Chapter 1: Sets

The set on the right is $$\{P, R, I, N, C, A, L\}$$, a collection of letters.

Among the four set-builder descriptions on the left, the only one that describes a set of letters is (d) $$\{x : x \text{ is a letter of the word PRINCIPAL}\}$$.

The word PRINCIPAL is made up of the letters P, R, I, N, C, I, P, A, L; removing the repeated letters gives the distinct letters $$\{P, R, I, N, C, A, L\}$$, which is exactly this set.

Hence this set matches with (d).

The set on the right is $$\{0\}$$, whose only element is the integer $$0$$.

Test the set-builder description (c) $$\{x : x \text{ is an integer and } x + 1 = 1\}$$: solving $$x + 1 = 1$$ gives $$x = 0$$, which is an integer, so (c) describes exactly the set $$\{0\}$$.

The remaining descriptions give $$\{1, 2, 3, 6, 9, 18\}$$ for (a), $$\{3, -3\}$$ for (b) and a set of letters for (d) — none of which equals $$\{0\}$$.

Hence this set matches with (c).

The set on the right is $$\{1, 2, 3, 6, 9, 18\}$$.

These are precisely the positive integers that divide $$18$$: since $$18 = 1 \times 18 = 2 \times 9 = 3 \times 6$$, its divisors are $$1, 2, 3, 6, 9, 18$$.

This matches description (a) $$\{x : x \text{ is a positive integer and is a divisor of } 18\}$$.

Hence this set matches with (a).

The set on the right is $$\{3, -3\}$$.

Consider description (b) $$\{x : x \text{ is an integer and } x^2 - 9 = 0\}$$. Solving $$x^2 - 9 = 0$$ gives $$x^2 = 9$$, so $$x = 3$$ or $$x = -3$$; both are integers, giving the set $$\{3, -3\}$$.

Hence this set matches with (b).

The months beginning with the letter J are well-defined: January, June and July. Membership can be decided without ambiguity for every month.

Hence the collection is a set.

``Most talented'' is a subjective criterion — different people will pick different writers. Therefore the collection is not well-defined.

Hence it is not a set.

``Best cricket batsmen'' is subjective; different people would name different players. So the collection is not well-defined.

Hence it is not a set.

Whether or not a given person is a boy in the class is a definite, well-defined property.

Hence the collection is a set.

For any given natural number we can clearly decide whether it is less than $$100$$ or not. The criterion is well-defined.

Hence it is a set, namely $$\{1, 2, 3, \ldots, 99\}$$.

It can be unambiguously decided whether a particular novel was written by Munshi Prem Chand. The criterion is well-defined.

Hence the collection is a set.

For any integer we can unambiguously decide whether it is even (divisible by $$2$$) or not. The criterion is well-defined.

Hence the collection is a set, $$\{\ldots, -4, -2, 0, 2, 4, \ldots\}$$.

Each question printed in the chapter can be identified unambiguously. The criterion is well-defined.

Hence the collection is a set.

``Most dangerous'' is subjective; different people will disagree on which animals belong. So the collection is not well-defined.

Hence it is not a set.

$$A = \{1, 2, 3, 4, 5, 6\}$$ and $$5$$ is listed in $$A$$, so $$5$$ belongs to $$A$$.

Hence $$5 \in A$$.

$$8$$ is not listed among the elements $$1, 2, 3, 4, 5, 6$$ of $$A$$, so $$8$$ does not belong to $$A$$.

Hence $$8 \notin A$$.

$$0$$ is not in the list $$\{1, 2, 3, 4, 5, 6\}$$, so $$0$$ does not belong to $$A$$.

Hence $$0 \notin A$$.

$$4$$ is listed in $$A = \{1, 2, 3, 4, 5, 6\}$$, so $$4$$ belongs to $$A$$.

Hence $$4 \in A$$.

$$2$$ is listed in $$A = \{1, 2, 3, 4, 5, 6\}$$, so $$2$$ belongs to $$A$$.

Hence $$2 \in A$$.

$$10$$ is not in the list $$\{1, 2, 3, 4, 5, 6\}$$, so $$10$$ does not belong to $$A$$.

Hence $$10 \notin A$$.

We list all integers strictly greater than $$-3$$ and strictly less than $$7$$:

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

Hence in roster form,

$$A = \{-2, -1, 0, 1, 2, 3, 4, 5, 6\}.$$

Natural numbers strictly less than $$6$$ are $$1, 2, 3, 4, 5$$.

Hence in roster form, $$B = \{1, 2, 3, 4, 5\}$$.

A two-digit number has the form $$\overline{ab}$$ with $$a \in \{1, 2, \ldots, 9\}$$ and $$b \in \{0, 1, \ldots, 9\}$$. We need $$a + b = 8$$.

Possible $$(a, b)$$ pairs: $$(1, 7), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (7, 1), (8, 0)$$.

The corresponding numbers are $$17, 26, 35, 44, 53, 62, 71, 80$$.

Hence $$C = \{17, 26, 35, 44, 53, 62, 71, 80\}$$.

Factorise: $$60 = 2^2 \times 3 \times 5$$.

The prime divisors of $$60$$ are $$2, 3, 5$$.

Hence $$D = \{2, 3, 5\}$$.

The letters of ``TRIGONOMETRY'' are T, R, I, G, O, N, O, M, E, T, R, Y. Remove repeated letters: T, R, I, G, O, N, M, E, Y.

Hence $$E = \{T, R, I, G, O, N, M, E, Y\}$$.

The letters of ``BETTER'' are B, E, T, T, E, R. Remove repetitions: B, E, T, R.

Hence $$F = \{B, E, T, R\}$$.

The elements $$3, 6, 9, 12$$ are multiples of $$3$$, specifically $$3 \cdot 1, 3 \cdot 2, 3 \cdot 3, 3 \cdot 4$$.

Hence in set-builder form, $$\{3, 6, 9, 12\} = \{x : x = 3n,\ n \in \mathbf{N},\ 1 \leq n \leq 4\}$$.

The elements are powers of $$2$$: $$2 = 2^1,\ 4 = 2^2,\ 8 = 2^3,\ 16 = 2^4,\ 32 = 2^5$$.

