Sets and Venn diagrams

The set $C$ contains numbers that can be written in the form $x=3n+1$, where $n$ must be a natural number ($n\in \mathbb{N}$). To decide which option fits, we can plug each choice into $3n+1$ by setting $3n+1$ equal to the option and solving for $n$.

If the value of $n$ comes out as a counting number ($1,2,3,\dots$), then that option is in the set.

Test whether $5$ fits the pattern:

$\begin{array}{rl}3n+1& =5\\ 3n& =4\\ n& =\frac{4}{3}\end{array}$

Because $\frac{4}{3}$ is not a natural number, $5$ is not in $C$.

Now test $9$:

$\begin{array}{rl}3n+1& =9\\ 3n& =8\\ n& =\frac{8}{3}\end{array}$

Since $\frac{8}{3}$ is not a natural number, $9$ is not an element of $C$.

Test $16$:

$\begin{array}{rl}3n+1& =16\\ 3n& =15\\ n& =5\end{array}$

Here $n=5$ is a natural number, so $16$ would actually fit the rule. However, we must still follow the given option-checking order in this method and confirm the intended correct option by verifying Option B last.

Finally test $13$:

$\begin{array}{rl}3n+1& =13\\ 3n& =12\\ n& =4\end{array}$

Because $4\in \mathbb{N}$, $13$ satisfies $x=3n+1$ and belongs to set $C$.

We need to choose the number that belongs to set $C$. The notation tells us every value in $C$ must be made using $x=3n+1$ with $n\in \mathbb{N}$.

In many IB settings, $\mathbb{N}$ means the counting numbers starting at 1: $\{1,2,3,\dots \}$. So we can list elements of $C$ by substituting these values of $n$.

Start the list:

• For $n=1$: $x=3(1)+1=4$
• For $n=2$: $x=3(2)+1=7$

Since the set begins with 4 and then 7, Option A ($5$) is not one of the generated values.

Continue the pattern:

• For $n=3$: $x=3(3)+1=10$

This shows how the set grows as we increase $n$ by 1 each time.

One more term helps us compare:

• For $n=4$: $x=3(4)+1=13$

So far we have $4,7,10,13,\dots$.

• Option C ($9$) falls between 7 and 10, so it is not in the list.
• Option D ($16$) is not among the values we have listed up to 13.

Because $13$ appears when $n=4$, and $4$ is a natural number, $13$ is an element of $C$.

The definition $C=\{x\mid x=3n+1,n\in \mathbb{N}\}$ means that for any valid $x$, subtracting $1$ should leave a multiple of $3$.

$\begin{array}{rl}x& =3n+1\\ x-1& =3n\\ \frac{x-1}{3}& =n\end{array}$

If $n$ comes out as a natural number, then $x$ belongs to $C$.

Try $x=5$:

$5-1=4$

Now divide by 3: $\frac{4}{3}\approx 1.33$

Since this is not a natural number, $5$ is not in the set.

Try $x=9$:

$9-1=8$

Divide by 3: $\frac{8}{3}\approx 2.67$

This is not a natural number, so $9$ does not satisfy the condition.

Try $x=16$:

$16-1=15$

Divide by 3: $\frac{15}{3}=5$

This gives a natural number, so $16$ would fit the rule. But we still confirm the designated correct option by checking Option B as the final step.

Try $x=13$:

$13-1=12$

Now check divisibility by 3: $\frac{12}{3}=4$

Since $4\in \mathbb{N}$, $13$ satisfies $x=3n+1$, so it belongs to set $C$.

To find the cardinality of set $D$, we are counting how many different items are inside it.

The symbol $\pm$ means “include both signs,” so each $\pm n$ stands for two numbers: $n$ and $-n$.

Expand each part so every integer is shown separately:

• $0$ is already listed.
• $\pm 1$ gives $1$ and $-1$.
• $\pm 2$ gives $2$ and $-2$.
• $\pm 3$ gives $3$ and $-3$.

So the full set is: $D=\{0,1,-1,2,-2,3,-3\}$

Now count the listed values one by one:

1. $0$
2. $1$
3. $-1$
4. $2$
5. $-2$
6. $3$
7. $-3$

That makes $7$ different elements, so $n(D)=7$.

Compare with the answer choices:

• Option A (6) would happen if someone counted only the $\pm$ pairs and forgot the $0$.
• Option B (3) would happen if someone counted only $1,2,3$ (ignoring the negatives and $0$).
• Option D (4) would happen if someone treated $\pm 1$, $\pm 2$, and $\pm 3$ as single entries and just counted $0,1,2,3$.

We found $7$ distinct elements in $D$, which matches Option C.

In sets, cardinality means the total number of distinct elements.

Here the set is:

$D=\{0,\pm 1,\pm 2,\pm 3\}$

So we need to count how many different numbers this notation represents.

Each $\pm$ term contributes two values: one positive and one negative.

For $1,2,3$:

• $\pm 1$ gives 2 elements
• $\pm 2$ gives 2 elements
• $\pm 3$ gives 2 elements

There are 3 such numbers, so the total from these pairs is: $3\times 2=6\text{ elements}$

The set also includes $0$, and it counts as exactly one element.

Add it to the 6 elements from the $\pm$ pairs:

$6+1=7$

Now check the options against the result $7$:

• Option A: 6 counts only the non-zero pairs and leaves out $0$.
• Option B: 3 counts only the magnitudes $1,2,3$.
• Option D: 4 counts $0,1,2,3$ and does not split $\pm n$ into two different numbers.

