Russell’s Paradox and the Law of Excluded Middle

On 31 Jan 1997, Chris wrote:

> On 30 Jan 1997, Mark wrote:
>
>> Beware of a priori, non-controvertible claims. Once upon a time, each of
>> these was considered to be known to be true a priori, and uncontrovertibly; they
>> are so no longer.
>>
<snip>
>>         The Law of the Excluded Middle
>
> How would this not be a certain, necessary truth? Something is either
> A or non-A seems like a rather uncontestable claim to me.

>

In addition to the two cases Mark mentioned – systems of logic with more than two truth-values and fuzzy concepts that aren’t sufficiently crisp to decide for all cases that something either is or is not “A,” there’s another interesting issue.

Suppose that your term, A, is crisply defined. Still, to sensibly say that everything is either A or not-A, you need, at least implicitly, some kind of restriction to a domain or universe of discourse within which it applies. Whatever is not within that domain will be neither A nor not-A.

Why can’t you just say, “I mean the domain to cover everything ”? Because you have to impose some restrictions on what gets included to avoid falling into various sorts of paradoxes. (Are impossibilities part of “everything”?) The most famous is Russell’s Paradox which deals with sets of sets that do or do not include themselves. Briefly, he proved that if you allow a set of all sets that do not include themselves, you can prove both that if it does include itself, then it doesn’t, and if it doesn’t, then it does.

But to avoid Russell’s Paradox, you have to say that some things that are “sayable” don’t count as part of everything. So – back to the more restricted point – you always, whether explicitly or not, have to refer to a domain for “A or not-A” to have determinate sense. Then, anything outside that domain won’t count as either A or not-A.



Rob
_____
Rob Bass
rhbass@gmail.com





****

On 3 Feb 1997, Lance wrote:

> Rob wrote:
>
> No. “Not-A” is a denial of A. Anything which is not A is not-A.
>
>> Why can’t you just say, “I mean the domain to cover everything”?
>> Because you have to impose some restrictions on what gets included
>> to avoid falling into various sorts of paradoxes. (For example, are
>> impossibilities part of “everything”?)
>
> I don’t see how this clashes with the idea of A and not-A as exhaustive.
>
> If I claim that my car is either Blue or not-Blue, and you come along
> and say “is it a round circle?” I simply say: “No” (not-A).

I assume you meant to be talking about round squares (“squounds”) – I’d never use a round circle as an example of an impossibility!

Besides, once you substitute squounds for round circles, I still wouldn’t use it quite the way you suggest. The question I’d try to get you to answer is: “Is a squound blue or not-blue?” Or to make things simpler: “Is a squound blue?”

I suppose you’ll say No – but it doesn’t really matter. The same basic argument goes through whether you say Yes or No, that is, whether you say it’s true or false.

What does “squounds are not blue” mean?

Interpretation I: There’s a kind of thing called a squound, and none of them are blue. I assume that’s not what you want, because it leaves you admitting that round squares exist.

So, Interpretation II: For anything, if it is a squound, then it’s not blue. You see, that one doesn’t commit you to saying that there are any squounds.

So far, so good? Now, remember, “squound” is defined as something that has contradictory properties. If you plug Interpretation II into a standard logic with the contradiction in “squound” symbolized somehow, it turns out to be true, provably true. (Hurray for your side! – Better hold the cheering ….)

Is there a problem? Well, one little detail. If you plug in an analogue of Interpretation II for “squounds are blue” – then that turns out to be provably true, also!

So, by insisting that the Law of Excluded Middle ranges over everything without restrictions, you end up being able to prove “squounds are blue and squounds are not blue”! In order to save Excluded Middle, you have to give up Non-Contradiction!^[§]

None of this means, of course, that you can’t have bivalence and Excluded Middle relative to a domain.

This has been somewhat long, but I’m going to make it even longer. I’m going to take off from your example again, not even mention round squares, and try to show you how easy it is to get pushed into Russell’s Paradox.

So back to your car. I shall suppose that it’s blue and that “blue” is crisply defined in the sense that, for anything, it either definitely is, or definitely is not, blue – no messy, border-line cases to worry about.

Your car, then, is a member of the class or set of blue things. Is the set of blue things also blue? I’m not sure what you’d say about this, but my guess is that you’d say “No, the set of blue things is not a blue thing.” (Compare: the set of coffee cups is not a coffee cup; the set of keyboards is not a keyboard, etc.)

If that’s the answer you’d give (and it doesn’t really matter – I can run a parallel argument if you give the other answer), then it looks like the set of blue things doesn’t include itself as a member. Lots of sets are like that – sets of coffee cups, keyboards, bald people, Canadians and on and on. Since all these sets have something in common, namely, not including themselves, it looks like we can talk about a set of sets that don’t include themselves.

Is there any other kind of set? Well, it sure looks like it. What about the set of things that are not blue? That’s not a blue thing. Or the set of things that aren’t coffee cups? That’s not a coffee cup.

So, we’ve also got a set of sets that do include themselves. Let’s label the ones that don’t include themselves, “ordinary sets” and the ones that do include themselves, “extraordinary sets.”

Now, you’re ready for the Russell question:

Is the set of all ordinary sets itself ordinary or extraordinary?

Suppose it’s ordinary. Then, it’s included in itself after all – so it’s extraordinary. On the other hand, suppose it’s extraordinary. Then, it’s included in itself – which means it’s ordinary!

That is Russell’s Paradox.

Unless you draw lines somewhere – that is, unless you restrict your domain in some way – you can get pushed into saying that there are some sets which both are and are not members of themselves – and it can be done in ways that seem perfectly straightforward.

If you don’t want to end up there, you should rethink the mistaken commitment to the unrestricted application of the Law of Excluded Middle.



Rob
_____
Rob Bass
rhbass@gmail.com

[§] I slipped here. That’s not really a contradiction, even though it sounds like it. (I pointed it out in later correspondence.) “Squounds are blue” and “squounds are not blue” (on the current interpretation) are consistent with one another, provided there aren’t any squounds. Or, in general, (x) (Fx  Bx) and (x) (Fx  ~Bx) are consistent with one another but together imply ~(∃x) (Fx).