## What you need to know

In maths, a set is a collection of things. The kind of sets you’ll need to be familiar with are sets of numbers. We express a set using ‘curly brackets’, i.e. if we wanted to write a set that contained the numbers 5, 13, 2, and 110, we would write

$\{5, 13, 2, 110\}$

A nice thing about sets is that order doesn’t matter. You could write those numbers in reverse order and the set would still be the same.

We usually name sets using capital letters. To call the above set A, we would simply write

$A=\{5, 13, 2, 110\}$

Suppose we have two different sets, A and B. Then, the important set notation is as follows.

$A \cap B$ means ‘A and B’. It’s a new set that contains only the elements that are both in A and in B.

$A \cup B$ means ‘A or B’. It’s a new set that contains any element that appears anywhere in either A or B.

$A’$ means ‘not in A’. It’s the set of all the elements that don’t appear in A.

You may be thinking that a set that contains elements not in A is going to be, well, a really big set. After all, there are a lot of the numbers that aren’t 5, 13, 2 or 110. However, we usually narrow down the numbers we consider with 1 more bit of notation.

$\xi$ is the universal set – it contains all the numbers we’re interested in at that time. In this context, $A’$ would only contain numbers from the universal set that aren’t in A (as opposed to every number ever that isn’t in A).

All of these things start to make a bit more sense when we see some examples.

Example: Let $A=\{1, 2, 3, 8, 10, 12\}$ and $B=\{6, 5, 4, 3, 2, 10\}$. Write down the set

a) $A\cap B$,

b) $A\cup B$.

a) To recall, $A\cap B$ contains every element that appears in both A and B. Looking at the sets, we see that there are three of them: 3, 2, and 10. Therefore, we have

$A\cap B=\{2, 3, 10\}$

b) $A\cup B$ contains all the elements that are in either A or B. Writing the elements in – and it helps to cross them off with a pencil as you go – we get

$A \cup B=\{1, 2, 3, 4, 5, 6, 8, 10, 12\}$

Note: when a number appears in both sets, we don’t need to write it in twice. The numbers 2, 3 and 10 appear in both but we only count them once when working out $A \cup B$.

Example: Below is a Venn diagram. State which numbers are in the sets:

a) $P \cap Q$, b) $Q’$, c) $P\cup Q$.

a) $P\cap Q$ means all elements that are in both P and Q. Looking at the Venn diagram, this must be where the circles intersect. So, we get

$P\cap Q=\{4, 5\}$

b) $Q’$ means all numbers not in Q. All the numbers inside the Q circle are: 10, 6, 14, 4, and 5 (the ones in the intersection count). The numbers not in there are then

$Q’=\{2, 8, 7, 1\}$

We only include numbers in the universal set, $\xi$, as seen on the top-left.

c) $P\cup Q$ means any numbers that appear in either P or Q. So, anything in either circle. We get

$P\cup Q=\{2, 8, 4, 5, 10, 6, 14\}$

Set Notation for Inequalities

The other time set notation appears is in working with inequalities. The good news is, it’s pretty straightforward! To write the inequality $x \leq 5$ in set notation, we write

$\{x : x\leq 5\}$

So, you have to add 3 things: curly brackets, the variable, and a colon.

Example: Solve the inequality $3x-7>11$. Write your answer using set notation.

We solve linear inequalities like we do linear equations. Adding 7 to both sides, we get

$3x>7+11=18$

Then, dividing both sides by 3, we get

$x>6$

Now, to express this in set notation, we want to put “$x :$” before it and wrap the whole thing in curly brackets:

$\{x : x>6\}$

## Example Questions

a) Firstly, we should write down A, the set of all even numbers.

$A=\{104, 110, 112, 114\}$

Now, we must combine this set with B to find $A\cup B$. Doing so, we get

$A\cup B=\{103, 104, 110, 112, 114\}$

b) We’ve already determined A, so now we should consider all the elements that appear in the universal set but not in A. This gives us

$A’=\{103, 105, 109\}$

Firstly, let’s write down V, the set of prime numbers (from the numbers in the universal set):

$V = \{2, 3, 5, 7\}$

Our Venn diagram should be two circles, labelled, with a rectangle around them both, which is also labelled with the Greek letter $\xi$, since the rectangle represents the universal set.

Now, we’re told that $V\cap W=\{3, 5, 7\}$. This means 3, 5 and 7 are in both circles, i.e. they must go in the small section where the circles intersect.

From this, we can establish that the only other element of V, 2, must go in the part of the V circle that doesn’t cross over with W.

Now we’ve sorted all elements of V out, everything else that appears in $V\cup W$ must go in the part of the W circle that doesn’t cross over with V. These values are: 1, 10, and 12.

Lastly, since we’ve covered all elements of V and W, anything left over in $\xi$ will have to go in the section outside the circles. Putting this all together, we get the Venn diagram shown below.

The process for all 4 will be the same. Put the variable followed by a colon before the inequality, and then wrap it all in curly brackets. Doing this, we get

a) $\{x : x \geq 12\}$

b) $\{z : z < -2\}$

c) $\{a : a > 0\}$

d) $\{x : 13 < x\}$