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Two Empty Setsby Xidike
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#1
Nov612, 12:07 AM

P: 72

I want to ask that, are you empty sets equal ????



#2
Nov612, 12:14 AM

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Yes, two empty sets are equal, even if you regard them as subsets of entirely different beasts. So any two power sets have a nonempty intersection. 


#3
Nov612, 09:09 AM

P: 72

How can two Empty Sets be equal ???
take a look at this example.. 1st set: Number of trees in the forest 2nd Set: Number of Student of in the class suppose that these are empty sets.. how are they equal ??? they both have different description.. 


#4
Nov612, 03:33 PM

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Two Empty Sets
"A whole number between 1 and 3." "Positive square root of 4." Two descriptions, equal answers. 


#5
Nov612, 04:34 PM

P: 337

Two sets are equal if every member of the first set is also a member of the second set, and vice versa. If two empty set were not equal, there would be one element in one of the sets which is not a member of the other set, and this is impossible, since an empty set has no member at all, whether member or not member of another empty set.



#6
Nov612, 06:01 PM

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If two empty sets were different, one of them would contain an element
not contained in the other.... 


#7
Nov712, 07:12 AM

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Just to put my oar in: two sets, A and B, are equal if and only both statements
"if x is in A then x is in B" and "if x is in B then x is in A" are true. If A is the empty set, then "if x is in A" is false so the statement "if x is in A then x is in B" is trivially true. If B is the empty set then "if x is in A" is false so the statement "if x is in B then x is in A" is trivially true. Therefore, if A and B are both empty, then they are equal. 


#8
Nov712, 07:43 PM

P: 35

The empty set is unique and the following proof ascertains that.
Suppose that there is another empty set denoted by ##\emptyset'## ,then we have: 1)##\forall A[A\cup\emptyset =A]## 2)##\forall A[A\cup\emptyset' =A]## In (1) we put ##A =\emptyset'## and we get: ##\emptyset'\cup\emptyset =\emptyset'## In (2) we put ## A =\emptyset## and we get :##\emptyset\cup\emptyset' =\emptyset## But since ,##\emptyset'\cup\emptyset = \emptyset\cup\emptyset' ## We can conclude :##\emptyset'=\emptyset ## Hence all the empty sets are equal 


#9
Nov712, 08:45 PM

P: 35

Here is another proof that the empty set is unique:
Suppose again that there is another empty set denoted by,##\emptyset'## Then we have: 1) ##\forall A[\emptyset\subseteq A]## 2)##\forall A[\emptyset'\subseteq A]## In (1) put : ##A=\emptyset'## and we get : ##\emptyset\subseteq \emptyset']##..................................................................... 3 In (2) put : ##A=\emptyset## and we get : ##\emptyset'\subseteq \emptyset]##..................................................................... 4 And from (3) and ( 4) we conclude :,##\emptyset' = \emptyset## 


#10
Nov712, 09:23 PM

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different copies of the empty set. But this is true for any two sets, since union is commutative; I don't see how this shows that A=B. 


#11
Nov812, 03:54 AM

P: 35

##A\cup B= B\cup A## for all A,B Then put :##A=\emptyset ## and ## B=\emptyset'## and we have: ##\emptyset\cup\emptyset' = \emptyset'\cup\emptyset ## But ,##\emptyset\cup\emptyset'=\emptyset ## and ##\emptyset'\cup\emptyset =\emptyset'## as i have shown in my previous proof Hence ##\emptyset'= \emptyset ## 


#12
Nov812, 07:30 AM

P: 159

Those proofs strike me as a bit convoluted. The fact that all empty sets are equal is a consequence of the axiom of extensionality, which is literally the most fundamental property of sets, which says that sets are equal if and only if they have the same members. If the set A has no members and the set B has no members, then by extensionality A = B.



#13
Nov812, 11:08 AM

P: 35

WE know that: ##A=B\Longleftrightarrow [(A\subseteq B)\wedge(B\subseteq A)##. Hence: ##A\neq B\Longleftrightarrow[\neg(A\subseteq B)\vee \neg(B\subseteq A)]##. OR ##A\neq B\Longleftrightarrow[\exists x(x\in A\wedge \neg x\in B)]\vee[\exists x(x\in B\wedge \neg x\in A)]## Now if we put : ##A =\emptyset## and ##B=\emptyset'## ,we get that: ##\emptyset\neq \emptyset'\Longleftrightarrow[\exists x(x\in \emptyset\wedge \neg x\in \emptyset')]\vee[\exists x(x\in \emptyset'\wedge \neg x\in \emptyset )]## And if we assume :##\emptyset\neq \emptyset'##, then ##[\exists x(x\in \emptyset\wedge \neg x\in \emptyset')]\vee[\exists x(x\in \emptyset'\wedge \neg x\in \emptyset )]##.That implies : ##[ (x\in \emptyset\wedge \neg x\in \emptyset')]\vee[(x\in \emptyset'\wedge \neg x\in \emptyset )]##.Which in turn implies: ##[ (x\in \emptyset\vee x\in \emptyset')]## 


#14
Nov812, 04:12 PM

P: 31




#15
Nov912, 08:11 AM

P: 35

Let : A= {1,2}, and B={1,2,3}. Can you prove that :## A\neq B## , using the axiom of extensionality??. Then you will find out that using the same or different quantifiers makes no difference 


#16
Nov912, 10:01 AM

P: 31

It is not legitimate to instanciate the first quantifier to x and then instanciate the second one to x and act like they are one and the same thing. 


#17
Nov912, 03:12 PM

P: 337

##\exists x[ (x\in \emptyset\wedge \neg x\in \emptyset')]\vee[(x\in \emptyset'\wedge \neg x\in \emptyset )]## and ##\exists x[ (x\in \emptyset\vee x\in \emptyset')]##. Then, the argument is correct, since ##\exists x(P(x)\vee Q(x))## and ##\exists x \,P(x)\vee\exists x\,Q(x)## are logically equivalent. 


#18
Nov912, 10:12 PM

P: 35

Because i know that it will be useless to ask you ,where did you get the: "and ##\exists x[ (x\in \emptyset\vee x\in \emptyset')]##" ,part, e.t.c ,e.t.c 


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