Hence in set-builder form, $$\{2, 4, 8, 16, 32\} = \{x : x = 2^n,\ n \in \mathbf{N},\ 1 \leq n \leq 5\}$$.

The elements are powers of $$5$$: $$5 = 5^1,\ 25 = 5^2,\ 125 = 5^3,\ 625 = 5^4$$.

Hence in set-builder form, $$\{5, 25, 125, 625\} = \{x : x = 5^n,\ n \in \mathbf{N},\ 1 \leq n \leq 4\}$$.

The elements are all positive even integers, i.e. multiples of $$2$$ starting from $$2$$.

Hence in set-builder form, $$\{2, 4, 6, \ldots\} = \{x : x \text{ is an even natural number}\}$$, or equivalently $$\{x : x = 2n,\ n \in \mathbf{N}\}$$.

The elements are perfect squares: $$1 = 1^2,\ 4 = 2^2,\ 9 = 3^2, \ldots, 100 = 10^2$$.

Hence in set-builder form, $$\{1, 4, 9, \ldots, 100\} = \{x : x = n^2,\ n \in \mathbf{N},\ 1 \leq n \leq 10\}$$.

Odd natural numbers are $$1, 3, 5, 7, \ldots$$

Hence $$A = \{1, 3, 5, 7, 9, \ldots\}$$.

We need integers $$x$$ with $$-\dfrac{1}{2} < x < \dfrac{9}{2}$$, i.e. $$-0.5 < x < 4.5$$.

The integers in this range are $$0, 1, 2, 3, 4$$.

Hence $$B = \{0, 1, 2, 3, 4\}$$.

$$x^2 \leq 4 \iff -2 \leq x \leq 2$$.

The integers satisfying this are $$-2, -1, 0, 1, 2$$.

Hence $$C = \{-2, -1, 0, 1, 2\}$$.

The letters of ``LOYAL'' are L, O, Y, A, L. Removing the repeated L, the distinct letters are L, O, Y, A.

Hence $$D = \{L, O, Y, A\}$$.

Months with $$31$$ days: January, March, May, July, August, October, December.

The remaining months (not having $$31$$ days) are February ($$28$$ or $$29$$), April, June, September, November (each having $$30$$ days).

Hence $$E = \{\text{February, April, June, September, November}\}$$.

The English letters that precede $$k$$ are $$a, b, c, d, e, f, g, h, i, j$$. Discarding the vowels $$a, e, i$$, the consonants are $$b, c, d, f, g, h, j$$.

Hence $$F = \{b, c, d, f, g, h, j\}$$.

The set on the right is $$\{1, 2, 3, 6\}$$.

These are exactly the natural numbers that divide $$6$$: since $$6 = 1 \times 6 = 2 \times 3$$, its divisors are $$1, 2, 3, 6$$.

This matches description (c) $$\{x : x \text{ is a natural number and divisor of } 6\}$$. (Description (a) asks only for the prime divisors, which would give just $$\{2, 3\}$$.)

Hence this set matches with (c).

The set on the right is $$\{2, 3\}$$.

The divisors of $$6$$ are $$1, 2, 3, 6$$, and among these the prime numbers are $$2$$ and $$3$$.

So $$\{2, 3\}$$ is the set of prime numbers that are divisors of $$6$$, which is description (a) $$\{x : x \text{ is a prime number and a divisor of } 6\}$$.

Hence this set matches with (a).

The set on the right is $$\{M, A, T, H, E, I, C, S\}$$, a collection of letters.

The word MATHEMATICS is made up of the letters M, A, T, H, E, M, A, T, I, C, S; removing the repeated letters gives the distinct letters $$\{M, A, T, H, E, I, C, S\}$$.

This matches description (d) $$\{x : x \text{ is a letter of the word MATHEMATICS}\}$$.

Hence this set matches with (d).

The set on the right is $$\{1, 3, 5, 7, 9\}$$.

These are exactly the odd natural numbers less than $$10$$.

This matches description (b) $$\{x : x \text{ is an odd natural number less than } 10\}$$.

Hence this set matches with (b).

$$(x - 1)(x - 2) = 0 \Rightarrow x = 1$$ or $$x = 2$$. Both are natural numbers.

The set is $$\{1, 2\}$$, which has only two elements.

Hence it is a finite set.

$$x^2 = 4 \Rightarrow x = \pm 2$$. Out of these only $$x = 2$$ is a natural number.

The set is $$\{2\}$$, which has one element.

Hence it is a finite set.

$$2x - 1 = 0 \Rightarrow x = \dfrac{1}{2}$$, which is not a natural number.

So the set has no elements; it is the empty set $$\phi$$, which is finite.

There are infinitely many prime natural numbers (Euclid's theorem), $$2, 3, 5, 7, 11, \ldots$$

Hence the set is an infinite set.

The odd natural numbers are $$1, 3, 5, 7, \ldots$$ — there are infinitely many.

Hence the set is an infinite set.

Write in roster form. Letters of ALLOY: A, L, L, O, Y; distinct letters: $$X = \{A, L, O, Y\}$$.

Letters of LOYAL: L, O, Y, A, L; distinct letters: $$B = \{L, O, Y, A\} = \{A, L, O, Y\}$$.

Since both sets consist of exactly the same elements, $$X = B$$.

For $$A$$: $$n^2 \leq 4 \iff -2 \leq n \leq 2$$. Integers in this range: $$A = \{-2, -1, 0, 1, 2\}$$.

For $$B$$: $$x^2 - 3x + 2 = 0 \Rightarrow (x - 1)(x - 2) = 0 \Rightarrow x = 1$$ or $$x = 2$$. So $$B = \{1, 2\}$$.

$$A$$ has $$5$$ elements and $$B$$ has $$2$$ elements, so $$A \neq B$$.

An odd number is not divisible by $$2$$ (by definition of ``odd''). So no natural number is both odd and divisible by $$2$$.

Hence the set has no elements; it is the null set.

The number $$2$$ is even and prime. So the set is $$\{2\}$$, which is non-empty.

Hence it is not a null set.

No number can simultaneously be less than $$5$$ and greater than $$7$$.

Hence the set is empty; it is the null set.

Two parallel lines (in a plane) never meet, so they have no common point.

Hence the set is empty; it is the null set.

