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Tuesday, June 24, 2014


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Two Blunders:

I suppose that Armstrong means here that the "empty set" belongs to every "set" and not class. The mistake here is to conflate classes with sets. While all sets are classes, the converse is not the case. Since a class is any collection of objects, whereas a set is not, conflating sets with classes will lead directly to Russell's paradox. Thus, if all classes were sets, then the set (collection) of all sets that are not members of themselves would be a set. But, of course, this will lead to the conclusion that this collection: i.e., set, is both a member of itself and it is not a member of itself. Therefore, not all classes are sets and Russell's class is not a set.

Second, it makes no sense to say, as Armstrong says in the above quotation, that logicians "make" the empty set a member of every set. I am not even sure what could such a phrase mean. It is like saying that mathematicians "make" the *number* two even; as if they could have alternatively "make" the *number* two odd.

The empty class is not a member of every class; it is a subset of every class.

I should have elaborated on my previous comment. The empty class is not a member of every class; it is a subset of every class. This is easily demonstrated. The set of all apples does not have the empty class as a member. Its only members are apples. But just as the set of Granny Smith apples is a subset of the set of all apples, so too is the empty set a subset of the set of all apples. Set membership is not the same relation as the subset relation. The latter is, for example, transitive; the former is not.

"Armstrong makes a bad mistake in that footnote. He writes, "Wade Martin has reminded me about the empty class which logicians make a member of every class." Explain the mistake in the ComBox."

I am even afraid of writing this, as the mistake is elementary enough that I suspect that the *real* one is located elsewhere (I do not have Armstrong's book by me, so cannot check).

If the empty set, denoted by 0, were a member of every set, it would also be a member of itself, and therefore 0 would have at least one member and be non-empty. Contradiction. Ergo 0 cannot be a member of every class.

The empty set is not a member of every set (for every x, 0 e x) but rather is contained in every set (for every x, 0 <= x). The membership and containment relations are very different. For one, the latter is a partial order, therefore it is transitive, but the membership relation is not transitive.


Exactly right.

G. R.,

You are quite right as well. And I like your reductio.


The context is very informal, so no distinction is being made between classes and sets. Both of your points are good, even if the second is a bit pedantic, but I think you missed the really glaring mistake pointed out by John and G. R.

Of course, Armstrong is well aware or perhaps I should say was well aware -- he died recently -- of the distinction between the membership relation and the subset relation. He was just having a 'senior moment.'

Bill, I enjoy reading your blog. I've felt like making a comment or two to some of your posts however very few of them have a comment box. Is that intentional? As you asked for comments on one post, and there was no available place for remarks, that seems unlikely. I'm putting this message in the most recent post I can find that accepts comments. Im using Chrome but have tried IE.


Yes, it is intentional. You can shoot me an e-mail if you like. Thanks for your interest.

Hmmm...obtuse man looks for an email link but finds none.

Go to right sidebar. Click on 'About' near the top.

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