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2011-01-27, 22:59   #331
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by Mr. P-1 You're almost there. If A is a 1-element set, then the only subsets of A are TEX]\empty[/TEX] and A itself. A cannot be an element of A. What could be an element of A?
anything but A or $\empty$ ? as long as A can be a collection or a set of sets

Last fiddled with by science_man_88 on 2011-01-27 at 23:00

2011-01-27, 23:07   #332
Mr. P-1

Jun 2003

7·167 Posts

Quote:
 Originally Posted by CRGreathouse The set S itself isn't a member of S -- ZF doesn't allow that, so that won't work.
I would argue that Z, rather than ZF is a closer formalisation of the informal set theory we've been doing here, because 1. nothing has been said which relies upon the axioms of replacement or regularity, and 2, we have admitted individual elements, which most treatments of ZF do not admit.

S$\in$S still doesn't work, even in Z, because while not forbidden, it is impossible to prove such a set exists.

2011-01-27, 23:10   #333
Mr. P-1

Jun 2003

116910 Posts

Quote:
 Originally Posted by science_man_88 anything but A or $\empty$ ? as long as A can be a collection or a set of sets
Stick with your idea of a 1-element set. A has two subsets: $\empty$ and A. A's sole element is a subset. The element is not A. What is the element?

Last fiddled with by Mr. P-1 on 2011-01-27 at 23:11

2011-01-27, 23:13   #334
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

838410 Posts

Quote:
 Originally Posted by Mr. P-1 Stick with your idea of a 1-element set. A has two subsets: $\empty$ and A. A's sole element is a subset. The element is not A. What is the element?
A[1] in PARI terms, last time I checked.

2011-01-27, 23:38   #335
Mr. P-1

Jun 2003

7×167 Posts

Quote:
 Originally Posted by science_man_88 A[1] in PARI terms, last time I checked.
A is a one-element set with two subsets: $\empty$ and A. One of the two subsets I have just named is an element of A. The element is not A. What is the element?

Last fiddled with by Mr. P-1 on 2011-01-27 at 23:38

2011-01-27, 23:45   #336
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by Mr. P-1 A is a one-element set with two subsets: $\empty$ and A. One of the two subsets I have just named is an element of A. The element is not A. What is the element?
$\empty$ is what you seem to imply.

2011-01-27, 23:52   #337
Mr. P-1

Jun 2003

7·167 Posts

Quote:
 Originally Posted by science_man_88 $\empty$ is what you seem to imply.
If $\empty$ is an element of a 1-element set. What is the set?

2011-01-27, 23:55   #338
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by Mr. P-1 If $\empty$ is an element of a 1-element set. What is the set?
$\empty$ by the sounds of it.

2011-01-28, 01:49   #339
CRGreathouse

Aug 2006

3×1,993 Posts

Quote:
 Originally Posted by Mr. P-1 I would argue that Z, rather than ZF is a closer formalisation of the informal set theory we've been doing here
Yep, agreed. But I didn't want to bring up issues of the foundation axiom and I wanted to help sm along the path so I wrote what was needed.

2011-01-28, 01:58   #340
CRGreathouse

Aug 2006

3·1,993 Posts

Quote:
 Originally Posted by science_man_88 $\empty$ by the sounds of it.
$\empty$'s not a member of that set, though; that has no elements.

2011-01-28, 12:48   #341
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

20C016 Posts

Quote:
 Originally Posted by science_man_88 A[1] in PARI terms, last time I checked.
this is the final thing I come up with if it's not that I have no idea because I obviously don't know the question.

Last fiddled with by science_man_88 on 2011-01-28 at 12:51

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