Friday, 5 June 2015

Bertrand Russell’s Paradox


What is the primary feature of Bertrand Russell’s Paradox? It's one of self-reference. Thus it's like the Liar Paradox in which a sentence (or speaker) speaks about itself (or himself). With Russell’s Paradox, it's a case of a set (or a class) referring to itself or including itself as a member of itself.

What about sets that contain themselves as members? Consider this set:

Set X = the set of all the sets I've written about today

Today I've written about the set of all the things owned by Paul Murphy, the set of all writing implements, the set of arseheads, among others. All these are subsets of the set of all the sets I have written about today. But, in that case, I've also written about the set of all I've written about today – today. Thus this set will be a set (or a subset) of itself. That is, the set of the sets I've written about today will be a member of the set of all the sets I've written about today.

Let’s call my previous sets S, H, B and L. We also have Set X – the set of all the sets I've written about today. Now take another set:

Set Z = the set of all sets that don’t contain themselves as elements

Well, Set S, the set of all things that are owned by Paul Murphy, isn't itself owned by Paul Murphy - so it's not a member of itself. Neither are sets H, B and L. Set X, the set of all sets I've written about today, is a member of itself because it too is a set I've written about today.

What about Set Z above: the set of all sets that don’t contain themselves as elements/members? The question is:

Is Set Z an element of itself?

If Set Z isn't an element of itself, then it's a member of the set of all sets that aren't members of themselves. If it is an element of itself, then it can’t be a member of itself. Though if it isn’t a set that doesn’t contain itself as a member, then it is indeed a member of the set of all sets that don’t contain themselves as elements (of itself).

That means that if it is, it isn’t. And if it isn’t, it is. Or, alternatively, Set Z above both is and is not a member of itself. This is a contradiction.

This is the same problem that we have with the Liar Paradox. If everything the Liar says is false, then what he's saying must be false. He's saying that what he's saying is false. Therefore what he is saying must be true. Though if what the liar is saying is true, then it must be false. (Because everything he says is false.) Thus what he says is both true and false. This is, again, a contradiction.

No comments:

Post a Comment