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I Have a question. Given only the definition of equality of two sets, how can we prove that there is one and only one empty set. I mean by equality of two sets the following: $$A=B \iff \forall x (x \in A \iff x \in B)$$

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  • $\begingroup$ Two sets are different if and only if there is an element of one that doesn't belong to the other one. Is “there is an element of the empty set” true? $\endgroup$ – egreg Oct 23 '14 at 14:35
  • $\begingroup$ For the existence, we need more than only the definition of equality for sets. The definition, however, gives the uniqueness, for if $A$ and $B$ are sets such that no $x$ is an element of either, they are equal per definition. $\endgroup$ – Daniel Fischer Oct 23 '14 at 14:35
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    $\begingroup$ Here is a marginally different perspective: You have $A \neq B$ iff there exists some $x$ such that either $x \in A$ and $x \notin B$ or vice versa. So, if two versions of the empty set are not equal, then one must contain an element which is a contradiction. $\endgroup$ – copper.hat Oct 23 '14 at 14:52
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Modern set theory conceives of a set as an abstraction of a property. Two properties might seem different, but be essentially the same because they are true of the same objects. For example, the property $\mathcal O_1$ of being a natural number of the form $2n+1$, and the property $\mathcal O_2$ of being a natural number that is the difference of two consecutive perfect squares $S_{n+1} - S_n$. These are not the same property, but one can prove that they are the same in a certain sense, namely that $\mathcal O_1$ holds for some object $x$ precisely when $\mathcal O_2$ also holds for $x$. Sets are a formalization of this idea: we say that the set of objects for which $\mathcal O_1$ holds is the same set as the set for which $\mathcal O_2$ holds. That is, $$\{ x \mid \mathcal O_1(x) \} = \{ x \mid \mathcal O_2(x) \}.$$

The idea here is that we want sets to be equal not if their defining conditions are the same (which is the complicated situation we are trying to simplify) but if they contain the same objects.

Suppose we have two empty sets, say $$\varnothing_1 = \{ n \mid \text{$n$ is an even prime number bigger than 10} \}$$ and $$\varnothing_2 = \{ n \mid \text{$n$ is a living crown prince of the Ottoman Empire} \}$$

These sets do have the same elements, so we want to consider them to be the same set, because that's what sets are for: to abstract away the confusing details of properties, and focus only on the things for which the properties hold or don't hold. So because there is no object by which we can distinguish these two properties—there is no living Crown Prince of the Ottoman Empire who is not also an even prime number bigger than 10, and vice versa—we say that the two sets are equal.

In some theories there is more than one empty set. For example, Bertrand Russell's theory of types (1913) has multiple empty sets. In addition, it has a family of empty relations, which are different from the empty sets. (In modern theories an empty relation is an empty set.) This proliferation of empty sets is one of the most criticized points of the theory of types.

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  • $\begingroup$ WOW! That's really a good abstract answer :D $\endgroup$ – M.darwish Oct 23 '14 at 14:59
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As $$\forall x(x\not\in\emptyset_i)$$ is obvious that $$x\in\emptyset_1\iff x\in\emptyset_2$$ because both conditions are false.

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In the case of two empty sets, $\emptyset_1, \emptyset_2$, we have that $\forall x(x\in\emptyset_1\Leftrightarrow x\in \emptyset_2)$ is vacuously true.

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