Sets • Topic 2 of 3

Subsets, Intervals and the Power Set

$A$ is a subset of $B$, written $A \subseteq B$, if every element of $A$ is also an element of $B$. Formally:

$$A \subseteq B \iff (x \in A \Rightarrow x \in B)$$

If $A \subseteq B$ but $A \ne B$ (so $B$ has at least one element not in $A$), then $A$ is a proper subset, written $A \subset B$. Two facts hold for every set $A$: the empty set is a subset of it ($\varnothing \subseteq A$), and every set is a subset of itself ($A \subseteq A$). A key consequence is the test for equality — $A = B$ if and only if $A \subseteq B$ and $B \subseteq A$ — which is the standard way to prove two sets are the same.

Element vs subset — do not confuse $\in$ and $\subseteq$. The symbol $\in$ relates an object to a set ($2 \in \{1,2,3\}$); $\subseteq$ relates a set to a set ($\{2\} \subseteq \{1,2,3\}$). Writing $2 \subseteq \{1,2,3\}$ or $\{2\} \in \{1,2,3\}$ is wrong.

Subsets of the real numbers. The number systems are nested as subsets: $\mathbb{N} \subset \mathbb{W} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}$. The irrationals $\mathbb{R} \setminus \mathbb{Q}$ (such as $\sqrt{2}$ and $\pi$) are also a subset of $\mathbb{R}$.

Universal set $U$ is the fixed "background" set containing all objects under discussion for a particular problem. Every other set in that problem is a subset of $U$. The choice of $U$ depends on context — it may be $\mathbb{R}$ in a number problem, or "all students of a class" in a survey problem.

Intervals are subsets of $\mathbb{R}$ describing continuous stretches of the number line. A square bracket includes the endpoint; a round bracket excludes it. The symbols $\infty$ and $-\infty$ are not numbers, so the bracket beside them is always round:

IntervalSet-builderEndpoints
$[a, b]$$\{x : a \le x \le b\}$both included (closed)
$(a, b)$$\{x : a < x < b\}$both excluded (open)
$[a, b)$$\{x : a \le x < b\}$left included only (half-open)
$(a, b]$$\{x : a < x \le b\}$right included only (half-open)
$(a, \infty)$$\{x : x > a\}$$\infty$ is never included
$(-\infty, b]$$\{x : x \le b\}$$-\infty$ is never included

The length of a bounded interval with endpoints $a$ and $b$ is $b - a$, regardless of which endpoints are included.

Power set: the set of all subsets of $A$, written $P(A)$. Its elements are themselves sets — note that $\varnothing \in P(A)$ and $A \in P(A)$ for every $A$. The count of subsets is fixed by the size of $A$:

$$\text{If } n(A) = m, \text{ then } n\big(P(A)\big) = 2^{m}$$

The reason is a counting argument: to build a subset you make one independent yes/no choice per element — "include it or not" — giving $2 \times 2 \times \dots \times 2 = 2^m$ possibilities. Of these, $2^m - 1$ are proper subsets (all except $A$ itself), and $2^m - 2$ are proper non-empty subsets (excluding $A$ and $\varnothing$). The number of subsets containing one fixed element is $2^{m-1}$ — exactly half — because the choice for that element is settled and the remaining $m-1$ elements are still free.

Deeper Insight — subset is the backbone, $2^m$ is its first dividend: The subset relation $\subseteq$ is the most important relation in set theory because nearly every later definition is phrased through it — equality ($A=B$ means $A\subseteq B$ and $B\subseteq A$), the complement, even the way a universal set frames a problem. Keep two edge cases permanently in mind: $\varnothing$ is a subset of everything (vacuously — there is no element of $\varnothing$ that could fail the test), and every set is a subset of itself. The power-set formula $2^m$ is your first real glimpse of why these ideas matter beyond bookkeeping: it is the same "independent binary choice" logic that powers the Binomial Theorem ($\sum \binom{m}{k} = 2^m$, the total number of subsets counted by size) and probability over finite outcomes. Once you see a subset as a string of yes/no decisions, $2^m$ stops being a formula to memorise and becomes something you can re-derive on the spot.

A as a subset of B inside the universal set Subset: A ⊆ B ⊆ U U B A Closed and open intervals on the number line Closed [a, b] vs Open (a, b) ab[a, b] includedab(a, b) excluded
1
Worked Example
List all subsets of $A = \{a, b, c\}$ and verify the count.
Solution
  1. Size $0$: $\varnothing$.
  2. Size $1$: $\{a\}, \{b\}, \{c\}$.
  3. Size $2$: $\{a,b\}, \{a,c\}, \{b,c\}$.
  4. Size $3$: $\{a,b,c\}$.
  5. Total $= 1+3+3+1 = 8 = 2^3$.

Answer: $P(A)$ has $8$ elements; $7$ of them are proper subsets.

2
Worked Example
If $n(A) = 5$, how many subsets and how many proper subsets does $A$ have?
Solution
  1. Subsets: $2^5 = 32$.
  2. Proper subsets exclude $A$ itself: $32 - 1 = 31$.

Answer: $32$ subsets, $31$ proper subsets.

3
Worked Example
Write $\{x : x \in \mathbb{R},\ -2 \le x < 5\}$ as an interval, and state whether $-2$ and $5$ belong to it.
Solution
  1. Left endpoint $-2$ uses $\le$ → included → square bracket.
  2. Right endpoint $5$ uses $<$ → excluded → round bracket.

