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In the Tractatus, Wittgenstein attempts a solution of Russells paradox

3.333 A function cannot be its own argument, because the functional sign already contains the prototype of its own argument and it cannot contain itself.

If, for example, we suppose that the function F(fx) could be its own argument, then there would be a proposition “F(F(fx))”, and in this the outer function F and the inner function F must have different meanings;

for the inner has the form g(fx), the outer the form h(g(fx)).

Common to both functions is only the letter “F”, which by itself signifies nothing. This is at once clear, if instead of “F(F(u))” we write “There exists g : F(gu). gu = Fu”.

Herewith Russell’s paradox vanishes.

Does this work, or be made to work? (Assuming of course that this the same paradox that we now know by the name of Russell).

some defintions from Russells introduction:

From Russells introduction:

a. A propositional function is a function whose values are propositions; for example 'x is human'.

b. A truth-function of a proposition p is a proposition containing p and such that its truth or falsehood depends only upon the truth or falsehood of p.

c. Wittgenstein shows every propositional function is a truth-function.

  • Neat question. I haven't seen this version of Russell's paradox before. Of the cuff, I'm fairly this Wittgenstein's paradox ends up being equivalent to Russell's. W's solution of stipulating the axiom "a function cannot be its own argument" is also highly reminiscent to how Russell initially tackled it by postulating that certain self-referential classes like {x|x∉x} are not sets. – David H Nov 27 '13 at 17:18
  • Wittgenstein's assertion that "function cannot be its own argument" needs precise clarification. There are many contexts where functions are routinely used as their own arguments, ranging from Universal Turing Machine in Computer Science to self-learning algorithms in Artificial Intelligence. Therefore the meaning of the word "function" needs to be made very precise and the context where it cannot apply to itself very explicit with the explanation why the recursion is forbidden. Without such clarification the argument does not stand. – Michael Nov 27 '13 at 18:41
  • @Michael: Wittgenstein was probably not thinking of the contexts that yo've suggested. The problem is as you've pointed out what does he mean by function. – Mozibur Ullah Nov 28 '13 at 2:31
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I think that for an understanding Wittgenstein's "solution", you must take in account the original (historical) context.

Wittgenstein was a student and disciple of Frege and Russell.

According to Frege (see his immediate response to Russell's letter communicating to him the discovery of the Paradox), Russell's form of the Paradox ( f(f) ) was not reproducible in Frege's logic, because syntax rules prescribe that the name of a function has an empty place that can be fileld only by the name of an object, and functions (in Frege's logic) are not objects. Immediately after, Frege was able to reproduce the Paradox in his own logic, with a more "complicated" formulation.

According to Whitehead & Russell Principia Mathematica, the solution of the paradoxes (included Russell's one) was Ramified Type Theory. Thos theory was from the beginning subjected to several objections, included a difficulty connected with the formal exposition of the theory itself in PM (PM does not has a neat separation between object language and meta-language, and so no precise demarcation between object- and meta-theory).

One of Tractatus' proposed improvements upon Frege's and Russel's logics is the idea that logical form cannot be described: it can only be showed. A plausible reading of this idea his that syntax rules are already enclosed into the signs themselves : the possibility of hanging together different signs by itself prevents paradoxical constructions (like f(f) ).

  • Is this letter of Frege to Russell online somewhere? – Mozibur Ullah Nov 28 '13 at 2:33
  • Online ... I don't know. In Jean van Heijenoort volume on tthe sources of Math Log [Form Frege to Godel, Harvard UP - 1967] you can find the first reply of Frege to Russel (June 22, 1902) [pag.126]. In the following weeks Frege added an Appendix to his 2nd vol of Grudgesetze (still in print) with the reproduction oof Russel's Paradox in his hown system and a proposed solution (that didn't work : see W.V.O.Quine, On Frege's Way Out - 1954). – Mauro ALLEGRANZA Nov 28 '13 at 11:23
  • You can try also with : Gregory Landini. Russell to Frege, 24 May 1903 (Russell: the Journal of the Bertrand Russel Archives, n°12 - winter 1992-93); the first part of Gregory Landini, The Ins and Outs of Frege’s Way Out (Philosophia Mathematica (III) 14 (2006))and Kevin Klement, Putting Form Before Function : Logical Grammar in Frege, Russell, and Wittgenstein (2004) (I have found it in Klement's website). – Mauro ALLEGRANZA Nov 28 '13 at 11:47
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Wittgenstein is alluding to how Russell himself solved the Paradox - the theory of ramified types. He alludes to this in:

3.332 No proposition can say anything about itself, because the propositional sign cannot be contained in itself (that is the “whole theory of types”).

And he reformulates as

3.333 A function cannot be its own argument, because the functional sign already contains the prototype of its own argument and it cannot contain itself.

A functional sign is simply the sign of the function; the function being what the sign signifies. He expands what he means by this:

If, for example, we suppose that the function F(fx) could be its own argument, then there would be a proposition “F(F(fx))”, and in this the outer function F and the inner function F must have different meanings;

for the inner has the form g(fx), the outer the form h(g(fx)).

That is F(F(fx)) is different from F(F(fx)) because in the expression they signify different things, that is they have different meanings or precisely functions; and only the sign 'F' is common to both, as he affirms:

Common to both functions is only the letter “F”, which by itself signifies nothing.

and by

This is at once clear, if instead of “F(F(u))” we write “There exists g : F(gu). gu = Fu”.

Herewith Russell’s paradox vanishes.

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