Re: Prove = is an equivalence relation for sets
- From: MoeBlee <jazzmobe@xxxxxxxxxxx>
- Date: Sun, 5 Oct 2008 15:30:17 -0700 (PDT)
On Oct 4, 2:11 am, hagman <goo...@xxxxxxxxxxxxx> wrote:
You biggest obstacle is indeed one's intuitively knowing what equality
means.
That is not an obstacle for the particular problem mentioned.
We are given the axiom of extensionality and are asked to prove that
equality is an equivalence. I.e., to prove:
Axyz(x=x & (x=y->y=x) & ((x=y & y=z)->x=z))
And that is easily done.
Of course, in ordinary set theory, equality is not an equivalence
relation since there is no set {p | Ex p=<xx>}, but equality ON any
given set, i.e., {<x x> | x in S}, is a set, a relation, and an
equivalence relation.
MoeBlee
.
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- Prove = is an equivalence relation for sets
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- From: hagman
- Prove = is an equivalence relation for sets
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