Sets · step 10 of 13

Characteristic (indicator) functions

The characteristic function 1_A maps each element of the universe to 1 if it lies in A and 0 otherwise. This converts set algebra into arithmetic: intersection becomes multiplication, complement becomes 1 − 1_A, and union follows by inclusion–exclusion. It is also the cleanest way to see why |𝒫(U)| = 2^|U| — subsets correspond exactly to functions U → {0, 1}.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Write each set as its indicatorReplace A by 1_A(x) throughout, so every membership statement becomes a 0/1 value.
Translate the operations1_{A∩B} = 1_A · 1_B; 1_{Aᶜ} = 1 − 1_A; 1_{A∪B} = 1_A + 1_B − 1_A·1_B; 1_{A\B} = 1_A(1 − 1_B).
Do ordinary algebraProve set identities by expanding both sides as polynomials in the indicators and comparing. Note 1_A² = 1_A, since 0 and 1 are idempotent.
Sum to countOver a finite universe, |A| = Σₓ 1_A(x). Summing an identity between indicators yields a counting identity for free.

Worked example

Use indicator functions to prove |A ∪ B| = |A| + |B| − |A ∩ B|.

For each x: 1_{A∪B}(x) = 1_A(x) + 1_B(x) − 1_A(x)·1_B(x). Check the four cases of membership to confirm.
Note 1_A(x)·1_B(x) = 1_{A∩B}(x).
Sum both sides over all x in the finite universe.
Σ 1_{A∪B} = Σ 1_A + Σ 1_B − Σ 1_{A∩B}.

Answer. |A ∪ B| = |A| + |B| − |A ∩ B|, obtained by summing a pointwise identity.

Where marks get dropped

These are the specific errors that cost credit on characteristic (indicator) functions questions — QED's rubric penalises each of them separately.

Writing 1_{A∪B} = 1_A + 1_B. That gives 2 for elements in both sets, which is not a 0/1 value — the correction term is essential.
Forgetting to state the universe. The indicator is a function on U, so complements need U fixed.
Treating 1_A² as something new. Since values are 0 or 1, squaring changes nothing, and this identity simplifies most algebra.

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Characteristic (indicator) functions — frequently asked questions

Why are indicator functions useful in probability?

Because E[1_A] = P(A). Writing a count as a sum of indicators and applying linearity of expectation is the standard trick for computing expected values without touching the distribution.

How do they show |𝒫(U)| = 2^|U|?

Subsets of U correspond bijectively to functions U → {0,1}, since each subset has exactly one indicator and each such function determines exactly one subset. There are 2^|U| such functions.

Is 1_A the same as the identity function?

No. The identity maps every element to itself; the indicator maps into {0, 1} and records membership only.

The rest of Sets

Set-builder notation, operations, and set identities. Each subtopic below has its own method, worked example and mark-losing traps.

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