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 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 $A \subset \emptyset$, then $A= \emptyset$. 1. My development: So we have to prove the following: $A \subset \emptyset \Rightarrow $$A \subset \emptyset \wedge \emptyset \subset A$$$ $A \subset \emptyset$ 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: $A \not\subset \emptyset \Rightarrow A= \emptyset$. or better: $A \not\subset \emptyset \Rightarrow (A \subset \emptyset \wedge \emptyset \subset A)$. 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 non-empty 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: $A \subset \emptyset \Rightarrow A \ne \emptyset$. or better: $A \subset \emptyset \Rightarrow (A \not\subset \emptyset \wedge \emptyset \not\subset A)$. 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 $A \subset B$ and $B \subset A$ My development: So we have to prove that: $A= B \Rightarrow $$A \subset B \wedge B \subset A$$$ 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

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Posts: 2,409
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Re: Some trivial proofs about sets.

Quote:
 Originally Posted by beesee 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 $A \subset \emptyset$, then $A= \emptyset$. 1. My development: So we have to prove the following: $A \subset \emptyset \Rightarrow $$A \subset \emptyset \wedge \emptyset \subset A$$$ $A \subset \emptyset$ 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 = ?.
That's rather "wordy"! If $x\in A$, then, since $A\subset\emptyset$, $x\in \emptyset$, a contradiction.

Quote:
 2. Proof by contradiction: assume the starting theorem as false: $A \not\subset \emptyset \Rightarrow A= \emptyset$. or better: $A \not\subset \emptyset \Rightarrow (A \subset \emptyset \wedge \emptyset \subset A)$.
No! A contradiction would be $A\subset \emptyset$ and $A\ne \emptyset$

Quote:
 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 non-empty 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: $A \subset \emptyset \Rightarrow A \ne \emptyset$. or better: $A \subset \emptyset \Rightarrow (A \not\subset \emptyset \wedge \emptyset \not\subset A)$.[quote:tzq7yvvx] Again, that is NOT the negation of the original statement. [quote:tzq7yvvx]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 $A \subset B$ and $B \subset A$ My development: So we have to prove that: $A= B \Rightarrow $$A \subset B \wedge B \subset A$$$ 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!!
[/quote:tzq7yvvx][/quote:tzq7yvvx]

March 2nd, 2013, 02:40 AM   #3
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Joined: Jan 2013
From: Italy

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Re: Some trivial proofs about sets.

Quote:
 Originally Posted by HallsofIvy Again, that is NOT the negation of the original statement.
thanks, please, can you explain me a little bit better?

 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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