There are exactly $$12$$ months in a year.

Hence the set is finite.

This is the set of all natural numbers, which continues without end.

Hence it is an infinite set.

The set contains exactly $$100$$ elements.

Hence it is finite.

The set is $$\{101, 102, 103, \ldots\}$$, which continues without end.

Hence it is an infinite set.

The primes less than $$99$$ form a finite list: $$2, 3, 5, 7, 11, \ldots, 97$$ (there are $$25$$ such primes).

Hence the set is finite.

For each real number $$c$$, the line $$y = c$$ is parallel to the $$x$$-axis. Since there are infinitely many real numbers $$c$$, there are infinitely many such lines.

Hence the set is infinite.

The English alphabet has $$26$$ letters.

Hence the set is finite.

The multiples of $$5$$ are $$\ldots, -10, -5, 0, 5, 10, 15, \ldots$$, an unending list.

Hence the set is infinite.

Although very large, the number of animals living on the earth at any given moment is a definite (finite) number.

Hence the set is finite.

For each positive radius $$r$$ and each centre $$(h, k)$$ with $$h^2 + k^2 = r^2$$, there is a distinct circle through the origin. There are infinitely many such choices.

Hence the set of such circles is infinite.

Order of listing does not matter in a set. $$A$$ and $$B$$ contain exactly the same elements $$a, b, c, d$$.

Hence $$A = B$$.

$$12 \in A$$ but $$12 \notin B$$. Also $$18 \in B$$ but $$18 \notin A$$.

Hence $$A \neq B$$.

The positive even integers $$x$$ with $$x \leq 10$$ are $$2, 4, 6, 8, 10$$, so $$B = \{2, 4, 6, 8, 10\}$$.

Hence $$A = B$$.

$$A = \{10, 20, 30, 40, \ldots\}$$ (multiples of $$10$$). $$B$$ also contains $$15$$, but $$15$$ is not a multiple of $$10$$. So $$15 \in B$$ while $$15 \notin A$$.

Hence $$A \neq B$$.

Solve $$x^2 + 5x + 6 = 0$$: $$(x + 2)(x + 3) = 0 \Rightarrow x = -2$$ or $$x = -3$$.

So $$B = \{-2, -3\}$$, whereas $$A = \{2, 3\}$$.

The elements are different, so $$A \neq B$$.

Distinct letters of FOLLOW: F, O, L, W. So $$A = \{F, O, L, W\}$$.

Distinct letters of WOLF: W, O, L, F. So $$B = \{W, O, L, F\} = \{F, O, L, W\}$$.

Both sets contain exactly the same letters.

Hence $$A = B$$.

The empty set is a subset of every set.

Hence $$\phi \subset B$$.

$$A = \{1, 3\}$$, $$B = \{1, 5, 9\}$$. The element $$3 \in A$$ but $$3 \notin B$$.

Hence $$A \not\subset B$$.

$$A = \{1, 3\}$$, $$C = \{1, 3, 5, 7, 9\}$$. Both elements $$1, 3$$ of $$A$$ are in $$C$$.

Hence $$A \subset C$$.

$$B = \{1, 5, 9\}$$, $$C = \{1, 3, 5, 7, 9\}$$. Every element of $$B$$ ($$1, 5, 9$$) is in $$C$$.

Hence $$B \subset C$$.

Each of $$2, 3, 4$$ is an element of $$\{1, 2, 3, 4, 5\}$$.

Hence $$\{2, 3, 4\} \subset \{1, 2, 3, 4, 5\}$$.

The element $$a \in \{a, b, c\}$$ but $$a \notin \{b, c, d\}$$.

Hence $$\{a, b, c\} \not\subset \{b, c, d\}$$.

Every Class XI student of the school is, in particular, a student of the school.

Hence the first set is a subset of the second; ``$$\subset$$''.

The first set contains circles of every radius (e.g. radius $$2$$). Such a circle is not of radius $$1$$, so it is not in the second set.

Hence the first set is not contained in the second; ``$$\not\subset$$''.

A triangle (a $$3$$-sided polygon) is never a rectangle (a $$4$$-sided polygon). So no triangle lies in the set of rectangles.

Hence ``$$\not\subset$$''.

Every equilateral triangle is, in particular, a triangle.

Hence ``$$\subset$$''.

Every even natural number ($$2, 4, 6, \ldots$$) is an integer.

Hence ``$$\subset$$''.

$$a \in \{b, c, a\}$$ and $$b \in \{b, c, a\}$$. So every element of $$\{a, b\}$$ lies in $$\{b, c, a\}$$, meaning $$\{a, b\} \subset \{b, c, a\}$$.

The given statement claims the opposite, so it is false.

The set of vowels is $$\{a, e, i, o, u\}$$. Both $$a$$ and $$e$$ are in this set.

Hence $$\{a, e\} \subset \{a, e, i, o, u\}$$ — the statement is true.

$$2 \in \{1, 2, 3\}$$ but $$2 \notin \{1, 3, 5\}$$.

Hence the statement is false.

The only element of $$\{a\}$$ is $$a$$, and $$a \in \{a, b, c\}$$.

Hence the statement is true.

The elements of $$\{a, b, c\}$$ are $$a, b, c$$ (not the set $$\{a\}$$).

Hence $$\{a\} \notin \{a, b, c\}$$; the statement is false.

Left set: even natural numbers $$< 6$$ are $$\{2, 4\}$$.

Right set: natural-number divisors of $$36$$ are $$\{1, 2, 3, 4, 6, 9, 12, 18, 36\}$$.

$$2 \in \{1, 2, 3, 4, 6, 9, 12, 18, 36\}$$ and $$4 \in \{1, 2, 3, 4, 6, 9, 12, 18, 36\}$$.

Hence the inclusion holds; the statement is true.

The elements of $$A$$ are $$1, 2, \{3, 4\}, 5$$. The numbers $$3$$ and $$4$$ are not themselves elements of $$A$$ — only the set $$\{3, 4\}$$ is.

For $$\{3, 4\} \subset A$$, we would need both $$3 \in A$$ and $$4 \in A$$, which is false.

Hence the statement is incorrect.

The set $$\{3, 4\}$$ is one of the listed elements of $$A = \{1, 2, \{3, 4\}, 5\}$$.

