Q.Let $U = \{1, 2, 3, 4, 5, 6, 7, 8\}$; $A = \{1, 2, 3\}$; $B = \{2, 4, 6\}$, $C = \{3, 6, 9, 12\}$. Verify:

Set Membership

The idea in plain words

Every set is defined by exactly one question: "does this object belong to the set, or not?" That yes/no relationship between an object and a set is called membership. If an object is in the set, it is a member (or element) of the set; if it isn't, it simply is not.

A set is only "well-defined" if this question always has a clear answer for every possible object — that's what makes membership testable.

The notation

For a set $A$ and an object $x$:

• $x \in A$ reads "$x$ belongs to $A$" or "$x$ is an element of $A$" — TRUE membership.
• $x \notin A$ reads "$x$ does not belong to $A$" — FALSE membership.

Example. Let $A = \{2, 4, 6, 8\}$.

Testing membership: roster form vs. set-builder form

Roster form — just look for the object in the list.

Set-builder form — plug the candidate into the defining rule and check if it's satisfied.

Is $7 \in C$? Check: is 7 prime and less than 10? Yes → $7 \in C$.

Is $9 \in C$? Check: 9 is less than 10 but not prime ($9 = 3 \times 3$) → $9 \notin C$.

Properties every student must know

1. Each element is either in or out — never "partly in." Membership is binary, not a matter of degree.
2. Repetition doesn't affect membership. $\{1, 1, 2\} = \{1, 2\}$ — asking "is 1 a member?" gives the same YES either way.
3. Order never affects membership. $2 \in \{1, 2, 3\}$ is exactly the same fact as $2 \in \{3, 2, 1\}$.
4. Nothing belongs to the empty set. For any $x$, $x \notin \emptyset$ — there is nothing inside to belong to.
5. A set can be an element of another set. If $D = \{1, \{2, 3\}\}$, then $1 \in D$ is true, and $\{2,3\} \in D$ is true, but $2 \in D$ is false — 2 is not directly listed in $D$; it is only inside the set that is listed.

A common slip

Students often confuse $x \in A$ (membership: is the object present?) with $B \subseteq A$ (subset: is every element of $B$ also in $A$?). The two symbols compare different kinds of things:

• $\in$ / $\notin$ compares an element to a set.
• $\subseteq$ / $\not\subseteq$ compares a set to a set.

So for $A = \{1, 2, 3\}$: writing $1 \in A$ is correct, but $1 \subseteq A$ is technically wrong notation (1 is an element, not a set) — although $\{1\} \subseteq A$ is correct, because now both sides being compared are sets.

Why it matters

Membership is the single test every other set idea is built on — union, intersection, subset, and complement are all ultimately defined by asking "$x \in A$?" and "$x \in B$?" for every candidate $x$. Get comfortable with $\in$/$\notin$ first, and every later operation becomes just a combination of membership questions.

Takeaway

Membership answers one question only: is this exact object listed in this exact set? Everything else in set theory is built from asking that question, over and over, about different sets.

"Set membership symbol meaning" and "element vs subset difference class 11" are common queries around this idea, which forms the very first building block of the Sets chapter in the NCERT/CBSE Class 11 Mathematics curriculum. Getting the $\in$ vs $\subseteq$ distinction right is a small but frequent trap in board and competitive exam questions.