20110201, 19:57  #364 
Jun 2003
7×167 Posts 
Not really, though I've had to spend quite a while thinking about what I do "want", why what you've written doesn't satisfy, and how to explain it to you. Plus I've been distracted...
Basically mathematicians try to express their ideas using a very limited vocabulary. The reason for this is so they can define very precisely how these terms interact, and how one can derive new statements (theorems) from previously proved theorems or unprovenbutassumed statements (axioms). One consequence of this limited vocabulary as it applies to set theory is that it isn't necessary or even really possible to talk about "repeats". Given a set S and an object a, the only thing you can say about the latter's membership of the former is that either aS or aS. It isn't possible to even discuss how many times aS. The vocabulary doesn't permit it. So what I "want" I guess, is for you to define these terms using the following vocabulary only: Variables: a, b, c, X, Y, Z, etc. Quantifiers: all, at least one, exactly one, no (as in no set) Logical operators: and, or, not, neither ... nor, if ... then, iff (short for "if and only if".) Set membership: member of Punctuation: brackets and commas. Other terms already defined using these terms. Unambiguous synonyms, for example: "each", "every", instead of "all"; "implies" instead of "if ... then", etc. Any additional words needed to make your sentences into grammatical English, so long as they don't carry substantive meaning. For example. I can define "subset of" using the above by saying that "A is a subset of B" means the following: For every x in A, x is in B. Here "For" and "is" don't carry substantive meaning. (You could omit them and still understand the statement), "every" means "all"; "in" means "member of". Another example: Define : For every x, x is not in . Binary union: AB means: For every x, x is in AB iff (x is in A or x is in B) Union of a collection: C: For every x, x is in C iff x is in at least one X in C. Do you think you could define binary intersection and intersection of a collection like this? 
20110202, 00:13  #365  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
Last fiddled with by science_man_88 on 20110202 at 00:26 

20110202, 17:56  #366  
Aug 2006
5979_{10} Posts 
Yes.
Yes. Quote:


20110204, 12:01  #367  
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
Quote:
their's(pasted) > Quote:


20110204, 12:15  #368 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
10939_{10} Posts 
That's to be commended!
If your statements are both correct and "booklike" then others will be able to understand you relatively easily. If they are wrong and booklike, it will be much easier to work out where you are going wrong and to provide corrections. Paul 
20110204, 12:55  #369 
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 

20110204, 12:58  #370  
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
Quote:
technically I think I could get that second one to align with: For all x, x is in iff(x is a one element subset of all X in C). 

20110204, 13:48  #371 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

20110204, 13:51  #372 
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 

20110204, 13:54  #373 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
oh doh I forgot that A!=B it's important because a set is always equal to itself and so no collection could be called pairwise disjoint under my definition, thanks for catching that.

20110204, 14:18  #374  
Aug 2006
3·1,993 Posts 
Quote:
There are sets which are "smdisjoint", that is, collections C for which all a,b in C are such that a β© b = {}. Can you find them? 

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