Hence $$\{3, 4\} \in A$$ is correct.

The only element of $$\{\{3, 4\}\}$$ is $$\{3, 4\}$$, and $$\{3, 4\} \in A$$.

Hence every element of $$\{\{3, 4\}\}$$ lies in $$A$$, so $$\{\{3, 4\}\} \subset A$$ is correct.

$$1$$ is listed as an element of $$A = \{1, 2, \{3, 4\}, 5\}$$.

Hence $$1 \in A$$ is correct.

``$$\subset$$'' is a relation between two sets, but $$1$$ is a number, not a set. So $$1 \subset A$$ is meaningless.

Hence the statement is incorrect.

Each of $$1, 2, 5$$ is a listed element of $$A = \{1, 2, \{3, 4\}, 5\}$$.

Hence $$\{1, 2, 5\} \subset A$$ is correct.

The set $$\{1, 2, 5\}$$ is not one of the listed elements of $$A$$ (the elements are $$1, 2, \{3, 4\}, 5$$).

Hence $$\{1, 2, 5\} \notin A$$; the statement is incorrect.

$$3 \in \{1, 2, 3\}$$ but the bare number $$3$$ is not a listed element of $$A$$ (only $$\{3, 4\}$$ is).

Hence $$\{1, 2, 3\} \not\subset A$$; the statement is incorrect.

The empty set $$\phi$$ is not one of the listed elements of $$A$$.

Hence $$\phi \notin A$$; the statement is incorrect.

The empty set is a subset of every set.

Hence $$\phi \subset A$$ is correct.

The only element of $$\{\phi\}$$ is $$\phi$$. For $$\{\phi\} \subset A$$ we need $$\phi \in A$$, but $$\phi$$ is not listed as an element of $$A$$.

Hence $$\{\phi\} \not\subset A$$; the statement is incorrect.

A set with $$n$$ elements has $$2^n$$ subsets. Here $$n = 1$$, so there are $$2^1 = 2$$ subsets: the empty subset and the set itself.

Subsets of $$\{a\}$$: $$\phi,\ \{a\}$$.

With $$n = 2$$ elements there are $$2^2 = 4$$ subsets.

Subsets of $$\{a, b\}$$: $$\phi,\ \{a\},\ \{b\},\ \{a, b\}$$.

With $$n = 3$$ elements there are $$2^3 = 8$$ subsets.

List them by size:

Size $$0$$: $$\phi$$.

Size $$1$$: $$\{1\},\ \{2\},\ \{3\}$$.

Size $$2$$: $$\{1, 2\},\ \{1, 3\},\ \{2, 3\}$$.

Size $$3$$: $$\{1, 2, 3\}$$.

Total: $$1 + 3 + 3 + 1 = 8$$ subsets.

The empty set has $$0$$ elements and hence $$2^0 = 1$$ subset — namely itself.

The only subset is $$\phi$$.

``$$<$$'' (strict) corresponds to an open endpoint ``$$($$''; ``$$\leq$$'' (non-strict) corresponds to a closed endpoint ``$$]$$''.

Hence the interval is $$(-4, 6]$$.

Both endpoints are strict, so the interval is open at both ends.

Hence the interval is $$(-12, -10)$$.

$$0$$ is included ($$\leq$$) and $$7$$ is excluded ($$<$$).

Hence the interval is $$[0, 7)$$.

Both endpoints are included.

Hence the interval is $$[3, 4]$$.

Round brackets mean the endpoints are excluded.

Hence $$(-3, 0) = \{x : x \in \mathbf{R},\ -3 < x < 0\}$$.

Square brackets mean the endpoints are included.

Hence $$[6, 12] = \{x : x \in \mathbf{R},\ 6 \leq x \leq 12\}$$.

The left endpoint $$6$$ is excluded; the right endpoint $$12$$ is included.

Hence $$(6, 12] = \{x : x \in \mathbf{R},\ 6 < x \leq 12\}$$.

The left endpoint $$-23$$ is included; the right endpoint $$5$$ is excluded.

Hence $$[-23, 5) = \{x : x \in \mathbf{R},\ -23 \leq x < 5\}$$.

The set of right triangles is a particular sub-collection of all triangles in a plane. A natural universal set that contains every right triangle is the set of all triangles in a plane (or, more broadly, the set of all polygons / plane figures).

Every isosceles triangle is a triangle. A suitable universal set is the set of all triangles in a plane.

For a universal set $$U$$ we need $$A \subset U$$, $$B \subset U$$ and $$C \subset U$$.

$$C = \{0, 2, 4, 6, 8\}$$ has $$8$$ as an element, but $$8 \notin \{0, 1, 2, 3, 4, 5, 6\}$$. So $$C \not\subset \{0, 1, 2, 3, 4, 5, 6\}$$.

Hence this is not a universal set for $$A, B, C$$.

The empty set has no elements, so it cannot contain the non-empty sets $$A$$, $$B$$, $$C$$ as subsets in a useful way (in fact $$A \not\subset \phi$$ since $$1 \in A,\ 1 \notin \phi$$).

Hence $$\phi$$ cannot serve as the universal set.

Check each set: $$A = \{1, 3, 5\} \subset \{0, 1, \ldots, 10\}$$ ✓, $$B = \{2, 4, 6\} \subset \{0, 1, \ldots, 10\}$$ ✓, $$C = \{0, 2, 4, 6, 8\} \subset \{0, 1, \ldots, 10\}$$ ✓.

All three are subsets, so this can serve as a universal set.

$$C = \{0, 2, 4, 6, 8\}$$ has $$0$$, but $$0 \notin \{1, 2, 3, 4, 5, 6, 7, 8\}$$.

So $$C \not\subset \{1, 2, 3, 4, 5, 6, 7, 8\}$$; it cannot serve as a universal set for all three.

$$X \cup Y$$ contains every element appearing in $$X$$ or $$Y$$, without repetition. Combining $$1, 3, 5$$ and $$1, 2, 3$$ gives $$1, 2, 3, 5$$.

Hence $$X \cup Y = \{1, 2, 3, 5\}$$.

