What
about sets that contain themselves as members? Consider this set:
Set
X = the set of all the sets I've written about today
Today
I've written about the set of all the things owned by Paul Murphy,
the set of all writing implements, the set of arseheads,
among others. All these are subsets of the set of all the sets I
have written about today. But, in that case, I've also written
about the set of all I've written about today – today. Thus
this set will be a set (or a subset) of itself. That is, the set of
the sets I've written about today will be a member of the set
of all the sets I've written about today.
Let’s
call my previous sets S, H, B and L. We also have Set X – the set
of all the sets I've written about today. Now take another set:
Set
Z = the set of all sets that don’t contain themselves as
elements
Well,
Set S, the set of all things that are owned by Paul Murphy,
isn't itself owned by Paul Murphy - so it's not a member of itself.
Neither are sets H, B and L. Set X, the
set of all sets I've written about today, is a member of itself
because it too is a set I've written about today.
What
about Set Z above: the set of all sets that don’t contain
themselves as elements/members? The question is:
Is
Set Z an element of itself?
If
Set Z isn't an element of itself, then it's a member of the
set of all sets that aren't members of themselves. If it is an
element of itself, then it can’t be a member of itself. Though if
it isn’t a set that doesn’t contain itself as a member, then it is
indeed a member of the set of all sets that don’t contain
themselves as elements (of itself).
That
means that if it is, it isn’t. And if it isn’t, it is. Or,
alternatively, Set Z above both is and is not a member of
itself. This is a contradiction.
This
is the same problem that we have with the Liar Paradox. If everything
the Liar says is false, then what he's saying must be false. He's saying that what he's saying is false. Therefore what he is
saying must be true. Though if what the liar is saying is true, then
it must be false. (Because everything he says is false.) Thus what he
says is both true and false. This is, again, a contradiction.
No comments:
Post a Comment