A Set With Three Elements Hides Eight Sets Inside It

Eight — and there is a formula rather than a need to count. A set with $n$ elements has $2^n$ subsets, so three elements give $2^3=8$.

That includes two that are easy to forget: the empty set and the whole set itself. This page covers everything in the ICSE Class 7 Mathematics chapter's second part: subsets and proper subsets, listing all subsets, the universal set and complement, and union, intersection and difference.

$A$ is a subset of $B$, written $A\subseteq B$, if every element of $A$ is also in $B$. It is a proper subset, written $A\subset B$, if it is a subset and $A\ne B$ — so $B$ has at least one element that $A$ does not.

Take $A=\{1,2\}$ and $B=\{1,2,3\}$. Both 1 and 2 lie in $B$, so $A\subseteq B$. And since $B$ also holds 3, $A\subset B$ as well.

But with $C=\{1,2,3\}$ and $B=\{1,2,3\}$: $C\subseteq B$ is true, while $C\subset B$ is false, because the two sets are equal.

Two facts follow from the definition and are asked directly:

- Every set is a subset of itself, so $A\subseteq A$ always holds.
- The empty set is a subset of every set, so $\varnothing \subseteq A$ always holds — there is no element of $\varnothing$ that could fail the test.

The cricket team is a subset of the whole class, and a proper subset so long as somebody in the class is not in the team.

The symbol to watch is $\in$ against $\subseteq$. Membership joins an element to a set, as $2\in \{1,2,3\}$; subset joins a set to a set, as $\{2\}\subseteq \{1,2,3\}$. Writing $2\subseteq \{1,2,3\}$ or $\{2\}\in \{1,2,3\}$ is marked wrong.

Work up by size — the empty set, then every single element, then every pair, and so on up to the whole set.

For $A=\{a,b,c\}$:

- 0 elements: $\varnothing$
- 1 element: $\{a\}$, $\{b\}$, $\{c\}$
- 2 elements: $\{a,b\}$, $\{a,c\}$, $\{b,c\}$
- 3 elements: $\{a,b,c\}$

That is $1+3+3+1=8$ subsets, matching

$$2^n=2^3=8$$

The proper subsets are all of them except $A$ itself, so there are

$$2^n-1=7$$

Checking the rule on other sizes: a set of 2 elements has $2^2=4$ subsets, and a set of 4 elements has $2^4=16$. A set of 5 has 32.

The reason the rule works is that each element faces one independent choice — in the subset or out of it. Three elements give $2\times 2\times 2=8$ combinations of those choices, which is exactly the subset count.

So listing them in order of size is not just tidiness. It guarantees you produce all 8 rather than stopping at 6, which is what happens when people forget $\varnothing$ and $A$ itself.

The universal set, written $U$ or $\xi$, is the set containing all the elements under discussion in a particular problem. Every other set in that problem is a subset of it.

The complement of $A$, written $A^{\prime }$, is the set of all elements of $U$ that are not in $A$.

Worked example. Let $U=\{1,2,3,4,5,6,7,8,9,10\}$ and $A=\{2,4,6\}$. Then

$$A^{\prime }=\{1,3,5,7,8,9,10\}$$

Count them: $n(A)=3$ and $n(A^{\prime })=7$, and

$$n(A)+n(A^{\prime })=3+7=10=n(U)$$

That relation always holds, and it is a quick way to check a complement — if the two counts do not add to $n(U)$, an element has been dropped or duplicated.

Two boundary cases are examined: $U^{\prime }=\varnothing$, since nothing in $U$ is outside $U$; and ${\varnothing }^{\prime }=U$. Also $(A^{\prime }{)}^{\prime }=A$, because excluding twice returns you to the start.

In a class survey, the universal set is the whole class, and the complement of "pupils who play cricket" is "pupils who do not".

The complement has no meaning without a stated universal set. The complement of $\{2,4,6\}$ is $\{1,3,5,7,8,9,10\}$ for the $U$ above, but it would be $\{1,3,5\}$ if $U$ were $\{1,\dots ,6\}$ — so always write down $U$ first.