Combining the elements of $$A$$ and $$B$$ without repetition:

$$A \cup B = \{a, b, c, e, i, o, u\}.$$

$$A = \{3, 6, 9, 12, 15, \ldots\}$$ (positive multiples of $$3$$).

$$B = \{1, 2, 3, 4, 5\}$$ (natural numbers $$< 6$$).

$$A \cup B$$ contains $$1, 2, 4, 5$$ (from $$B$$ only) together with every multiple of $$3$$. Note that $$3 \in A \cap B$$, so it appears only once.

$$A \cup B = \{1, 2, 3, 4, 5, 6, 9, 12, 15, \ldots\}$$.

In set-builder form: $$A \cup B = \{x : x \in \mathbf{N},\ x < 6 \text{ or } x \text{ is a multiple of } 3\}$$.

$$A = \{2, 3, 4, 5, 6\}$$ (natural numbers with $$1 < x \leq 6$$).

$$B = \{7, 8, 9\}$$ (natural numbers with $$6 < x < 10$$).

Combining gives $$A \cup B = \{2, 3, 4, 5, 6, 7, 8, 9\}$$.

In set-builder form: $$A \cup B = \{x : x \in \mathbf{N},\ 1 < x < 10\}$$.

The empty set has no elements, so adding it changes nothing: $$A \cup \phi = A$$.

Hence $$A \cup B = \{1, 2, 3\}$$.

Combine elements of $$A = \{1, 2, 3, 4\}$$ and $$B = \{3, 4, 5, 6\}$$ without repetition.

$$A \cup B = \{1, 2, 3, 4, 5, 6\}.$$

$$A = \{1, 2, 3, 4\}$$, $$C = \{5, 6, 7, 8\}$$. They share no element.

$$A \cup C = \{1, 2, 3, 4, 5, 6, 7, 8\}.$$

$$B = \{3, 4, 5, 6\}$$, $$C = \{5, 6, 7, 8\}$$. Common elements $$5, 6$$ are listed once.

$$B \cup C = \{3, 4, 5, 6, 7, 8\}.$$

$$B = \{3, 4, 5, 6\}$$, $$D = \{7, 8, 9, 10\}$$. They share no element.

$$B \cup D = \{3, 4, 5, 6, 7, 8, 9, 10\}.$$

From (i), $$A \cup B = \{1, 2, 3, 4, 5, 6\}$$. Now form union with $$C = \{5, 6, 7, 8\}$$ (adding the new elements $$7, 8$$).

$$A \cup B \cup C = \{1, 2, 3, 4, 5, 6, 7, 8\}.$$

From (i), $$A \cup B = \{1, 2, 3, 4, 5, 6\}$$. Union with $$D = \{7, 8, 9, 10\}$$ adds $$7, 8, 9, 10$$.

$$A \cup B \cup D = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}.$$

From (iii), $$B \cup C = \{3, 4, 5, 6, 7, 8\}$$. Union with $$D = \{7, 8, 9, 10\}$$ adds $$9, 10$$.

$$B \cup C \cup D = \{3, 4, 5, 6, 7, 8, 9, 10\}.$$

$$A = \{3, 5, 7, 9, 11\}$$, $$B = \{7, 9, 11, 13\}$$. Common elements: $$7, 9, 11$$.

$$A \cap B = \{7, 9, 11\}.$$

$$B = \{7, 9, 11, 13\}$$, $$C = \{11, 13, 15\}$$. Common elements: $$11, 13$$.

$$B \cap C = \{11, 13\}.$$

$$A \cap C$$: common elements of $$\{3, 5, 7, 9, 11\}$$ and $$\{11, 13, 15\}$$ is $$\{11\}$$.

$$(A \cap C) \cap D = \{11\} \cap \{15, 17\} = \phi$$.

Hence $$A \cap C \cap D = \phi$$.

$$A = \{3, 5, 7, 9, 11\}$$, $$C = \{11, 13, 15\}$$. Common element: $$11$$.

$$A \cap C = \{11\}.$$

$$B = \{7, 9, 11, 13\}$$, $$D = \{15, 17\}$$. No common element.

$$B \cap D = \phi.$$

First, $$B \cup C = \{7, 9, 11, 13\} \cup \{11, 13, 15\} = \{7, 9, 11, 13, 15\}$$.

Then $$A \cap (B \cup C) = \{3, 5, 7, 9, 11\} \cap \{7, 9, 11, 13, 15\} = \{7, 9, 11\}$$.

$$A = \{3, 5, 7, 9, 11\}$$, $$D = \{15, 17\}$$. No common element.

$$A \cap D = \phi.$$

First, $$B \cup D = \{7, 9, 11, 13\} \cup \{15, 17\} = \{7, 9, 11, 13, 15, 17\}$$.

Then $$A \cap (B \cup D) = \{3, 5, 7, 9, 11\} \cap \{7, 9, 11, 13, 15, 17\} = \{7, 9, 11\}$$.

From (i), $$A \cap B = \{7, 9, 11\}$$. From (vi), $$B \cup C = \{7, 9, 11, 13, 15\}$$.

$$(A \cap B) \cap (B \cup C) = \{7, 9, 11\} \cap \{7, 9, 11, 13, 15\} = \{7, 9, 11\}.$$

(Alternatively: $$A \cap B \subset B \subset B \cup C$$, so the intersection equals $$A \cap B$$.)

$$A \cup D = \{3, 5, 7, 9, 11\} \cup \{15, 17\} = \{3, 5, 7, 9, 11, 15, 17\}$$.

From (vi), $$B \cup C = \{7, 9, 11, 13, 15\}$$.

Common elements of $$\{3, 5, 7, 9, 11, 15, 17\}$$ and $$\{7, 9, 11, 13, 15\}$$ are $$7, 9, 11, 15$$.

$$(A \cup D) \cap (B \cup C) = \{7, 9, 11, 15\}.$$

$$A$$ is all natural numbers and $$B$$ is all even natural numbers, so $$B \subset A$$.

Hence $$A \cap B = B = \{x : x \text{ is an even natural number}\}$$.

$$C$$ (odd natural numbers) is a subset of $$A$$ (all natural numbers).

Hence $$A \cap C = C = \{x : x \text{ is an odd natural number}\}$$.

$$D$$ (primes) is a subset of $$A$$ (all natural numbers), since every prime is a natural number.

