Empty set

Definition and notation. A set with no elements is the empty set, written $\varnothing$ or $\{\ \}$. Never write it as $\{\varnothing\}$ or $\{0\}$ — each of those is a set containing one element (respectively the symbol $\varnothing$, and the number $0$), so both have $n = 1$, not $n = 0$.

When a set-builder rule gives $\varnothing$. Whenever no object in the stated universe satisfies the condition, the result is empty. For instance:

where: $\mathbb{N}$ is the set of natural numbers under the NCERT convention $\{1, 2, 3, \dots\}$, and the condition asks for a natural number strictly between $1$ and $2$ — there isn't one, so $A$ has no elements.

Cardinality and finiteness. The empty set has a definite count, $n(\varnothing) = 0$, so it is classified as finite — a small but common exam point, since "empty" is sometimes mistaken for "undefined" or "infinite."

Uniqueness. There is only one empty set, no matter which failed condition produced it. The picture below shows three completely different rules — no real square root of $-1$, no natural number below $1$, and no natural number strictly between $1$ and $2$ — all collapsing onto the same object.

The most striking property of $\varnothing$ is that it is a subset of every set $A$, including itself:

where: $\subseteq$ means "every element of the left set is also an element of the right set." This looks strange the first time you meet it — how can "nothing" be inside every set? The reasoning is a classic case of a vacuously true statement, and it is worth walking through carefully, since exams often ask you to justify it rather than just state it. To disprove $\varnothing \subseteq A$, you would have to produce an element of $\varnothing$ that is not an element of $A$ — that single counterexample would break the subset claim. But $\varnothing$ has no elements at all, so there is nothing you could ever point to. The statement "if $x \in \varnothing$ then $x \in A$" is an implication whose starting condition, $x \in \varnothing$, is never true for any $x$. An implication with a condition that is never satisfied cannot ever be false — it is automatically, vacuously, true. Since this holds for every choice of $A$, including $A = \varnothing$ itself, the empty set is a subset of every set without exception. This is exactly why $\varnothing$ behaves like a universal "starting point" in set theory: it is the one set guaranteed to sit inside any other set you write down.

Determine whether $C = \{x : x \text{ is an even prime number greater than } 2\}$ is the empty set, compare it with $D = \{0\}$, and confirm that $C \subseteq D$ holds regardless.

Step 1. List the even prime numbers. A prime number has exactly two factors, $1$ and itself. Among even numbers, every one except $2$ is divisible by $2$ and by some other factor too, so $2$ is the only even prime.

Step 2. Apply the condition "greater than $2$." The set $C$ needs an even prime that is also strictly greater than $2$. The sole even prime, $2$, is not greater than $2$, so no number satisfies both requirements together.

Step 3. Conclude $C = \varnothing$, so $n(C) = 0$. Compare with $D = \{0\}$: here $n(D) = 1$, since $D$ contains exactly one element, the number $0$. Because $n(C) \ne n(D)$, $C \ne D$ — being empty and containing zero are entirely different ideas.

Step 4. Check the subset claim anyway. Even though $C \ne D$, the general fact $\varnothing \subseteq A$ says $C \subseteq D$ must still hold, since $C = \varnothing$ and $D$ is some set. There is no element of $C$ to fail the test, so the inclusion is automatic — the same vacuous-truth reasoning as above, now applied to this specific pair.

Final answer: $C = \varnothing$ with $n(C) = 0$; $C \ne D$ since $n(D) = 1$; and $C \subseteq D$ holds because $\varnothing$ is a subset of every set.

A quick sanity habit. Before declaring a set-builder description empty, restate the condition in words and ask whether the domain rules it out for a reason other than the algebra. $\{x \in \mathbb{Z} : x + 3 = 3\}$ is not empty — it is $\{0\}$, a singleton — because $\mathbb{Z}$ includes zero. Change the domain to $\mathbb{N}$ under the NCERT convention and the very same equation, $x + 3 = 3$, gives $\varnothing$, since $0 \notin \mathbb{N}$ there. The algebra is identical; only the domain changed the answer. This is why checking the domain first, before searching for a counterexample, saves time and avoids a wrong conclusion.

• Writing the empty set as $\{0\}$ or $\{\varnothing\}$ — both have $n = 1$, not $n = 0$; only $\varnothing$ or $\{\ \}$ is correct.
• Assuming $\varnothing$ being "nothing" means it cannot be a subset of anything — in fact $\varnothing \subseteq A$ for every set $A$, including $\varnothing$ itself.
• Forgetting that $\mathbb{N}$ under the NCERT convention starts at $1$, which can decide whether a condition like $x + 3 = 3$ (needing $x = 0$) produces the empty set.
• Treating $\varnothing$ as several different "empty sets" depending on how they arose — there is only one empty set, regardless of the condition that produced it.

Ask the tutor: if a set-builder condition you're working with looks empty but you're not fully sure, ask the AI tutor to check it and the subset reasoning with you step by step.