Equal sets and subsets

If two sets describe the same collection, we want to say so. If one set sits inside another, we want a clean notation for that too. This subtopic introduces equality, subset, proper subset, and the intervals of $\mathbb{R}$ , the most important examples of subsets you will use this year.

Definitions

Two sets $A$ and $B$ are equal, written $A = B$, if they have exactly the same elements: $A = B \iff (\forall x)\,(x \in A \iff x \in B).$

$A$ is a subset of $B$, written $A \subseteq B$, if every element of $A$ is also an element of $B$: $A \subseteq B \iff (\forall x)\,(x \in A \Rightarrow x \in B).$

$A$ is a proper subset of $B$, written $A \subsetneq B$ (or $A \subset B$ in some texts), if $A \subseteq B$ but $A \ne B$.

If $A \subseteq B$ we say $B$ is a superset of $A$, written $B \supseteq A$.

The antisymmetry of $\subseteq$ gives the standard recipe for proving equality: $A = B \iff A \subseteq B \text{ and } B \subseteq A.$

Basic facts you must know

• $\varnothing \subseteq A$ for every set $A$. (Vacuously: there is no $x \in \varnothing$ to check.)
• $A \subseteq A$ for every set $A$ (reflexivity).
• If $A \subseteq B$ and $B \subseteq C$ then $A \subseteq C$ (transitivity).
• $A = B$ iff $A \subseteq B$ and $B \subseteq A$ (antisymmetry).

For finite sets, $A \subseteq B$ implies $|A| \le |B|$, with equality iff $A = B$.

Two-way containment in practice

To prove $A = B$ you almost always do it in two steps:

1. Take a typical $x \in A$ and show $x \in B$. This proves $A \subseteq B$.
2. Take a typical $x \in B$ and show $x \in A$. This proves $B \subseteq A$.
3. Conclude $A = B$.

This double inclusion technique is universal , you will use it in every later chapter from inequalities to probability.

Intervals as subsets of $\mathbb{R}$

The real line is so important that we have a special notation for its subsets:

Note that $\infty$ never gets a square bracket , it is not a real number.

Length of a bounded interval $(a, b), [a, b], [a, b), (a, b]$ is $b - a$ in every case.

Worked examples

Example 1. Are $A = \{x \in \mathbb{R} : x^2 = 1\}$ and $B = \{-1, 1\}$ equal?

The real solutions of $x^2 = 1$ are exactly $\pm 1$. So $A = \{-1, 1\} = B$.

Example 2. Decide $A \subseteq B$ or not, where $A = \{1, 2, 3\}$, $B = \{x \in \mathbb{N} : x \le 5\}$.

$B = \{1,2,3,4,5\}$. Every element of $A$ is in $B$, so $A \subseteq B$. Since $4 \in B \setminus A$, the inclusion is proper.

Example 3. Show $\{x \in \mathbb{R} : x^2 - 3x + 2 = 0\} = \{1, 2\}$.

Factor: $x^2 - 3x + 2 = (x-1)(x-2) = 0$ gives $x = 1$ or $x = 2$. Both are real, so the solution set is $\{1, 2\}$.

Example 4. Write the set $\{x \in \mathbb{R} : -2 \le x < 5\}$ as an interval.

This is $[-2, 5)$.

Example 5 (harder). Prove that the set of multiples of $6$ is a subset of the set of multiples of $3$.

Let $A = \{6k : k \in \mathbb{Z}\}$ and $B = \{3m : m \in \mathbb{Z}\}$. Take any $x \in A$. Then $x = 6k$ for some $k \in \mathbb{Z}$, so $x = 3(2k)$. Since $2k \in \mathbb{Z}$, we have $x \in B$. Hence $A \subseteq B$. The inclusion is proper because $3 \in B \setminus A$.

Try it yourself

1. Decide equal or not: $A = \{x \in \mathbb{Z} : x^2 \le 4\}$ and $B = \{-2, -1, 0, 1, 2\}$.
2. Decide $A \subseteq B$: $A = \{2, 4, 6\}$ and $B = \{x \in \mathbb{N} : x \le 10,\ x \text{ even}\}$.
3. Write as interval: $\{x \in \mathbb{R} : 0 < x \le 7\}$.
4. Write in set-builder form: $(-3, 5]$.
5. Show that $\mathbb{N} \subsetneq \mathbb{Z} \subsetneq \mathbb{Q} \subsetneq \mathbb{R}$.
6. Find all subsets of $\{a, b\}$.
7. Is $\{1, 2\} \subseteq \{\{1, 2\}, 3, 4\}$? Explain.
8. Decide whether each is true: (a) $\varnothing \in \varnothing$; (b) $\varnothing \subseteq \varnothing$; (c) $\varnothing \in \{\varnothing\}$; (d) $\varnothing \subseteq \{\varnothing\}$.
9. Show $\{x \in \mathbb{R} : x = \tfrac{p}{q},\ p, q \in \mathbb{Z}, q \ne 0\} = \mathbb{Q}$.
10. If $A \subseteq B$ and $B \subseteq A$, what can you conclude?
11. List all subsets of $\{1, 2, 3\}$. How many are there?
12. The set of perfect squares in $\mathbb{N}$ is a subset of which familiar sets?

Pitfalls / Tricks

• $\in$ vs $\subseteq$. "$1 \in \{1, 2\}$" is true. "$\{1\} \in \{1, 2\}$" is false. "$\{1\} \subseteq \{1, 2\}$" is true. Membership and containment are not the same.
• $\varnothing \subseteq A$ always , even if you cannot picture it. Use the implication-with-no-instance argument.
• To prove $A = B$, the double inclusion is the cleanest tool: prove $A \subseteq B$ and $B \subseteq A$ separately.
• Insight. Subset is the mathematician's less-than-or-equal-to for sets. It is reflexive, transitive and antisymmetric , exactly the three properties that make $\le$ an order on numbers.