Textbook
A set $A$ is a subset of $B$ when every element of $A$ belongs to $B$. Inclusion uses the notation $\subseteq$ when equality is permitted:
$$A\subseteq B\quad\Longleftrightarrow\quad\forall x\,(x\in A\Rightarrow x\in B).$$
Equality is allowed: $A\subseteq A$. If also $A\ne B$, the inclusion is proper, written $A\subsetneq B$. Authors use $\subset$ with different conventions, so it is avoided here.
One element $x\in A$ with $x\notin B$ disproves inclusion. The empty set is a subset of every set: it contains no element that could violate the condition. If $A\subseteq B$ and $B\subseteq C$, then $A\subseteq C$: an arbitrary element of $A$ belongs successively to $B$ and $C$.
Inclusion points from the set whose elements are checked to the set where they must occur. Additional elements on the right do not violate it; an extra element on the left requires another check. Thus $A\subseteq B$ does not imply the reverse inclusion. If the reverse also holds, the sets are equal. For finite sets, inclusion implies $|A|\le|B|$, but comparing sizes alone cannot recover inclusion: a smaller set may contain entirely different objects.
Related concepts: Set membership.
In plain language
Inclusion means that every item in one inventory occurs in another; extra items in the second do not spoil the check. One missing sock refutes a complete transfer, even if the new cupboard is much larger. Membership checks an object, while inclusion checks a set: $2\in\{1,2,3\}$, whereas $\{2\}\subseteq\{1,2,3\}$. The cupboards represent inventories rather than capacity; extra shelves cannot establish the presence of objects.
Example
In this example, $B=\{1,2,3\}$. The first check concerns $A=\{1,2\}$, followed by adding 4 to the set being checked.
Set inclusion
The first element
The element 1 from A is present in B.
Step 1 of 3: The first element
Read all steps
The first element
The element 1 from A is present in B.
- A
- 1current2
- B
- 1included23
Every element checked
The element 2 is also present. The extra element 3 in B is allowed.
- A
- 1included2included
- B
- 1included2included3
A counterexample
The set {1,2,4} contains 4, which is absent from B. One violation is enough.
- Checking
- 1included2included4excluded
- B
- 1included2included3
The first two steps inspect every element of $\{1,2\}$, so this exhaustive finite check proves inclusion. The unchecked three on the right is harmless and shows that inclusion is proper. Once four appears on the left, its absence on the right refutes the new claim. A common mistake is to search for right-hand elements absent from the left and call them violations. That checks the reverse direction, which the original statement never required.