Hence $$A \cap D = D = \{x : x \text{ is a prime number}\}$$.

$$B$$ is the set of even natural numbers and $$C$$ is the set of odd natural numbers. A natural number cannot be both even and odd.

Hence $$B \cap C = \phi$$.

$$B$$ = even natural numbers, $$D$$ = primes. A prime is even only if it equals $$2$$ (since any other even number has $$2$$ as a proper divisor).

Hence $$B \cap D = \{2\}$$.

$$C$$ = odd natural numbers, $$D$$ = primes. The common elements are the odd primes: $$3, 5, 7, 11, 13, \ldots$$ (every prime except $$2$$).

Hence $$C \cap D = \{x : x \text{ is an odd prime}\} = \{3, 5, 7, 11, 13, \ldots\}$$.

The second set is $$\{4, 5, 6\}$$.

$$\{1, 2, 3, 4\} \cap \{4, 5, 6\} = \{4\} \neq \phi$$.

Hence the sets are not disjoint.

The letter $$e$$ appears in both sets, so $$\{a, e, i, o, u\} \cap \{c, d, e, f\} = \{e\} \neq \phi$$.

Hence the sets are not disjoint.

No integer is simultaneously even and odd, so the intersection is empty.

Hence the sets are disjoint.

Elements of $$A$$ not in $$B$$: from $$\{3, 6, 9, 12, 15, 18, 21\}$$ drop $$12$$ (the only one also in $$B$$).

$$A - B = \{3, 6, 9, 15, 18, 21\}.$$

Elements of $$A$$ not in $$C = \{2, 4, 6, 8, 10, 12, 14, 16\}$$: drop $$6$$ and $$12$$ from $$A$$.

$$A - C = \{3, 9, 15, 18, 21\}.$$

Elements of $$A$$ not in $$D = \{5, 10, 15, 20\}$$: drop $$15$$.

$$A - D = \{3, 6, 9, 12, 18, 21\}.$$

Elements of $$B = \{4, 8, 12, 16, 20\}$$ not in $$A$$: drop $$12$$.

$$B - A = \{4, 8, 16, 20\}.$$

Elements of $$C = \{2, 4, 6, 8, 10, 12, 14, 16\}$$ not in $$A = \{3, 6, 9, 12, 15, 18, 21\}$$: drop $$6, 12$$.

$$C - A = \{2, 4, 8, 10, 14, 16\}.$$

Elements of $$D = \{5, 10, 15, 20\}$$ not in $$A$$: drop $$15$$.

$$D - A = \{5, 10, 20\}.$$

Elements of $$B = \{4, 8, 12, 16, 20\}$$ not in $$C = \{2, 4, 6, 8, 10, 12, 14, 16\}$$: drop $$4, 8, 12, 16$$ (all common with $$C$$); leftover $$20$$.

$$B - C = \{20\}.$$

Elements of $$B$$ not in $$D = \{5, 10, 15, 20\}$$: drop $$20$$.

$$B - D = \{4, 8, 12, 16\}.$$

$$C - B$$ consists of the elements of $$C$$ that are not in $$B$$.

Here $$C = \{2, 4, 6, 8, 10, 12, 14, 16\}$$ and $$B = \{4, 8, 12, 16, 20\}$$. The elements common to both sets are $$4, 8, 12, 16$$, so we drop these from $$C$$.

$$C - B = \{2, 6, 10, 14\}.$$

Elements of $$D = \{5, 10, 15, 20\}$$ not in $$B$$: drop $$20$$.

$$D - B = \{5, 10, 15\}.$$

$$C - D$$ consists of the elements of $$C$$ that are not in $$D$$.

Here $$C = \{2, 4, 6, 8, 10, 12, 14, 16\}$$ and $$D = \{5, 10, 15, 20\}$$. The only element common to both sets is $$10$$, so we drop it from $$C$$.

$$C - D = \{2, 4, 6, 8, 12, 14, 16\}.$$

Elements of $$D = \{5, 10, 15, 20\}$$ not in $$C = \{2, 4, 6, 8, 10, 12, 14, 16\}$$: drop $$10$$.

$$D - C = \{5, 15, 20\}.$$

Elements of $$X = \{a, b, c, d\}$$ not in $$Y = \{f, b, d, g\}$$: drop $$b, d$$.

$$X - Y = \{a, c\}.$$

Elements of $$Y = \{f, b, d, g\}$$ not in $$X$$: drop $$b, d$$.

$$Y - X = \{f, g\}.$$

Common elements of $$X = \{a, b, c, d\}$$ and $$Y = \{f, b, d, g\}$$: $$b$$ and $$d$$.

$$X \cap Y = \{b, d\}.$$

$$3$$ appears in both sets, so $$\{2, 3, 4, 5\} \cap \{3, 6\} = \{3\} \neq \phi$$.

Hence the statement is false.

$$a$$ appears in both sets, so the intersection is $$\{a\} \neq \phi$$.

Hence the statement is false.

Comparing elements, no number appears in both sets — one consists of even numbers, the other of odd numbers. So the intersection is empty.

Hence the statement is true.

The first set has only even numbers; the second has only odd numbers. No element is common.

Hence the statement is true.

$$A' = U - A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{1, 2, 3, 4\} = \{5, 6, 7, 8, 9\}$$.

$$B' = U - B = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2, 4, 6, 8\} = \{1, 3, 5, 7, 9\}$$.

$$A \cup C = \{1, 2, 3, 4\} \cup \{3, 4, 5, 6\} = \{1, 2, 3, 4, 5, 6\}$$.

$$(A \cup C)' = U - (A \cup C) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{1, 2, 3, 4, 5, 6\} = \{7, 8, 9\}$$.

$$A \cup B = \{1, 2, 3, 4\} \cup \{2, 4, 6, 8\} = \{1, 2, 3, 4, 6, 8\}$$.

$$(A \cup B)' = U - (A \cup B) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{1, 2, 3, 4, 6, 8\} = \{5, 7, 9\}$$.

From (i), $$A' = \{5, 6, 7, 8, 9\}$$.

$$(A')' = U - A' = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{5, 6, 7, 8, 9\} = \{1, 2, 3, 4\} = A$$.

