Re: Is the Empty Set a member of itself?

On Mar 24, 6:30 pm, Chris Menzel <cmen...@xxxxxxxxxxxxxxxxxxxx> wrote:
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"?

"There are no elements of x satisfying that condition. "

This is a fallacy.

"an empty set may test to set that it does not exist."

Membership exists for empty state.

A nonexistence.

.

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