Probability · Venn diagrams & set notation

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Set notation

The formal language of sets — the universal set ξ, n(A) for a count, union A∪B ("in either, or both"), intersection A∩B, the complement A′, and combined expressions such as A∩B′ and (A∪B)′ — read as names for regions of a Venn diagram.

Probability · Venn diagrams & set notation

Set notation

Why it works

Set notation looks like a pile of unfamiliar symbols, but every symbol is a name for a region of a picture. Draw two overlapping circles inside a rectangle and you have already made four regions: in

A

only, in both, in

B

only, and in neither. The notation exists so you can say "that region" in writing, without pointing at it.

The box: the universal set $\xi$ . Everything the question is about lives in the rectangle — the

30

students in the class, the whole numbers from

1

12

, the people who answered the survey. The Greek letter

\xi

("xi") names that whole collection. It matters more than it looks, because

A'

below means "everything else", and "everything else" is meaningless until you have said what everything is.

Elements, lists, and $n$ . A set is a collection of things written inside curly brackets:

A = \{2, 3, 5, 7, 11\}

. Each thing in it is an element, and

\in

means "is an element of", so

5 \in A

while

6 \notin A

. The notation

n(A)

is not another set — it is a count: "how many elements are in

A

". Here

n(A) = 5

. So "list the members of

A

" wants a set in brackets, and "find

n(A)

" wants a single number; getting those two mixed up throws marks away before any real thinking has happened.

Union $A \cup B$ — and the everyday-"or" trap. In ordinary English, "tea or coffee?" means one or the other, and not both. Mathematics cannot afford that ambiguity, so

\cup

is defined once and for all as: in $A$ , or in $B$ , or in both. It is everything inside either circle, and the overlap is very definitely included. The symbol helps —

\cup

is a cup, and it scoops up everything from both circles. This is why

n(A \cup B) = n(A) + n(B) - n(A \cap B),

because adding the two totals counts everyone in the overlap twice, so you subtract that overlap exactly once to put it right.

Intersection $A \cap B$ . Only the things that are in

A

and in

B

— the overlap alone. The peak of

\cap

is the "n" of "and", and it is always the smaller region: nothing can be in the intersection without also being in the union.

Complement $A'$ . The dash means "not".

A'

is everything in

\xi

that is outside circle

A

— including things that are in

B

, and things that are in neither. Because every element is either in

A

or not in

A

n(A) + n(A') = n(\xi).

Combined expressions. Read them left to right, treating the dash as "not":

A \cap B'

is "in

A

and not in

B

", which is the part of circle

A

outside the overlap — the region a question would call "

A

only". That is the standard exam way of writing "only", so it is worth recognising instantly. Likewise

A' \cap B

is "

B

only", and

A \cup B'

is a big region: everything except the part of

B

outside

A

.

Why $(A \cup B)' = A' \cap B'$ — read, don't memorise. Take

(A \cup B)'

first. The bracket comes first, so build the union: both circles, overlap included. Then the dash says "everything else", so you are left with the region outside both circles — the corner of the box. Now take

A' \cap B'

A'

is everything outside circle

A

;

B'

is everything outside circle

B

; the intersection asks for what is in both of those at once — outside

A

and outside

B

. That is the same corner of the box. Two names, one region, no memorising required.

Notice what changed on the way: the dash moved inside the bracket and the

\cup

flipped to a

\cap

. The flip is the whole point. The tempting

(A \cup B)' = A' \cup B'

is wrong, and badly so:

A' \cup B'

means "outside

A

or outside

B

", which is everything except the overlap — nearly the whole diagram, rather than one small corner. The same swap works the other way round:

(A \cap B)' = A' \cup B'

.

Two last symbols.

\varnothing

is the empty set — the set with no elements at all, so

n(\varnothing) = 0

. You meet it whenever two sets have nothing in common, e.g. the odd numbers and the multiples of

4

. And

A \subset B

says every element of

A

is also an element of

B

: circle

A

drawn entirely inside circle

B

.

Once you can name a region, you can count it — and every set question at GCSE is either "name this region" or "count that one".