TITLE: Yablo's paradox? a paradox without self-reference
QUESTION [7 upvotes]: 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?

REPLY [4 votes]: 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.