Re: Kuratowski Ordered Pair

On Dec 17, 6:01 pm, noel etters wrote:

Imagine that (a, b) means that a is in one slot and b is in
another
slot. Call it a slotted pair. We could call the slots S and S', but there
is no need for this to be explicit. The slot is created, as it were,
by 'a'
being in it. S and S' would merely be interchangeable labels. If we had
called them S1 and S2, there would be nothing 'firstish' about the slot
S1.
The important thing is that they are distinct slots. So we allow (a, a),
the same object or term or element or whatever being in both slots. If
however a not= b, then clearly (a, b) is going to be different from (b,
a). 'a' being in one slot and 'b' being in the other is different from 'b'
being in that 'first' slot and 'a' in the other. It is also pretty clear
that another slotted pair, (p, q), can only be equal to the first pair by
having the same elements in the same slots, that is by a=p (in slot S) and
b=q (in slot S'). So, everything is ready, I think, for a Kuratowski
definition of the slotted pair.

(a, b) = {{a}, {a, b}}

We can instead have:

(a, b) = {{b}, {a, b}}

And it should be noted that the expression : {{a}, {a, b}},
which equals (a, b) in the standard Kuratowski, is equal to (b, a) in the
reverse Kuratowski. This corresponds to the arbitrary, alternative
labellings: SS' and S'S.

Yes, the ordered pairing operation is not commutative. Just like we
could have defined 'x^y' to be y to the power x rather than the way we
defined it as x to the power y. The operation of exponentiation is not
commuatative and the operation of ordered pairing is also not
commutative.

I think it's clear that the slots are not ordered. This is even
more obvious if you consider n-tuples. All that's required is as many
distinct slots as there are elements (counting repetitions, of course). As
it were, S, S''', S'', S'''' etc. all in a jumble. We could only count (a,
b) as an ordered pair if we agreed in the first place on an order for S,
S'.
But this is not the situation for an ordered pair. In an ordered
pair, (a, b), 'a' is in the first slot; alternatively it is before 'b'.
There is no first and second, nor before and after with a slotted pair, or
Kuratowski set; there are only distinct slots.

You went through all that rigmarole just to say what can be stated as
simply as "the ordered paring operation is not commutative". Yes, it's
not commutative. Lots of operations are not commutative. That's not
reason to think they're not adequately defined.

MoeBlee
.