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I haven't been able to figure out if the following reasoning is correct, so I'd like to have the opinion of other people on that.

The goal is to prove that the empty set is unique.

In order to do that, let $E$ be a set, $A$ be an empty set and $B$ be an empty set. I want to prove that $A = B$.

First, I can try proving that $A \subset B$.

I know that $\forall x \in E, x \notin A$. Now I can consider this proposition : $x \in A \Rightarrow x \in B$

Since $x \notin A$, I can safely say that that the proposition above is true.

Hence, I can conclude that $A \subset B$

Am I right in my conclusion?

Heffebaycay
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5 Answers5

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Your conclusion is correct. You can also resort directly to the Axiom of Extension

$A=B\iff \forall z\colon(z\in A\leftrightarrow z\in B)$

For the given conditions give us $z\notin A$ and (equivalently) $ z\notin B$ for arbitrary $z$, hence the right hand side in the axiom is true.

hmakholm left over Monica
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Hagen von Eitzen
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  • Using the symmetry and transitivity of "=", we can conclude $A=B$ from $A=\emptyset$ and $B=\emptyset$. Why bother using the Axiom of Extension? – William Nov 02 '24 at 08:09
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I think you might be overlooking the importance (or unaware) of the Axiom of Extensionality, which simply says two sets are equal if they have the same members. Thus two empty sets are equal by this axiom.

Added: In some variations of set theory, we may want to have more than one thing without members, and these are called ur-elements. The Axiom of Extensionality can then be modified to allow for more than one "empty set", while still allowing us to compare and identify things that have members belonging to them.

hardmath
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    The OP apparently uses a "lemma" ($A\subseteq B\land B\subseteq A\implies A=B$). The Axiom of Extensionality of course hides in the proof of this lemma. – Hagen von Eitzen Oct 13 '14 at 20:31
  • Interesting. Are ur-elements sort of like letters from an alphabet, in that letters can have distinct identities but cannot contain any objects i.e. a letter cannot be considered a set? – Display name Aug 11 '20 at 21:38
  • @Nicholas: Yes, that's a good way to think about it. The Axiom of Extensionality compels us to treat all "empty" sets as equal, so we modify it to allow ur-elements to have distinct identities. See my link in the Answer above. – hardmath Aug 11 '20 at 23:02
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It will be very helpful to start form the definition of implies to (in other words how we define $P \implies Q$). This will be helpful to understand vacuously true statements. Then it will be easy to undersand how Axiom of Extenstionality $\implies$ uniqueness of null set.

Definition($P \implies Q$):
$ P \implies Q := \neg(P \wedge \neg Q) = \neg P \vee Q\,$, in other words $P$ and not $\neg Q$ are not simultaneously true. This is our besic intuition of what 'implies to' mean.

Now, If $P$ is false, then regardless of whether $Q$ is true or not $P \implies Q\,$ is always true from the definition above. This statements are called vacuously true statements.

Now, let $\varnothing,\,\varnothing'\,$ be two null sets. We can say $x \in \varnothing$ and $x \in \varnothing'$ is false $\forall x$ \begin{align} &\implies \forall x(x\in \varnothing \implies x \in \varnothing')\, \big[ \text{vacuously true}\big]\\ & \text{and,}\quad \forall x(x\in \varnothing \implies x \in \varnothing') \big[ \text{also vacuously true}\big]\\ &\implies \forall x(x \in \varnothing \iff x \in \varnothing') \end{align}

Axiom of Extentionality: $ \forall A \forall B\: \forall x((x \in A \iff x \in B) \implies (A = B))$

Since, $\forall x(x \in \varnothing \iff x \in \varnothing')\quad \therefore \varnothing = \varnothing'$.

Pritam Jatuya
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  • The use of the phrase 'vacuously true' warrants a +1 smile in the empty set... Mind you, of course, voidology - i.e., the study of the empty set - is innately entertaining: I've commented on this here before, but on youtube one can find Benedict Gross describing an alg. top. course ["a great course - I learnt nothing"] he had taken, taught by his supervisor, John Tate: motivated throughout by the categorical pov, Tate posed the likes of "is the empty set connected?" [Pause] "is the empty set simply connected?" – peter a g Jun 12 '25 at 18:23
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Here's an approach that's probably cleaner:

Per Axiom of Existence, there exists a set which has no elements. Let $A$ and $B$ each be a set which has no elements. Since $ x \in A \iff x \in B $ is vacuously true, by Axiom of Extensionality $A=B$.

Equivalently, that $x \in A \iff x \in B $ is true can be established using the truth table of the biconditional, by recognizing that $x \in A$ and $x \in B$ are in each case false (ex falso quodlibet).

winvicta
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  1. $A=\emptyset$
  2. $B=\emptyset$
  3. $\emptyset=B$ "=" symmetry applied to 2
  4. $A=B$ "=" transitivity applied to 1,3
William
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