1.5 - Set notation and counting elements

Algebraically defined sets (Higher)

Can $40$ people produce club membership totals of $22$, $20$ and $17$ without a mistake? Those totals add to $59$. The answer depends on who belongs to more than one club; you will resolve it in the last part.

Sometimes a rule is shorter than a list. In $\{x:x\text{ is an integer},\ -2<x\leq3\}$, read the colon as “such that”. The variable names a candidate; the words and inequalities decide whether it belongs.

State the domain, such as integers. A real-number interval would include numbers between the integers too. Neither set-builder notation nor set-size rules are supplied on the Higher formulae sheet; they are not Foundation requirements.

Change only the lower inequality from strict to inclusive. One extra integer now qualifies.

The set and its size answer different questions.

Try new bounds. Read each endpoint separately before you list anything.

Elements and subsets (Higher)

A single shared member does not make one set part of another. Compare $A=\{1,2,3,4\}$ with $B=\{2,4\}$: every member of $B$ belongs to $A$. That is a stronger statement than saying the two sets overlap.

To establish the relationship, check all elements of the smaller set. To disprove it, one counterexample is enough. For instance, $D=\{2,5\}$ is not a subset of $A$, because $5$ is missing from $A$. The shared member $2$ cannot rescue the claim.

Be precise about what you are comparing. The statement $2\in B$ compares an element with a set. The statement $\{2\}\subset B$ compares two sets. In a Venn diagram, a subset's curve lies wholly inside the containing set's curve; partial overlap alone does not show a subset.

The empty set is a subset of every set: it has no member that could fail the membership test. If two sets have exactly the same elements, say they are equal. If the distinction matters, state “proper subset” or “equal” explicitly, since symbol conventions can vary. Subset notation and these facts are not supplied on the Higher formulae sheet.

A set's size alone cannot settle the question either. The sets $\{1,2\}$ and $\{3,4\}$ have the same size but share no element. Look at membership, not just totals.

Counting disjoint Venn regions (Higher)

A number in a Venn region can be a frequency rather than an individual element. Read the caption. In this diagram, $9$ means nine learners in $A$ only; it does not mean that the number $9$ is an element of $A$.

[DIAGRAM: asset_slug: sets_venn_counts; caption: Numbers of learners in the four disjoint regions; description: Rectangle E and overlapping circles A and B. A only has count 9, intersection has count 5, B only count 6, neither count 3. These are frequencies not individual elements and areas are not to scale.; file: sets_venn_counts.png; recommended_method: matplotlib]

Read whole circles by including the overlap: here $n(A)=9+5=14$, $n(B)=5+6=11$, and $n(\mathrm{E})=9+5+6+3=23$. The region counts do not overlap, so you can add them directly.

Start with the shared region, subtract it from each circle total, then use the universal total for the outside region. Each resulting count must be a nonnegative integer.

Change only the overlap from $7$ to $8$. The circle totals stay fixed, so both “only” regions become smaller.

Language tells you which regions to add. “Exactly one” excludes the overlap; “at least one” includes it.

Use fresh survey totals and find the union before the outside count.

Practical problems with overlapping groups (Higher)

With three sets, begin where all three overlap. A pairwise total such as $n(A\cap B)$ normally includes the people in all three sets. Subtract the triple intersection to obtain “$A$ and $B$ only”. Then work outwards to the single-set regions.

For the opening puzzle, let $A,B,C$ be three clubs. A person can contribute to two or three club totals, but still occupies one region of the diagram. “Memberships” and “people” are different totals.

Read the diagram as frequencies. The central $3$ contributes to every circle total, but to the universal total only once.

[DIAGRAM: asset_slug: sets_three_club_counts; caption: Forty people sorted into the eight regions of three clubs; description: Rectangle E contains three overlapping circles A, B and C. A only 8, B only 7, C only 5, AB only 6, AC only 5, BC only 4, all three 3, and none 2. Counts are frequencies and circle areas are not to scale.; file: sets_three_club_counts.png; recommended_method: matplotlib]

Now change only the triple count, from $3$ to $4$. Keep the club totals and pairwise totals fixed. Recalculate every region affected by the change.

Here is a surprising consequence: with the same three club totals and pairwise totals, the number of distinct members rose from $38$ to $39$. The larger triple overlap forced changes in the other six regions. You cannot reason about one region in isolation. This counting method is not supplied on the Higher formulae sheet.

A useful final check is to reconstruct each original total. Checking only the grand total can miss a value placed in the wrong region.

The opening totals are possible: $38$ of the $40$ people can hold $59$ memberships between them, with $2$ people in no club. Apply the same reasoning to a new group.

Quick Recap

• Set-builder notation needs a domain and a rule; $n(A)$ is a number, not a set.
• A subset requires every member to belong; one counterexample disproves the claim.
• Count disjoint regions once. Whole-set totals include their overlaps.
• For three sets, place the triple intersection first, then pair-only regions, then single-only regions and none.

Yes, the opening totals $22,20,17$ are possible for $40$ people: the first club diagram has $38$ people in at least one club and $2$ in none. Adding memberships counts some people more than once.