(This illustrates the general identity $$(A')' = A$$.)

$$B - C = \{2, 4, 6, 8\} - \{3, 4, 5, 6\} = \{2, 8\}$$ (drop the elements $$4, 6$$ shared with $$C$$).

$$(B - C)' = U - (B - C) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2, 8\} = \{1, 3, 4, 5, 6, 7, 9\}$$.

$$A' = U - A = \{a, b, c, d, e, f, g, h\} - \{a, b, c\} = \{d, e, f, g, h\}$$.

$$B' = U - B = \{a, b, c, d, e, f, g, h\} - \{d, e, f, g\} = \{a, b, c, h\}$$.

$$C' = U - C = \{a, b, c, d, e, f, g, h\} - \{a, c, e, g\} = \{b, d, f, h\}$$.

$$D' = U - D = \{a, b, c, d, e, f, g, h\} - \{a, f, g, h\} = \{b, c, d, e\}$$.

With $$U = \mathbf{N}$$, the complement of the even natural numbers is the set of natural numbers that are not even, i.e. the odd natural numbers.

$$\{x : x \text{ is an even natural number}\}' = \{x : x \text{ is an odd natural number}\}.$$

The complement (in $$\mathbf{N}$$) of the odd natural numbers is the even natural numbers.

The complement in $$\mathbf{N}$$ consists of natural numbers that are not multiples of $$3$$.

Hence $$\{x : x \in \mathbf{N} \text{ and } x \text{ is not a multiple of } 3\}$$.

The complement in $$\mathbf{N}$$ is the set of natural numbers that are not prime — i.e. $$1$$ together with all composite numbers.

A natural number divisible by both $$3$$ and $$5$$ is divisible by $$\mathrm{lcm}(3, 5) = 15$$. So the given set is $$\{15, 30, 45, \ldots\}$$.

The complement in $$\mathbf{N}$$ is the set of natural numbers not divisible by $$15$$.

The given set is $$\{1, 4, 9, 16, 25, \ldots\}$$. Its complement in $$\mathbf{N}$$ consists of those natural numbers which are not perfect squares.

The given set is $$\{1, 8, 27, 64, 125, \ldots\}$$. Its complement in $$\mathbf{N}$$ consists of natural numbers which are not perfect cubes.

$$x + 5 = 8 \Rightarrow x = 3$$. So the given set is $$\{3\}$$.

Its complement in $$\mathbf{N}$$ is the set of natural numbers other than $$3$$.

$$2x + 5 = 9 \Rightarrow 2x = 4 \Rightarrow x = 2$$. So the given set is $$\{2\}$$.

Its complement in $$\mathbf{N}$$ is the set of natural numbers other than $$2$$.

In $$\mathbf{N}$$, the given set is $$\{7, 8, 9, 10, \ldots\}$$. Its complement consists of natural numbers $$< 7$$: $$\{1, 2, 3, 4, 5, 6\}$$.

In set-builder form: $$\{x : x \in \mathbf{N},\ x < 7\}$$.

$$2x + 1 > 10 \Rightarrow 2x > 9 \Rightarrow x > 4.5$$. So in $$\mathbf{N}$$ the given set is $$\{5, 6, 7, \ldots\}$$.

Its complement in $$\mathbf{N}$$ is $$\{1, 2, 3, 4\}$$, i.e. natural numbers with $$x \leq 4$$ (equivalently $$2x + 1 \leq 10$$).

$$A \cup B = \{2, 4, 6, 8\} \cup \{2, 3, 5, 7\} = \{2, 3, 4, 5, 6, 7, 8\}$$.

$$(A \cup B)' = U - (A \cup B) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2, 3, 4, 5, 6, 7, 8\} = \{1, 9\}$$.

$$A' = U - A = \{1, 3, 5, 7, 9\}$$.

$$B' = U - B = \{1, 4, 6, 8, 9\}$$.

$$A' \cap B' = \{1, 3, 5, 7, 9\} \cap \{1, 4, 6, 8, 9\} = \{1, 9\}$$.

Both expressions equal $$\{1, 9\}$$, so $$(A \cup B)' = A' \cap B'$$ is verified.

$$A \cap B = \{2, 4, 6, 8\} \cap \{2, 3, 5, 7\} = \{2\}$$.

$$(A \cap B)' = U - \{2\} = \{1, 3, 4, 5, 6, 7, 8, 9\}$$.

From the previous part, $$A' = \{1, 3, 5, 7, 9\}$$ and $$B' = \{1, 4, 6, 8, 9\}$$.

$$A' \cup B' = \{1, 3, 5, 7, 9\} \cup \{1, 4, 6, 8, 9\} = \{1, 3, 4, 5, 6, 7, 8, 9\}$$.

Both expressions equal $$\{1, 3, 4, 5, 6, 7, 8, 9\}$$, so $$(A \cap B)' = A' \cup B'$$ is verified.

Draw a rectangle to represent the universal set $$U$$ and two overlapping circles inside the rectangle labelled $$A$$ and $$B$$.

The set $$A \cup B$$ is the region covered by either circle (i.e. both circles together). The complement $$(A \cup B)'$$ is everything in the rectangle outside these two circles.

Shade the region of the rectangle that lies outside both circles. The shaded portion represents $$(A \cup B)'$$.

Draw a rectangle for $$U$$ with two overlapping circles $$A$$ and $$B$$. $$A'$$ is the region outside circle $$A$$; $$B'$$ is the region outside circle $$B$$. Their intersection $$A' \cap B'$$ is the region that lies outside both circles.

Shade the part of the rectangle lying outside both circles. This is the same region as $$(A \cup B)'$$, illustrating De Morgan's law $$(A \cup B)' = A' \cap B'$$.

Draw $$U$$ as a rectangle with overlapping circles $$A$$ and $$B$$. $$A \cap B$$ is the lens-shaped overlap of the two circles. $$(A \cap B)'$$ is everything in the rectangle except this overlap.

Shade the entire rectangle except the lens-shaped intersection of $$A$$ and $$B$$.

Draw $$U$$ as a rectangle with overlapping circles $$A$$ and $$B$$. $$A'$$ is everything outside $$A$$; $$B'$$ is everything outside $$B$$. Their union $$A' \cup B'$$ is everything that lies outside at least one of the two circles.

