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Empty set?by pivoxa15
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#1
Dec1006, 06:31 AM

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In my book, it says
"We agree to regard the empty set as a subset of every set. Thus any nonempty set S has just two improper subsets, the empty set and the set S itself; all other subsets of S are proper." Does this sound right? I thought the improper subset is the set which is the same as the original set. Why is the empty set also an improper subset? Later in the book it said the empty set is a subset of every set. So the empty set is a subset of the universal set. Than it said the complement of the universal set is the empty set. From this information it implies the complement of the universal set is a subset of the universal set. Which is a contradiction? 


#2
Dec1006, 06:48 AM

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It is called improper because it is useful to do so. You are looking for too much meaning in the adjective.
I don't know how you deduced that contradiction. It is perfectly possible for a set E to be a subset X and X^c. As long as E is the empty set anyway. Now do you see why it is important that we don't include the empty set as a proper subset? The empty set is different from other sets. In particular the statement If x is in the empty set then ANYTHING AT ALL is always true, since 'x in the empty set' is always false. This is one reason why we might choose to exclude it from the 'proper subsets'. 


#3
Dec1006, 09:39 AM

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Textbooks, unfortunately, vary on this. Some texts use the word "proper" subset to mean a set that is strictly contained in another: A is a proper subset of B if there exist a point x in B that is not in A. Using that definition, the empty set is a "proper" subset of any set (except, of course, the empty set). Other textbooks specifically define "proper" subset to mean a nonempty subset having that property.
Textbooks that use the first definition often have to use the phrase "nontrivial proper subset" where texts using the second can just say "proper subset"! 


#4
Dec1006, 09:49 AM

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Empty set?
if a set is a subset of any set and its complement then it is the empty set 


#5
Dec1006, 02:13 PM

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Just practicing...
P subset Q and P subset ~Q > for all x ((x elem P > x elem Q) & (x elem P > x elem ~Q)) > for all x (x elem P > (x elem Q & x elem ~Q)) > for all x (x elem P > (x elem Q & x notelem Q)) > for all x (x elem P > bottom) > for all x (x notelem P) > P = 0 > for all x ((x elem 0 > x elem U) & (x elem 0 > x elem ~U)) > for all x ((x elem 0 > x elem U) & (x elem 0 > x notelem U)) > for all x (x elem 0 > (x elem U & x notelem U)) > for all x (x elem 0 > bottom) > for all x (x notelem 0) > true (by definition of 0) 


#6
Dec1006, 03:01 PM

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It was late at night and I made a mistake in my opening post about the deduction. I have changed it now.
Here is the edited version. 'Later in the book it said the empty set is a subset of every set. So the empty set is a subset of the universal set. Than it said the complement of the universal set is the empty set. From this information it implies the complement of the universal set is a subset of the universal set. Which is a contradiction?' 


#7
Dec1006, 03:19 PM

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Why is that a contradiction? As has been explained by at least two people in this thread, the empty set is different from other sets owing to the fact x in E is false for all x if and only if E is the empty set, thus one can make any deduction one wishes vacuously.



#8
Dec1006, 04:35 PM

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(U is the universal set)
~U subset U > for all x (x notelem U > x elem U) > for all x (~(x notelem U) or (x elem U)) > for all x (x elem U or x elem U) > for all x (x elem U) > true (by definition of U) 


#9
Dec1006, 05:15 PM

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I jumped to the conclusion too quickly without rational thought.
I will try to prove the result that the complement of the universal set U' is a subset of U. U = universal set E = empty set Defn: The complement of a set A, is the set A' such that every element in A' is not in A. Proof: (for every x)(if x is in U' then x is not in U) where x is chosen from U because it's the universal set => E=U' is the only choice. Lemma: E is a subset of every set Proof: Suppose E is not a subset of some set A. Then there contains an element x in E that is not part of A. The consequent is false. So the antecedent must also be false in order for the conditional statement to be true. This leads to E is a subset of A. Hence E is a subset of every set. => E is a subset of U => U' is a subset of U QED Hence this statement is not a contradiction but can be proved. Correct proof? 


#10
Dec1006, 05:24 PM

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For the 4th time, any reasoning that starts 'for all x in X' is vacuously true when X is the empty set since the precedent is false. If x is in the empty set this implies x is not in the empty set, it implies x is not the square root of minus 1....



#11
Dec1006, 05:28 PM

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Whereas I have tried to prove my alleged contradiction to show it is not a contradiction. It's two different ways of doing the same thing isn't it. 


#12
Dec1006, 05:40 PM

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#13
Dec1006, 05:42 PM

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#14
Dec1006, 05:56 PM

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Well I'm learning but evidentally if showing that assuming some statement leads to a contradiction means that statement is false, then showing that no contradiction arises must corroborate it.
I wouldn't say I tried to prove or disprove it but rather that I merely reduced it to a more elementary form and in doing so it became evident that accepting it implies a necessarily true consequent, so by accepting it we can never derive a falsehood. 


#15
Dec1006, 05:57 PM

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I doubt you have used it correctly becuase you appear to have several lines of proof. It is not a definition that the empty set is a subset of everything, it is a (vacuously) true deduction (in any reasonable logic). If it weren't a subset of X, say, then there is an e in E (empty set) that is not in X. But that is false  there is no e in E since E is empty. I don't see that there is anything to prove at all. The complement of U is the set of u in U that are not in U  this is vacuously empty, and it is vacuously (and I am using the word deliberately, pun intended) a subset of U as well.



#16
Dec1006, 06:01 PM

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Erm, any deduction from a false premise can be true or false, so I wouldn't go round accepting things from which we can deduce *true* a consequence. You have made several leaps that aren't justified. if (A implies B) is true and B is true then we know nothing abuot A. 


#17
Dec1006, 06:21 PM

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Ok, let me try again.
Let U be the universal set. Then by definition: for all x (x elem U) for all x (x elem U) > for all x (x notelem U > x elem U) > ~U subset U So U is the universal set > ~U subset U. I now show the reverse implication: ~U subset U > for all x (x notelem U > x elem U) > for all x (not (x notelem U) or x elem U) > for all x (x elem U or x elem U) > for all x (x elem U) By definition, U is the universal set <> for all x (x elem U) Therefore ~U subset U > U is the universal set. So ~U subset U <> U is the universal set. U is the universal set. Therefore I have shown that ~U subset U is true. 


#18
Dec1006, 06:23 PM

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Ah, I only needed the first half.



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