Finite and infinite sets

Finite sets terminate; infinite sets do not. A set $A$ is called finite if it is either empty or its elements can be put in correspondence with $1, 2, 3, \dots, n$ for some fixed natural number $n$ and the counting stops there. Whenever that happens, $n$ itself is a well-defined quantity attached to $A$:

where: $n(A)$ is the cardinal number (or order) of the finite set $A$ — simply the count of distinct elements it contains, and $n$ is the natural number at which counting terminates. By definition $\varnothing$ is finite too, with $n(\varnothing) = 0$; it is not a separate, third category sitting outside "finite" and "infinite." A set is infinite precisely when it fails this test — no matter how large a natural number you reach while counting, there is always another element left over. $\mathbb{N} = \{1, 2, 3, \dots\}$ is the standard example: whatever number you count up to, $n+1$ is still in the set.

Dots signal a pattern, not automatically "infinite." Roster form is allowed to use "$\dots$" to shorten a long but finite list, as long as both the pattern and the stopping point are clear — $\{1, 2, 3, \dots, 100\}$ is a perfectly finite set of 100 elements. An infinite set, by contrast, uses "$\dots$" with no stopping point on at least one side: $\{1, 2, 3, \dots\}$ has no last term, and $\{\dots, -2, -1, 0, 1, 2, \dots\}$ (the integers) has no first or last term at all. Reading the dots correctly — pattern and whether they stop — is the actual skill being tested, not the mere presence of dots.

The domain decides it, not the shape of the condition. A set-builder condition can look open-ended and still describe a finite set, or look bounded and still be infinite, depending on what values the variable is allowed to range over:

The rule "domain plus bound decides finiteness" is not a memorised trick — it follows from what counting actually means, and reasoning it through avoids the two most common mix-ups.

Discrete domains have a next element; continuous ones do not. In $\mathbb{N}$ or $\mathbb{Z}$, every element has a definite next neighbour ($5$ is followed by $6$, nothing in between). So the moment a condition also fixes an upper limit — even one stated as an inequality like $x^2 < 50$ — you can march $1, 2, 3, \dots$ forward and you are guaranteed to hit a value that fails the condition after finitely many steps, because each step moves to the very next integer with nothing skipped. That guarantees termination, which is exactly the definition of finite. This is why $\{x : x \in \mathbb{N}, x^2 < 50\}$ is finite even though the inequality itself never explicitly names a stopping number.

Unbounded discrete domains never run out. If instead the condition places no upper (or no lower) limit on a discrete variable — $\{x : x \in \mathbb{Z}, x < 5\}$ has a top but no bottom — then in the unbounded direction there is always a next element to add ($4, 3, 2, 1, 0, -1, -2, \dots$), so the counting process can never be declared complete. One unbounded side is enough to make the whole set infinite; you do not need both sides open.

Continuous domains fail even when bounded on both sides. Between any two distinct real numbers there sits another real number (and in fact infinitely many), so a "next element" does not exist at all — counting cannot even begin in a meaningful step-by-step sense. That is why $\{x \in \mathbb{R} : 0 < x < 1\}$ is infinite despite being squeezed into a bounded interval: boundedness alone is not the deciding factor, discreteness with a limit is. So the working test becomes: identify the domain first (is it $\mathbb{N}$, $\mathbb{Z}$, or $\mathbb{R}$-like and continuous), then check whether the condition bounds it in every direction that matters for that domain. Skipping the first step — jumping straight to "there's an inequality, so it must be finite" — is exactly the trap the exam sets.

Classify each set as finite or infinite, and give the cardinal number where finite: $P = \{x : x \in \mathbb{N}, x \text{ is a factor of } 36\}$, $Q = \{x : x \in \mathbb{N}, x^2 < 50\}$, and $R = \{x : x \in \mathbb{R}, x^2 < 50\}$.

Step 1. Find $P$. The factor pairs of 36 are $1 \times 36$, $2 \times 18$, $3 \times 12$, $4 \times 9$, $6 \times 6$, giving the distinct factors $1, 2, 3, 4, 6, 9, 12, 18, 36$. That is a fixed list of 9 numbers with nothing more to add, so

$P$ is finite.

Step 2. Find $Q$. The domain is $\mathbb{N}$, which is discrete, and the condition $x^2 < 50$ bounds it above. Testing consecutive naturals: $1^2=1$, $2^2=4$, $3^2=9$, $4^2=16$, $5^2=25$, $6^2=36$, $7^2=49 < 50$, but $8^2 = 64 > 50$. So $Q = \{1,2,3,4,5,6,7\}$ and $n(Q) = 7$. $Q$ is finite — exactly as the "Why it works" reasoning predicts for a discrete domain with an upper bound.

Step 3. Find $R$. Here the domain is $\mathbb{R}$, which is continuous, so the same-looking condition $x^2 < 50$ describes the interval $-\sqrt{50} < x < \sqrt{50}$, roughly $(-7.07, 7.07)$. Between any two points of this interval there are infinitely many more real numbers, so no counting process can exhaust it. $R$ is infinite, even though it is a bounded interval and looks, on the surface, exactly like $Q$'s condition.

Answer: $P$ is finite with $n(P) = 9$; $Q$ is finite with $n(Q) = 7$; $R$ is infinite — the identical inequality $x^2 < 50$ gives opposite verdicts once the domain switches from $\mathbb{N}$ to $\mathbb{R}$.

• Assuming every inequality-defined set must be finite — check the domain ($\mathbb{N}$/$\mathbb{Z}$ vs $\mathbb{R}$) before deciding, since $\mathbb{R}$-conditions like $0 < x < 1$ stay infinite even when bounded.
• Treating "$\dots$" as automatic proof of infinity — $\{1, 2, \dots, 100\}$ is finite; only an unterminated "$\dots$" (or a continuous domain) signals infinite.
• Forgetting $\varnothing$ is finite with $n(\varnothing) = 0$, not a special exception outside the finite/infinite split.
• Declaring a set infinite just because it is "hard to write out" — an infinite set can have an exact, unambiguous set-builder description; being infinite is about the counting process, not about vagueness.

Ask the tutor: if you have a set-builder condition you're unsure about, ask the AI tutor to check the domain and bound with you and count out $n(A)$ step by step.