Answer: $[-2, 5)$. Here $-2 \in$ the set but $5 \notin$ the set.

4
Worked Example
A set has $63$ proper subsets. How many elements does it have?
Solution
  1. Proper subsets $= 2^m - 1$.
  2. $2^m - 1 = 63 \Rightarrow 2^m = 64 = 2^6$.
  3. So $m = 6$.

Answer: $6$ elements.

5
Worked Example
Insert $\subset$, $\subseteq$ or $\in$ between: (a) $\{1\}\ \_\ \{1,2,3\}$, (b) $1\ \_\ \{1,2,3\}$, (c) $\varnothing\ \_\ \{1,2\}$.
Solution
  1. (a) $\{1\}$ is a set contained in the bigger set, and it is not equal to it → $\{1\} \subset \{1,2,3\}$.
  2. (b) $1$ is an element, not a set → $1 \in \{1,2,3\}$.
  3. (c) $\varnothing$ is a subset of every set → $\varnothing \subset \{1,2\}$.

Answer: (a) $\subset$, (b) $\in$, (c) $\subset$.

6
Worked Example
Write the power set of $A = \{1, \{2\}\}$.
Solution
  1. $A$ has two elements: the number $1$ and the set $\{2\}$, so $n(A) = 2$ and $P(A)$ has $2^2 = 4$ members.
  2. List them: $\varnothing$, $\{1\}$, $\{\{2\}\}$, $\{1, \{2\}\}$.

Answer: $P(A) = \big\{\varnothing,\ \{1\},\ \{\{2\}\},\ \{1, \{2\}\}\big\}$.

7
Worked Example
Express each interval in set-builder form and give its length: (a) $[-3, 4]$, (b) $(0, 7)$, (c) $(2, 9]$.
Solution
  1. (a) Square brackets include both ends: $\{x : -3 \le x \le 4\}$; length $= 4-(-3) = 7$.
  2. (b) Round brackets exclude both: $\{x : 0 < x < 7\}$; length $= 7-0 = 7$.
  3. (c) Half-open: $\{x : 2 < x \le 9\}$; length $= 9-2 = 7$.

Answer: (a) $[-3,4]=\{x : -3 \le x \le 4\}$, (b) $(0,7)=\{x : 0 < x < 7\}$, (c) $(2,9]=\{x : 2 < x \le 9\}$; all three have length $7$.

8
Worked Example
How many subsets of $A = \{1,2,3,4\}$ contain the element $1$?
Solution
  1. Fix $1$ as "included" — that choice is now settled.
  2. The remaining $3$ elements $\{2,3,4\}$ are each free to be in or out: $2^3 = 8$ ways.
  3. (Equivalently, half of all $2^4 = 16$ subsets contain any given element.)

Answer: $2^{3} = 8$ subsets contain the element $1$.

9
Worked Example
Find the number of (a) subsets, (b) proper subsets, and (c) proper non-empty subsets of a set with $4$ elements.
Solution
  1. (a) Subsets $= 2^4 = 16$.
  2. (b) Proper subsets exclude the set itself: $16 - 1 = 15$.
  3. (c) Proper non-empty subsets also exclude $\varnothing$: $16 - 2 = 14$.

Answer: (a) $16$, (b) $15$, (c) $14$.

10
Worked Example
Find $n\big(P(P(\varnothing))\big)$.
Solution
  1. $n(\varnothing) = 0$, so $n(P(\varnothing)) = 2^0 = 1$, i.e. $P(\varnothing) = \{\varnothing\}$.
  2. That set has $1$ element, so $n(P(P(\varnothing))) = 2^1 = 2$.

Answer: $n\big(P(P(\varnothing))\big) = 2$.

Key Points

  • $A \subseteq B$ means every element of $A$ lies in $B$; a proper subset $A \subset B$ also has $A \ne B$.
  • $\varnothing \subseteq A$ and $A \subseteq A$ for every set $A$.
  • $A = B \iff A \subseteq B \text{ and } B \subseteq A$ — the standard way to prove two sets equal.
  • Use $\in$ between an element and a set, $\subseteq$ between two sets — never mix them up.
  • The universal set $U$ is the background set for a problem; every set in that problem is a subset of $U$.
  • Intervals are subsets of $\mathbb{R}$: square bracket $=$ endpoint included, round bracket $=$ excluded; $\pm\infty$ always take a round bracket.
  • Power set $P(A)$ collects all subsets; if $n(A) = m$ then $n(P(A)) = 2^m$, with $2^m - 1$ proper subsets and $2^{m-1}$ subsets containing any fixed element.
Tap an option to check your answer0 / 4
Q1.The number of subsets of a set with $3$ elements is:
Explanation: A set with $n$ elements has $2^n$ subsets; $2^3=8$.
Q2.If $A=\{a,b\}$, which is a subset of $A$?
Explanation: $\{a\}$ contains only elements of $A$.
Q3.The power set of $\varnothing$ has how many elements?
Explanation: $P(\varnothing)=\{\varnothing\}$, which has one element.
Q4.The interval $(2,5]$ in set-builder form is:
Explanation: A round bracket excludes, a square bracket includes the endpoint.