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February 27th, 2013, 02:51 AM   #1
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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 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:
.
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!!
beesee is offline  
 
March 1st, 2013, 07:32 AM   #2
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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 , 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 = ?.
That's rather "wordy"! If , then, since , , a contradiction.

Quote:
2.
Proof by contradiction:
assume the starting theorem as false:
.
or better:
.
No! A contradiction would be and

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:
.
or better:
.[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 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!!
[/quote:tzq7yvvx][/quote:tzq7yvvx]
HallsofIvy is offline  
March 2nd, 2013, 02:40 AM   #3
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Joined: Jan 2013
From: Italy

Posts: 154
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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?
beesee is offline  
March 3rd, 2013, 07:03 AM   #4
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Posts: 2,409
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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".
HallsofIvy is offline  
March 3rd, 2013, 10:28 AM   #5
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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.
johnr is offline  
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