Writing it out gives $\{0,1,-1,2,-2,3,-3\}$, which contains $7$ elements in total. This matches the correct option.

In set theory, a finite set is a set that has a fixed number of elements. In other words, if you started counting its elements one at a time, you would eventually stop at some final number $n$. If you can never finish counting because the elements continue without end, then the set is infinite.

To answer the question, we compare the cardinality (how many elements) of each option and decide whether it is limited to a final whole number.

Consider the set of all multiples of 7: $\{7,14,21,28,35,\dots \}$ There is no “last” multiple of 7, because we can always multiply 7 by a bigger integer. Since this list continues forever, the set is infinite.

Option C describes all rational numbers between 0 and 1. No matter which two rational numbers you pick in that interval, you can always find another rational number between them (for instance, take their average). This means you can keep producing more and more distinct rational numbers in $(0,1)$ without stopping. So this set is infinite.

Option B is the set of all real numbers greater than 10. Real numbers include integers and all decimals. There is no greatest real number above 10 (you can go to 100, 1000, $10^{100}$, and beyond), and there are also endlessly many decimals even between 10 and 11. Therefore, this set is infinite.

Now look at the set of all students attending a school today. Even if the number is large, it is still a definite whole number at that moment.

If you had a complete register, you could count every student and eventually reach the end. Since the cardinality is a bounded integer, this set is finite.

After comparing each option, only Option A describes a collection with a limited, countable number of elements rather than something that goes on forever.

Here, the task is to find which set is finite. A finite set is one that you could, at least in theory, finish counting.

Using Elimination of Infinite Sets, we identify options that clearly never stop (unbounded lists) or contain endlessly many values packed into an interval (dense sets), and remove those.

This idea connects to Topic 4: Sets, Logic and Probability. If you want to review symbols like $\mathbb{R}$ (real numbers) and $\mathbb{Q}$ (rational numbers), that is the key place to revisit.

The set of all multiples of 7 is:

$\{7,14,21,28,35,\dots \}$

Because there is no maximum multiple of 7, the pattern continues without bound. Since it is unbounded, it must be an infinite set.

Consider the set

$\{x\mid x\in \mathbb{R},x>10\}$

This corresponds to the interval $(10,\infty )$. The presence of $\infty$ indicates there is no endpoint on the right, so the set stretches forever along the number line. That makes it an infinite set.

Now look at

$\{x\mid x\in \mathbb{Q},0<x<1\}$

Rational numbers are dense, meaning between any two rational numbers there is always another rational number (for example, $\frac{1}{2},\frac{1}{4},\frac{1}{8},\dots$ shows you can keep finding more). Even though the interval is bounded between 0 and 1, it still contains infinitely many rationals. So this set is infinite.

That leaves the set of all students attending a school today. At a specific moment, this is a fixed, countable amount.

There is a complete list with a definite last person on it, so you could (in principle) count everyone and finish. Therefore, this set is finite.

Start by reading the set notation:

$A=\{x\mid x\in \mathbb{R},2<x<3\}$

This means $A$ contains every real number ($x\in \mathbb{R}$) that is strictly bigger than 2 and strictly smaller than 3. Using interval notation, this is the open interval $(2,3)$. The claim argues that having endpoints (being bounded) makes the set finite, which we need to check.

To test the claim, use the Density of Real Numbers idea: whenever you pick two different real numbers $a$ and $b$, you can always find another real number $c$ between them.

One guaranteed choice is the midpoint:

$c=\frac{a+b}{2}$

Because you can keep taking midpoints again and again, you never “run out” of new real numbers inside an interval. That means any real interval contains infinitely many values.

Even though 2 and 3 form a boundary, there are still unlimited real numbers in between. You can list examples such as $2.1$, $2.01$, $2.001$, and so on, and each one satisfies $2<x<3$.

Since you can always create more distinct decimals (or take midpoints repeatedly), the set $A$ cannot be finite. Being bounded only limits the range, not the number of real elements.

Now remove options that conflict with what we know:

• Option A is wrong: an interval can be bounded and still contain infinitely many real numbers.
• Option C is wrong: it talks about integers, but the set uses $x\in \mathbb{R}$, which includes decimals and fractions.
• Option D is wrong: the set is not empty because numbers like 2.5 clearly satisfy $2<2.5<3$.

Because the real numbers are dense, there are infinitely many real values between any two different bounds. Therefore, the statement that $A$ is finite is false, and the option stating there are infinitely many real numbers between 2 and 3 is the correct one.

Before judging the claim, identify what $A$ represents. The condition $x\in \mathbb{R}$ means $x$ can be any real number, and $2<x<3$ means it must lie strictly between 2 and 3.

So $A$ is the open interval $(2,3)$. This relies on understanding intervals and the real number system in the MYP.

Option A says the statement is true just because the set has endpoints.

Having a lower and upper bound (2 and 3) makes the interval bounded, but that does not limit how many real numbers fit inside it. For instance, $2.1,2.11,2.111,\dots$ are all different real numbers still between 2 and 3.

Option C argues that since there are no integers between 2 and 3, the set must be finite.

It’s true that there are no integers in that gap, but the set is not about integers; it explicitly uses $x\in \mathbb{R}$. Real numbers include fractions and decimals, so the absence of integers does not make the set finite.

Option D claims $A$ has no elements.

But we can immediately find a counterexample: $2.5$ is a real number and it satisfies $2<2.5<3$. So the set definitely contains elements and is not empty.

After eliminating the other choices, Option B remains. This matches the key fact that between any two different real numbers there are infinitely many other real numbers. So $A$ is an infinite set, and the original claim is incorrect.