February 27th, 2013, 02:51 AM  #1 
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7  Some trivial proofs about sets.
Hi! Here there are some theorems I have to prove. I know that they are trivial but since they are very important, I want to toggle all doubt on these fondamental things. So I will post here my development, and I would want to receive correction and suggestions from you. Thanks. Prove : If , then . 1. My development: So we have to prove the following: means that: let A, ? (empty set) are two sets, A is subset of ?, if and only if, for each element x, if x is in A, then x is in ?. But by definition, the empty set is the set containing no elements. So we have that the proposition: x is in A AND x is in ? is false. To obtain a true proposition, the set A it must have no elements. hence: (x is not in A AND x is not in ?) implies (x is not in A AND x is not in ?) AND (x is not in A AND x is not in ?) implies (x is not in ? AND x is not in A) Hence: A = ?. 2. Proof by contradiction: assume the starting theorem as false: . or better: . We have: A set called A containing an element x, and not subset of ?, it is subset of ?. And this contradicts the definition of empty set. Similarly an empty set ? with no elements it can't be subset of an nonempty set A. So the proposition I have formulated is false and the starting theorem is true. 3. Proof by contradiction: assume the starting theorem as false: . or better: . in order that A it can be subset of ?, A it must have no elements so we have: (x is not in A AND x is not in ?) implies (A is not subset of ? AND V is not subset of A) Hence, briefly we are saying that a set called A, with no elements, it is different from the empty set ?. it means that ? has some elements within. But this fact contradicts the definition of empty set ?. So the proposition I have done is false and the starting theorem is true. Another prove: If A = B, then and My development: So we have to prove that: If we consider A={x}=B so we have that the element x it is both in A and in B and viceversa. Hence: from the definition: "A it is subset of B, if and only if, for each element x, if x is in A, then x is in B". B contains only that one element x. It is valid the viceversa. I think that later I will put here the prove by contradiction. in the meantime I would want to see if my developments are right. Waiting ardently for your suggestions. Many thanks really!! 
March 1st, 2013, 07:32 AM  #2  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Some trivial proofs about sets. Quote:
Quote:
Quote:
 
March 2nd, 2013, 02:40 AM  #3  
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7  Re: Some trivial proofs about sets. Quote:
 
March 3rd, 2013, 07:03 AM  #4 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Some trivial proofs about sets.
The negation of the statement "if A then B" is NOT "If not A then B", it is "A and not B". For example, the negation of the statement "If it rains today, I will take my umbrella" is NOT "if it does not rain today I will take my umbrella". It is "It rains today and I do not take my umbrella". 
March 3rd, 2013, 10:28 AM  #5 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Re: Some trivial proofs about sets.
Recall that A>B = A v B. So the negations of A>B, ie (A>B) = (AvB), which in turn is equivalent to A&B. Perhaps more intuitively, the negation of a conditional is true if and only if the conditional is false. A conditional P>Q is only false if the antecedent P is true and the consequence is false. The consequence is false if and only if the negation of the consequence is true. So P>Q is false, therefore its negation true, iff P&Q is true. 

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