Subsets of the set of real numbers

The nested number systems. Every real number belongs to one of several progressively larger families:

where $\mathbb{N} = \{1, 2, 3, \dots\}$ is the natural numbers, $\mathbb{Z} = \{\dots, -2, -1, 0, 1, 2, \dots\}$ adds zero and the negatives, $\mathbb{Q}$ is every number expressible as $\frac{p}{q}$ with $p, q \in \mathbb{Z}$ and $q \ne 0$, and $\mathbb{R}$ additionally contains the irrationals — numbers like $\sqrt{2}$ and $\pi$ that cannot be written as such a fraction. Each inclusion is proper (the symbol $\subsetneq$, not just $\subseteq$): $0$ is in $\mathbb{Z}$ but not $\mathbb{N}$, $\frac{1}{2}$ is in $\mathbb{Q}$ but not $\mathbb{Z}$, and $\sqrt{2}$ is in $\mathbb{R}$ but not $\mathbb{Q}$. A common slip is assuming rationals already "fill up" the number line — they do not, since between any two rationals sits an irrational, and vice versa.

Intervals as subsets of $\mathbb{R}$. An interval collects every real number lying between two fixed endpoints $a$ and $b$, with $a < b$. There are four bounded types, distinguished only by whether each endpoint is included:

• Open: $(a, b) = \{x \in \mathbb{R} : a < x < b\}$, both endpoints excluded.
• Closed: $[a, b] = \{x \in \mathbb{R} : a \le x \le b\}$, both endpoints included.
• Half-open, left-excluded: $(a, b] = \{x \in \mathbb{R} : a < x \le b\}$.
• Half-open, right-excluded: $[a, b) = \{x \in \mathbb{R} : a \le x < b\}$.

A round bracket always means "strictly beyond, endpoint not counted"; a square bracket always means "up to and including." Even though $(a,b)$ looks like a short segment on paper, it is an infinite set — between any two of its points sit infinitely many more real numbers.

Unbounded intervals. Some intervals extend forever in one direction, using $\infty$ and $-\infty$ as boundary markers. Since infinity is not an actual number that a set can "contain," it always sits beside a round bracket: $(a, \infty) = \{x \in \mathbb{R} : x > a\}$, $[a, \infty) = \{x \in \mathbb{R} : x \ge a\}$, $(-\infty, b) = \{x \in \mathbb{R} : x < b\}$, $(-\infty, b] = \{x \in \mathbb{R} : x \le b\}$, and the whole real line is $(-\infty, \infty) = \mathbb{R}$. The length of a bounded interval with endpoints $a$ and $b$ is simply $b - a$, regardless of whether the brackets are round or square.

Let $A = [3, 8)$ and $B = [0, 10]$. Using the rule just derived, determine whether $A \subseteq B$, and separately determine whether $A$ is a proper subset.

Step 1. Write $A$ in set-builder form to identify its endpoints: $[3, 8) = \{x \in \mathbb{R} : 3 \le x < 8\}$, so its left endpoint is $3$ (included) and its effective right boundary is $8$ (excluded). Compare against $B = [0, 10] = \{x \in \mathbb{R} : 0 \le x \le 10\}$.

Step 2. Apply the endpoint test. The left-endpoint condition needs $0 \le 3$, which holds. The right-boundary condition needs every $x < 8$ to also satisfy $x \le 10$; since $8 \le 10$, this holds too. Both conditions are satisfied, so $A \subseteq B$.

Step 3. Check for a proper subset by finding an element of $B$ that is not in $A$. Take $x = 0$: it satisfies $0 \le 0 \le 10$, so $0 \in B$, but $0 < 3$ means $0 \notin A$. Since $B$ has at least one element $A$ lacks, $A$ is a proper subset of $B$: $A \subsetneq B$.

Step 4 (competitive stretch). Now find the interval $A \cap B$ (their common elements) and $A \cup B$ (everything in either). Since $A \subseteq B$, every element of $A$ is already in $B$, so $A \cap B = A = [3, 8)$ and $A \cup B = B = [0, 10]$. This is a general fact: whenever $A \subseteq B$, $A \cap B = A$ and $A \cup B = B$ — a shortcut worth remembering instead of recomputing from scratch.

• Swapping round and square brackets: a round bracket always excludes its endpoint, a square bracket always includes it — mixing them up flips the whole answer.
• Writing $[a, \infty]$ or $(-\infty, b]$-style errors with a square bracket touching an infinity symbol — infinity is never an included endpoint, so it must always sit next to a round bracket.
• Checking only one endpoint when testing $[c,d] \subseteq [a,b]$ — both $a \le c$ and $d \le b$ must hold; satisfying only one is not enough.
• Reversing the inclusion chain, e.g. writing $\mathbb{Q} \subsetneq \mathbb{Z}$ — the correct order is smallest to largest: $\mathbb{N} \subsetneq \mathbb{Z} \subsetneq \mathbb{Q} \subsetneq \mathbb{R}$.

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