# Halmos Naive Set Theory - Set Inclusion Antisymmetric

In Chapter 1 of Halmos's Naive Set Theory, he mentions that

If $$A$$ and $$B$$ are sets such that $$A \subset B$$ and $$B \subset A$$, then A and B have the same elements and therefore, by the axiom of extension, A = B. This fact is described by saying that set inclusion is antisymmetric.

Now the first sentence makes perfect sense. Indeed, it is the standard method for proving that two sets are equal. I am consused by the second sentence, which seems to say that the first sentence implies that set inclusion is antisymmetric. My understanding is that a relation $$R$$ is antisymmetric iff for any $$a$$ and $$b$$ in the set on which R is defined, if $$aRb$$ then it is not true that $$bRa$$. But clearly we can have $$A\subset B$$ and $$B\subset A$$, so how is the inclusion relation antisymmetric? Indeed, if we had've defined the $$\subset$$ symbol as denoting proper inclusion, I would agree with the assertion. But we did not.

Just to be clear, Halmos is using the $$\subset$$ symbol in the standard way, and not to denote a proper subset (which he mentions but does not give a different symbol for). In particular, he says earlier that

If A and B are sets and if every element of A is an element of B, we say that A is a subset of B, or B includes A, and we write $$A \subset B$$ or $$B \supset A$$.

Obviously, Halmos was a superb mathematician, so I am wondering where I am missing his point here?

"Antisymmetric" doesn't mean what you think. A relation $$R$$ is antisymmetric if $$aRb$$ and $$bRa$$ implies that $$a = b$$. (If $$aRb$$ implies that $$bRa$$ cannot hold, the relation is called "irreflexive", which, confusingly is much stronger than "not reflexive".)
• Ah, I see I read a little bit too quickly through Wikipedia's definition. I missed the crucial $a \neq b$ hypothesis there in the definition of antisymmetry for a relation. Thanks very much for the info! – 1729_SR Sep 15 '20 at 14:35