that $k\in S$, we know $\forall n>k, n\notin S$. But since $k+1\notin S$, this implies $\exists n>k+1$ such that $n\in S$, which is a contradiction.

In this form, the circularity is apparent: The statements are about conditions on $S$, which are quite nontrivial and obviously contradictory. Even if you treat the individual sentences (elements of $S$) separately, the structure of the implications forms a complete graph $K_{\Bbb N}$, and even though the implications "only go one way", we can easily turn them around via contrapositives to get (many) circular implication cycles.

One way to understand the implications is to note that if $i<j$, then $\forall n>i\ n\notin S$ implies $\forall n>j\ n\notin S$, so there is a forward implication $S_i\to S_j$ for all $i<j$. At the same time, $S_i$ also implies $\neg S_j$ for all $i<j$ by the virtue of its statement, so by contrapositive, $S_i\to S_j\to \neg S_i$ for any two integers $i<j$. Thus Yablo's paradox is riddled with circular paradoxes, and anyone who tries to convince you that it is not circular is pulling your chain.

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Good, it is informative comment. your sentence isn't predicative. – logician Jun 13 '13 at 6:08
@logician True, but I needn't take it as a definition; another approach is to define $T=\{S\mid S=\{n\in{\Bbb N}\mid S\subseteq [n]\}\}$, so that the above proves that $T=\emptyset$. – Mario Carneiro Jun 13 '13 at 6:28
+1 Thanks for that remark given to me. + – Babak S. Jun 13 '13 at 7:10

First, note the definition of $S$ is quite explicitly self-referential. But that's okay, since it's a recursive definition, and recursive definitions are fine if set up properly.

The problem is that this isn't a well-founded recursion: there is no base case. So we can't invoke the theorem that says that recursive definitions are well-defined.

So we can't even say that what you've written is a definition at all. A priori, all we can say is that it is an infinite set of conditions on a sequence $S$ of propositions, which may or may not have a solutions, and if it does, the solution may or may not be unique. The situation really isn't all that different from solving systems of "ordinary" equations involving real variables and arithmetic.

Your analysis shows the system of "equations" that $S$ must satisfy is inconsistent; therefore, the system has no solution for $S$. In particular, this proves that this system does not implicitly define a sequence $S$ of propositions.

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Short answer: Self-reference isn't the problem; reference is the problem.

Longer answer: Let's examine this statement S1. What does it mean?

S1 means "S2 is false and S3 is false and S4 is false and..."

OK, let's expand that out some more, using the definition of S2:

"(S3 is true or S4 is true or S5 is true or...) and S3 is false and S4 is false and..."

OK, we can then start expanding this out using the definition of S3:

"((S4 is false and S5 is false and...) or S4 is true or S5 is true...) and (S4 is true or S5 is true or....) and S4 is false and S5 is false..."

I think you can see, this expansion will never terminate. It is impossible to fully write down what S1 actually means.

In modern mathematics, we require that it be possible to write down a statement in full, without referring to other statements -- statements should have meanings by themselves. Technically speaking, statements are not allowed to refer to each other at all. Now of course, if I have statements like

T1: 1>0

T2: 2>1

T3: 1>0 and 2>1

Then I can sum up T3 as "T1 is true and T2 is true"; but "T1 is true and T2 is true" is just a human-readable summary; it's not the actual formal statement. Formal statements, as I said above, must stand on their own and are not allowed to refer to each other.

This is why Yablo's paradox is not a problem in modern mathematics.

(Note, by the way, that Russell's paradox does not use this sort of self-reference -- or reference of any sort. The liar paradox can't even be formulated in the modern framework of first-order logic; Russell's paradox can be, and just requires the appropriate axioms to appear. It does definitely have a sort of self-referential quality to it, but it's not the same thing.)

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-1. This is the worst sort of non-answer. It sums up centuries of philosophical investigation by saying it's not a real problem. – dfeuer Jun 13 '13 at 6:27
@Harry This is not true; there is no need to use infinitary logic to write down the statements precisely in a formal language. See my answer. – Mario Carneiro Jun 13 '13 at 6:44
I'm not sure I agree. You made it about a sets; sets are objects. Once you make it about a set, it's just "Yup, here are a bunch of conditions on a set, but they're contradictory, so no such set exists." In order for it to seem paradoxical, it has to be about not conditions but definitions. Then you can get rid of the paradox by pointing out why the defintions don't make sense (because they don't bottom out). – Harry Altman Jun 13 '13 at 10:53

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