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Xidike
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I want to ask that, are you empty sets equal ?
I reply, therefore I am not an empty set.Xidike said:I want to ask that, are you empty sets equal ?
You've defined two numbers, not two sets, but passing over that...Xidike said: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..
stauros said: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
Bacle2 said:I don't get it: you showed ##A\cup\ B=B\cup\ A## for A,B being (supposedly)
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.
Erland said:. 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.
stauros said:##[\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')]##
MLP said:This argument is fallacious. You cannot infer that one and the same x is referred to by the two existential quantifers. If you could, it would be easy to prove that something is a Dodge and something is a Toyota implies that some one thing is both a Dodge and a Toyota. Maybe it would help to see this if you make the variable used in the second existential claim 'y'.
stauros said: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
Your argument shows that I was right, but perhaps that is what you meant?stauros said:This is wrong as the followinf argument shows:
To be clear, you should have written the two last lines asstauros said:##[\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')]##
Erland said:Your argument shows that I was right, but perhaps that is what you meant?
To be clear, you should have written the two last lines as
##\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.
Oops, the parenteses here are partially wrong. It should be:Erland said:##\exists x[ (x\in \emptyset\wedge \neg x\in \emptyset')]\vee[(x\in \emptyset'\wedge \neg x\in \emptyset )]##
Well, how did you get the corresponding line, in your previous post?stauros said:Can you support that ,by writing a complete formal proof??
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
Erland said:Well, how did you get the corresponding line, in your previous post?
If you want a complete formal proof, you must specify which formal system that should be used: which are the axioms and the rules of inference? Is a Hilbert style axiom system (and which variant in this case) or a natural deduction system (and which variant in this case) or some other kind of system?
And whatever system is used, complete formal proofs tend to be extremely lengthy. One almost always takes shortcuts. But you have a habit of questioning all possible shortcuts.
stauros said:There is no other way of checking whether your argument is right or wrong.
An empty set, also known as a null set, is a set that contains no elements. It is represented by the symbol ∅ or {}.
In mathematics, equality is defined as the state of being equal or having the same value or quantity. It is represented by the symbol = and is used to compare two or more expressions or sets.
Yes, empty sets are equal to each other because they both have no elements. This can be demonstrated through set notation as ∅ = ∅ or {} = {}.
No, an empty set cannot be equal to a non-empty set. This is because a non-empty set contains at least one element, while an empty set contains none. Therefore, they do not have the same value or quantity and are not equal.
The concept of equality is important in mathematics because it allows us to compare and evaluate different mathematical expressions and sets. It also forms the basis for solving equations and proving mathematical theorems.