display all the ideas for this combination of philosophers
11 ideas
| 7557 | To solve Zeno's paradox, reject the axiom that the whole has more terms than the parts [Russell] |
| Full Idea: Presumably Zeno appealed to the axiom that the whole has more terms than the parts; so if Achilles were to overtake the tortoise, he would have been in more places than the tortoise, which he can't be; but the conclusion is absurd, so reject the axiom. | |
| From: Bertrand Russell (Mathematics and the Metaphysicians [1901], p.89) | |
| A reaction: The point is that the axiom is normally acceptable (a statue contains more particles than the arm of the statue), but it breaks down when discussing infinity (Idea 7556). Modern theories of infinity are needed to solve Zeno's Paradoxes. |
| 14149 | The Achilles Paradox concerns the one-one correlation of infinite classes [Russell] |
| Full Idea: When the Achilles Paradox is translated into arithmetical language, it is seen to be concerned with the one-one correlation of two infinite classes. | |
| From: Bertrand Russell (The Principles of Mathematics [1903], §321) | |
| A reaction: Dedekind's view of infinity (Idea 9826) shows why this results in a horrible tangle. |
| 21585 | The tortoise won't win, because infinite instants don't compose an infinitely long time [Russell] |
| Full Idea: The idea that an infinite number of instants make up an infinitely long time is not true, and therefore the conclusion that Achilles will never overtake the tortoise does not follow. | |
| From: Bertrand Russell (Our Knowledge of the External World [1914], 6) | |
| A reaction: Aristotle spotted this, but didn't express it as clearly as Russell. |
| 21565 | Richard's puzzle uses the notion of 'definition' - but that cannot be defined [Russell] |
| Full Idea: In Richard's puzzle, we use the notion of 'definition', and this, oddly enough, is not definable, and is indeed not a definite notion at all. | |
| From: Bertrand Russell (On 'Insolubilia' and their solution [1906], p.209) | |
| A reaction: The background for this claim is his type theory, which renders certain forms of circular reference meaningless. |
| 15895 | Russell discovered the paradox suggested by Burali-Forti's work [Russell, by Lavine] |
| Full Idea: Burali-Forti didn't discover any paradoxes, though his work suggested a paradox to Russell. | |
| From: report of Bertrand Russell (The Principles of Mathematics [1903]) by Shaughan Lavine - Understanding the Infinite I |
| 13365 | Russell's Paradox is a stripped-down version of Cantor's Paradox [Priest,G on Russell] |
| Full Idea: Russell's Paradox is a stripped-down version of Cantor's Paradox. | |
| From: comment on Bertrand Russell (Letters to Frege [1902]) by Graham Priest - The Structure of Paradoxes of Self-Reference §2 |
| 10711 | Russell's paradox means we cannot assume that every property is collectivizing [Potter on Russell] |
| Full Idea: Russell's paradox showed that we cannot consistently assume what is sometimes called the 'naïve comprehension principle', namely that every property is collectivizing. | |
| From: comment on Bertrand Russell (Letters to Frege [1902]) by Michael Potter - Set Theory and Its Philosophy 03.6 |
| 6407 | The class of classes which lack self-membership leads to a contradiction [Russell, by Grayling] |
| Full Idea: The class of teaspoons isn't a teaspoon, so isn't a member of itself; but the class of non-teaspoons is a member of itself. The class of all classes which are not members of themselves is a member of itself if it isn't a member of itself! Paradox. | |
| From: report of Bertrand Russell (Mathematical logic and theory of types [1908]) by A.C. Grayling - Russell Ch.2 | |
| A reaction: A very compressed version of Russell's famous paradox, often known as the 'barber' paradox. Russell developed his Theory of Types in an attempt to counter the paradox. Frege's response was to despair of his own theory. |
| 21564 | Vicious Circle: what involves ALL must not be one of those ALL [Russell] |
| Full Idea: The 'vicious-circle principle' says 'whatever involves an apparent variable must not be among the possible values of that variable', or (less exactly) 'whatever involves ALL must not be one of ALL which it involves. | |
| From: Bertrand Russell (On 'Insolubilia' and their solution [1906], p.204) | |
| A reaction: He offers this as a parallel to his 'no classes' principle. That referred to classes, but this refers to propositions, and specifically the Liar Paradox (which he calls the 'Epimenedes'). |
| 21567 | 'All judgements made by Epimenedes are true' needs the judgements to be of the same type [Russell] |
| Full Idea: Such a proposition as 'all the judgements made by Epimenedes are true' will only be prima facie capable of truth if all his judgements are of the same order. | |
| From: Bertrand Russell (The Theory of Logical Types [1910], p.227) | |
| A reaction: This is an attempt to use his theory of types to solve the Liar. Tarski's invocation of a meta-language is clearly in the same territory. |
| 16475 | A 'heterological' predicate can't be predicated of itself; so is 'heterological' heterological? Yes=no! [Russell] |
| Full Idea: A predicate is 'heterological' when it cannot be predicated of itself; thus 'long' is heterological because it is not a long word, but 'short' is homological. So is 'heterological' heterological? Either answer leads to a contradiction. | |
| From: Bertrand Russell (An Inquiry into Meaning and Truth [1940], 5) | |
| A reaction: [Grelling's Paradox] Yes: 'heterological' is heterological because it isn't heterological; No: it isn't, because it is. Russell says we therefore need a hierarchy of languages (types), and the word 'word' is outside the system. |