Re: Is the Empty Set a member of itself?

On Mar 25, 12:44 am, Chris Menzel <cmen...@xxxxxxxxxxxxxxxxxxxx>
wrote:
On Sat, 24 Mar 2007 18:31:33 -0700, Russell Easterly
<logic...@xxxxxxxxxxx> said:







"Chris Menzel" <cmen...@xxxxxxxxxxxxxxxxxxxx> wrote in message
news:slrnf0b9jm.1de3.cmenzel@xxxxxxxxxxxxxxxxxxxx
On Sat, 24 Mar 2007 14:53:17 -0700, Russell Easterly
<logic...@xxxxxxxxxxx> said:

Let x = {{}}
Let y be an element of x such that y =/= {}.

There are no elements of x satisfying that condition. You might as well
try to define y as "an even prime such that y =/= 2".

Is y an element of x?

You've not got yourself a value for y, mate. The question is
meaningless.

In ZFC, is there a difference between "doesn't exist" and the empty
set?

Er, well, yeah. For instance, in ZFC, it is trivial to prove that there
doesn't exist a y satisfying "y is an element of {{}} and y =/= {}".
Perhaps your question is whether, in ZFC, there is a difference between
"doesn't exist" and "is a member of the empty set"?

OK

Let x = {{}}
Let y be the set of all elements of x not equal to {}.

Ok, so y = {}.

Is the following statement true?

Az(zey -> zex)

Sure. Think of it this way. "Az(zey -> zex)" is equivalent to
"~Ez(zey & ~zex)". And surely that's true, since there is no z in y,
hence no z in y that fails to be in x.-

In other words:

x is the set containing {} and only {}.

y is the subset of x, the elements of which are not = {}.

Therefore, y = {} and y is a subset of x, as required.


Here is a formal version of this proof generated using my proof
software (30 lines):

http://www.dcproof.com/EmptySet.html

Dan
Download my DC Proof software at http://www.dcproof.com

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