This consists of: (i) the part of circle $$A$$ not in $$B$$, (ii) the part of circle $$B$$ not in $$A$$, and (iii) the region of the rectangle outside both circles. Equivalently, it is everything in the rectangle except the lens-shaped overlap $$A \cap B$$ — the same picture as $$(A \cap B)'$$, illustrating De Morgan's law $$(A \cap B)' = A' \cup B'$$.

Every element of $$U$$ is either in $$A$$ or in $$A'$$ (and not both). So their union covers all of $$U$$.

Hence $$A \cup A' = U$$.

$$\phi' = U - \phi = U$$.

Then $$\phi' \cap A = U \cap A = A$$ (intersection with the universal set is the set itself).

No element can simultaneously be in $$A$$ and in $$A'$$ (the complement of $$A$$). So their intersection has no elements.

Hence $$A \cap A' = \phi$$.

$$U' = U - U = \phi$$.

Then $$U' \cap A = \phi \cap A = \phi$$.

False. Being an element of $$B$$ ($$A \in B$$) is different from saying every element of $$A$$ is in $$B$$ ($$A \subset B$$).

Counter-example. Let $$A = \{1, 2\}$$ and $$B = \{\{1, 2\}, 3\}$$. Then $$1 \in A$$ and $$A \in B$$ (since $$A = \{1, 2\}$$ is itself one of the listed elements of $$B$$). But $$1 \notin B$$, because the elements of $$B$$ are $$\{1, 2\}$$ and $$3$$.

False. $$A \subset B$$ does not mean $$A = B$$, so $$A$$ need not be a listed element of $$C$$ just because $$B$$ is.

Counter-example. Let $$A = \{1\}$$, $$B = \{1, 2\}$$ and $$C = \{\{1, 2\}, 3\}$$. Then $$A \subset B$$ and $$B \in C$$. But $$A = \{1\}$$ is not one of the listed elements of $$C$$, so $$A \notin C$$.

True. Let $$x \in A$$. Since $$A \subset B$$, we have $$x \in B$$. Since $$B \subset C$$, this gives $$x \in C$$.

Thus every element of $$A$$ is in $$C$$, i.e. $$A \subset C$$. (This is the transitivity of $$\subset$$.)

False. The relation $$\not\subset$$ is not transitive.

Counter-example. Let $$A = \{1, 2\}$$, $$B = \{2, 3\}$$, $$C = \{1, 2, 4\}$$.

Then $$1 \in A,\ 1 \notin B$$, so $$A \not\subset B$$. And $$3 \in B,\ 3 \notin C$$, so $$B \not\subset C$$. But $$1, 2 \in C$$, so $$A \subset C$$.

False. $$A \not\subset B$$ means at least one element of $$A$$ is not in $$B$$, but it says nothing about a particular $$x \in A$$.

Counter-example. Let $$A = \{1, 2\}$$, $$B = \{3, 4\}$$, $$x = 1$$. Then $$x \in A$$, and $$A \not\subset B$$ (since $$1 \notin B$$). But $$x = 1 \notin B$$.

True. Suppose, for contradiction, that $$x \in A$$. Since $$A \subset B$$, we would have $$x \in B$$, contradicting the hypothesis $$x \notin B$$.

Hence $$x \notin A$$.

(This is just the contrapositive of $$A \subset B$$: if every element of $$A$$ is in $$B$$, then anything outside $$B$$ is also outside $$A$$.)

We prove the cycle (i) $$\Rightarrow$$ (ii) $$\Rightarrow$$ (iii) $$\Rightarrow$$ (iv) $$\Rightarrow$$ (i). This subpart records the proof of (i) $$\Rightarrow$$ (ii).

(i) $$\Rightarrow$$ (ii): Assume $$A \subset B$$. Suppose for contradiction that $$A - B \neq \phi$$; pick $$x \in A - B$$. Then $$x \in A$$ and $$x \notin B$$, contradicting $$A \subset B$$.

Hence $$A - B = \phi$$.

(ii) $$\Rightarrow$$ (iii): Assume $$A - B = \phi$$, i.e. there is no element of $$A$$ outside $$B$$ — every element of $$A$$ is in $$B$$.

We show $$A \cup B = B$$. Clearly $$B \subset A \cup B$$. Conversely, any $$x \in A \cup B$$ is in $$A$$ or in $$B$$; in either case $$x \in B$$ (because every element of $$A$$ is in $$B$$). So $$A \cup B \subset B$$.

Hence $$A \cup B = B$$.

(iii) $$\Rightarrow$$ (iv): Assume $$A \cup B = B$$.

Then $$A \cap B \subset A$$ (always). Conversely, let $$x \in A$$. Then $$x \in A \cup B = B$$, so $$x \in B$$. Combined with $$x \in A$$, this gives $$x \in A \cap B$$. So $$A \subset A \cap B$$.

Both inclusions give $$A \cap B = A$$.

(iv) $$\Rightarrow$$ (i): Assume $$A \cap B = A$$.

Let $$x \in A$$. Since $$A = A \cap B$$, $$x \in A \cap B$$, so $$x \in B$$. Thus every element of $$A$$ lies in $$B$$, i.e. $$A \subset B$$.

This closes the cycle (i) $$\Rightarrow$$ (ii) $$\Rightarrow$$ (iii) $$\Rightarrow$$ (iv) $$\Rightarrow$$ (i), so all four conditions are equivalent.

We prove the equality by showing that each side is contained in the other.

($$\subset$$) Since $$A \cap B \subset A$$, taking the union with $$A$$ on both sides gives $$A \cup (A \cap B) \subset A \cup A = A$$.

($$\supset$$) A set is always contained in its union with another set, so $$A \subset A \cup (A \cap B)$$.

From these two inclusions, $$A \cup (A \cap B) = A$$. This is the absorption law.

($$\subset$$) $$A \cap (A \cup B) \subset A$$ trivially (intersection of $$A$$ with anything is a subset of $$A$$).

($$\supset$$) Let $$x \in A$$. Then $$x \in A \cup B$$ (since $$A \subset A \cup B$$), and also $$x \in A$$. So $$x \in A \cap (A \cup B)$$.

Hence $$A \cap (A \cup B) = A$$. (Absorption law.)