Venn diagrams with two sets

The Venn diagram

A Venn diagram is a picture of sets. The universal set $U$ is drawn as a rectangle: it holds every element we are talking about. Each set inside it is drawn as a circle (or oval).

Sets can be written by listing their elements, like $\{2, 4, 6\}$, or by a rule. The set-builder form $\{x \mid x \in \mathbb{N}, \ x \le 12\}$ is read "the set of all $x$ such that $x$ is in $\mathbb{N}$ and $x$ is less than or equal to $12$". The symbol $\in$ means "is an element of", and $\mathbb{N} = \{1, 2, 3, \dots\}$ is the set of natural numbers. So this set is $\{1, 2, 3, \dots, 12\}$.

When there are two sets $A$ and $B$, the circles overlap, so the diagram has four regions:

• A only: in $A$ but not in $B$. We also write this as $A \setminus B$.

• Both: in $A$ and in $B$. This is the intersection $A \cap B$.

• B only: in $B$ but not in $A$, written $B \setminus A$.

• Neither: inside the rectangle but outside both circles. This is $(A \cup B)'$.

The union $A \cup B$ is everything in at least one circle: A only, both and B only. The complement $A'$ is everything in $U$ that is not in $A$: that is B only and neither.

Placing elements

When the sets are given as lists, place each element of $U$ by asking two questions: "Is it in $A$?" and "Is it in $B$?"

Answer to the "tu turno": $A \cup B = \{2, 3, 4, 6, 8, 9, 10\}$, so $\#(A \cup B) = 7$. $A' = \{1, 3, 5, 7, 9\}$.

For example, if $U = \{x \mid x \in \mathbb{N}, \ x \le 8\}$, then $U = \{1, 2, 3, 4, 5, 6, 7, 8\}$. If $A$ is the set of multiples of $4$ in $U$, then $A = \{4, 8\}$.

Counting with numbers

Often a problem gives numbers, not lists. Write the number of elements in each region.

Answers: $7$ play football only; $25$ play at least one.

Notice what happens when you are told the total of a circle: you must subtract the middle to get the "only" part.

Problems with unknowns

Sometimes a region is given using a letter. Then you write each region in terms of $x$, add them to make $\#U$, and solve the equation.

Another useful idea is the range of possible values. If $\#A = 30$ and $\#B = 25$ in a group of $50$, the middle region cannot be bigger than the smaller set, so $\#(A \cap B) \le 25$. Also $A \cup B$ cannot be bigger than $50$, so $30 + 25 - \#(A \cap B) \le 50$, giving $\#(A \cap B) \ge 5$.

Summary

Back to the start: Seán's class is shown with the middle region for those who play both, and the number who put up no hand goes outside the circles.