Proof by axiomatic system of set theory & Venn Diagrams
According to the axioms of set theory, a set is defined as a collection of distinct objects, which can include other sets as well. As stated by @IraeVid, the null set, denoted by $\emptyset$ or $\{\}$ satisfies this definition because it is a well-defined and distinct mathematical object, even if it lacks elements.
Venn Diagrams
Certainly! Here's a kind of visual proof to demonstrate that the null set is considered a set. We can represent sets using Venn diagrams, which use circles or regions to represent sets and their relationships.
In this case, we'll use a simple Venn diagram with one circle to represent the null set ($\emptyset$). Since the null set has no elements, the circle representing it will be empty.
1
As you can see, the circle representing the null set is distinct and well-defined, despite not containing any elements. It fulfills the basic criteria of being a set, as it is a well-defined collection of objects (in this case, no objects) enclosed within a boundary.
By its very nature, the null set ($\emptyset$) is different from the empty space outside the circle. It represents a specific concept within set theory, and its inclusion as a set is essential for the consistency and completeness of mathematical reasoning.
Axiom of empty set
Speaking in the same tone as earlier,
One of the fundamental axioms in set theory is the axiom of empty set, which states the existence of a set with no elements. Represented as:
$\exists \emptyset\ \forall x\ x\notin\emptyset$, this axiom asserts the existence of the null set, $\emptyset$, such that for every element $x$, $x$ is not a member of $\emptyset$.
Axiom schema of specification
Furthermore, we can justify this classification of null set as a set by utilizing the comprehension axiom.
The comprehension axiom states that, for any condition P(x), there exists a set $\{x | \text{P}(x)\}$ that contains all elements x satisfying the condition. Applying this axiom to the null set we get $\{ x\ |\ x ∉ x \}$, that represents the set of all elements $x$ such that $x$ is not a member of itself.
In simple words, as well-demonstrated by Vercassivelaunos, there exists a set of elements satisfying $\text{P}(x)$ even if there's not a single element satisfying the same, i.e. the case of a null set.
Moreover, since the null set $\emptyset$ has no elements, it follows that any set with no elements must be equal to $\emptyset$. Therefore, the null set is unique and well-defined by the Axiom of Extensionality.
Further proving by Contradiction
Now looking to this problem with a different and easier approach, let us pave our ways to a proof by contradiction.
To show that the null set is indeed a set, we can prove it indirectly using the assumption that it is not a set. Let's assume that the null set, $\emptyset$, is not a set. In such a case, we would have: $\nexists \emptyset\ \forall x\ x\notin\emptyset$
This statement asserts that there is no set, $\emptyset$, such that for every element $x$, $x$ is not a member of itself. However, this contradicts the comprehension axiom and the definition we derived earlier.
By contradiction, we can conclude that the null set must be considered a set according to the axioms of set theory.
However, further proving by contradiction seems redundant tho. "-"