Sets, Venn diagrams and number sets

What a set is

For example, if $P$ is the set of vowels in the word "example", then $P=\{a,e\}$, $a\in P$, $x\notin P$ and $n(P)=2$. Although the "e" appears twice in the word, it is written only once in the set.

There are two ways to describe a set:

• By extension: listing the elements, $A=\{1,3,5,7\}$.
• By comprehension: giving the condition the elements satisfy, $A=\{x\mid x\in \mathbb{Z},x\text{ odd},0<x<8\}$, read as "the $x$ such that $x$ is an odd integer between $0$ and $8$".

Two special sets:

• The empty set, $\varnothing$ (or $\{\}$), has no elements: $n(\varnothing )=0$.
• The universal set, $U$, contains all the elements under consideration in the problem. You must always know what $U$ is in order to talk about the complement.

Relations and operations between sets

The "or" in the union is inclusive: an element that is in both sets is also in the union. And since $A$ and $A^{\prime }$ do not overlap and together make up $U$, $A\cup A^{\prime }=U$ and $A\cap A^{\prime }=\varnothing$.

How to calculate with elements

With $U=\{1,2,3,4,5,6\}$, $A=\{1,2,3,4\}$ and $B=\{3,4,5\}$:

• $A\cap B=\{3,4\}$ (what appears in both lists).
• $A\cup B=\{1,2,3,4,5\}$ (everything that appears in at least one list, without repetition).
• $A^{\prime }=\{5,6\}$ and $B^{\prime }=\{1,2,6\}$ (what each one is missing to reach $U$).

Venn diagrams

A Venn diagram draws $U$ as a rectangle and each set as a closed region (almost always a circle) inside it. With two sets that overlap, there are four regions:

1. only in $A$: $A\cap B^{\prime }$;
2. in both: $A\cap B$;
3. only in $B$: $A^{\prime }\cap B$;
4. outside both: $(A\cup B{)}^{\prime }$.

With three sets there are eight regions (seven inside the circles and one outside). To shade an operation, shade the region that satisfies the condition: the union is everything inside at least one of the circles; the intersection is the common part; the complement is everything in the rectangle that lies outside the circle.

Why. When you add $n(A)+n(B)$, the elements of the intersection are counted twice, once in each set. Subtracting $n(A\cap B)$ once, every element is counted exactly once. If the sets are disjoint, $n(A\cap B)=0$ and the cardinality of the union is simply the sum.

Also, within $U$, $n(A^{\prime })=n(U)-n(A)$ holds.

Intuition: not being in $A$ nor in $B$ (outside the union) is the same as not being in $A$ and not being in $B$. And not being in both at once (outside the intersection) means failing in $A$ or failing in $B$. When you "break" the bracket of the complement, union is swapped for intersection and vice versa.

Number sets

Numbers have been extended so that operations that previously had no answer can be solved.

The irrational numbers are the real numbers that are not rational, that is, those that cannot be written as a fraction of integers. Their set can be denoted $\mathbb{R}\setminus \mathbb{Q}$. Typical examples are $\sqrt{2}$, $\sqrt{7}$ and $\pi$.

Each set contains the previous one, which is written

$${\mathbb{Z}}^{+}\subset \mathbb{N}\subset \mathbb{Z}\subset \mathbb{Q}\subset \mathbb{R}$$

(in a Venn diagram they are circles nested inside one another, with the irrationals in the part of $\mathbb{R}$ that lies outside $\mathbb{Q}$). So the same number can belong to several sets at once: $-4\in \mathbb{Z}$, and therefore also $-4\in \mathbb{Q}$ and $-4\in \mathbb{R}$, but $-4\notin \mathbb{N}$.

How to recognise a rational number from its decimal

• A terminating decimal (such as $-2.5$) is rational: $-2.5=-\frac{5}{2}$.
• A recurring decimal (such as $0.\overline{6}=0.666\dots$ or $0.\overline{34}$) is rational: $0.\overline{6}=\frac{2}{3}$.
• A decimal with infinitely many digits that do not repeat in any period is irrational.

To convert a recurring decimal into a fraction, multiply by a power of $10$ that shifts the repeating block and subtract the two expressions. With $x=0.\overline{27}$: $100x=27.\overline{27}$, so $100x-x=27$, that is, $99x=27$ and $x=\frac{27}{99}=\frac{3}{11}$.