# Not closed under equality?

One of the Peano axioms state that

"For any $a \in \Bbb N : a = b, b \in \Bbb N$."

An example where transition is not closed under equality is the relation to friends. C may not be B's friend, but A may be B's friend, even though A isn't C's friend; that is : A = B =/= C.

Is there any case for any set (not limited to $\Bbb N$) where the set is not closed under equality, that is, it produces a member from another set? I would assume this to mean "open" (not in the topological definition), but correct me if I am wrong.

From Wikipedia: For all a and b, if a is a natural number and a = b, then b is also a natural number. That is, the natural numbers are closed under equality

• Not under the common definition of equality as finest equivalence relation. As a matter of fact, I am not used to seeing your axiom among the Peano axioms. Instead, equality $x=y$ is understood as $P(x)\leftrightarrow P(y)$ for arbitrary predicates $P$. – Hagen von Eitzen Aug 19 '13 at 20:33
• I'd be careful about using the word 'open' to mean 'not closed'. In maths they often mean different things. – Clive Newstead Aug 19 '13 at 20:40
• I don't recall this axiom. – Asaf Karagila Aug 19 '13 at 20:42
• @AsafKaragila see edit – MathApprentice Aug 19 '13 at 20:43
• You need to read that axiom in its historical context. It was before logic was developed to its current form. Today everything is closed under equality; but back in 1879 things were different. – Asaf Karagila Aug 19 '13 at 20:47