KindiakMath

The Indicator Function

Definition 1. Let \Omega be any set. For any K \subseteq \Omega, define the indicator function \mathbb I_K : \Omega \to \{0, 1\} \subseteq \mathbb N_0 on K by

\mathbb I_K(\omega) = \begin{cases} 1, & \omega \in K, \\ 0, & \omega \notin K. \end{cases}

Example 1. We have \mathbb I_{\Omega} = 1 and \mathbb I_{\emptyset} = 0.

Problem 1. Prove that the following propositions:

  • \mathbb I_K^{-1}(\{1\}) = K.
  • K = L \iff \mathbb I_K = \mathbb I_L.
(Click for Solution)

Solution. By definition,

\omega \in K \iff \mathbb I_K(\omega) = 1 \iff \omega \in \mathbb I_K^{-1}(\{1\}).

Therefore,

\begin{aligned} K = L \quad &\Rightarrow \quad \mathbb I_K = \mathbb I_L \\ &\Rightarrow \quad \mathbb I_K^{-1}(\{1\}) = \mathbb I_L^{-1}(\{1\}) \\ &\Rightarrow \quad K = L.\end{aligned}

Problem 2. Prove the following propositions:

  • \mathbb I_{K \cap L} = \mathbb I_{K} \cdot \mathbb I_{L},
  • \mathbb I_{\overline K} = 1 - \mathbb I_{K}, where \overline K:= \Omega \backslash K,
  • \mathbb I_{K \cup L} =\mathbb I_{K} + \mathbb I_{L} - \mathbb I_{K \cap L}.
(Click for Solution)

Solution. The proofs of the various equalities are straightforward applications of the definition of the indicator function.

For the first result, fix \omega \in \Omega. If \omega \in K \cap L \subseteq K, then \mathbb{I}_{K \cap L}(\omega) = 1 and \mathbb{I}_{K}(\omega) = 1. Similarly, \mathbb I_L(\omega) = 1. Hence,

\mathbb{I}_{K \cap L}(\omega) = 1 = 1 \cdot 1 = \mathbb{I}_{K}(\omega)\mathbb{I}_{L}(\omega).

The second result follows from \omega \in \overline K \iff \omega \notin K \iff \mathbb I_K(\omega) = 0.

The third result follows from case splitting:

Suppose \omega \in K and \omega \in L. Then \omega \in K \cap L, so that

\mathbb I_{K}(\omega) + \mathbb I_{L}(\omega) - \mathbb I_{K \cap L}(\omega) = 1 + 1 - 1 = 1.

In all other cases, \mathbb I_{K \cap L}(\omega) = 0.

If \omega \in K and \omega \notin L, then

\mathbb I_{K}(\omega) + \mathbb I_{L}(\omega) - \mathbb I_{K \cap L}(\omega) = 1 + 0 - 0 = 1.

If \omega \notin K and \omega \in K, then

\mathbb I_{K}(\omega) + \mathbb I_{L}(\omega) - \mathbb I_{K \cap L}(\omega) = 0 + 1 - 0 = 1.

If \omega \notin K and \omega \notin L, then

\mathbb I_{K}(\omega) + \mathbb I_{L}(\omega) - \mathbb I_{K \cap L}(\omega) = 0 + 0 - 0 = 0.

Therefore, \mathbb I_{K \cup L} =\mathbb I_{K} + \mathbb I_{L} - \mathbb I_{K \cap L}.

Problem 3. Use the previous problems to deduce the following set identities.

  • K \cap \emptyset = \emptyset,
  • K \cap K = K,
  • K \cap \Omega = K,
  • K \cap L = L \cap K,
  • K \cap (L \cap M) = (K \cap L) \cap M.
(Click for Solution)

Solution. By Problem 1, for each equality of the form K_1 = K_2, it suffices to prove that \mathbb I_{K_1} = \mathbb I_{K_2}:

  • \mathbb I_{K \cap \emptyset} = \mathbb I_K \cdot \mathbb I_{\emptyset} = \mathbb I_K \cdot 0 = 0 = \mathbb I_{\emptyset},
  • \mathbb I_{K \cap \Omega} = \mathbb I_K \cdot \mathbb I_{\Omega} = \mathbb I_K \cdot 1 = \mathbb I_K,
  • \mathbb I_{K \cap K} = \mathbb I_K \cdot \mathbb I_K = \mathbb I_K since 0 \cdot 0 = 0 and 1 \cdot 1 = 1,
  • \mathbb I_{K \cap L} = \mathbb I_K \cdot \mathbb I_L = \mathbb I_L \cdot \mathbb I_K = \mathbb I_{L \cap K}.

The fifth result is slightly longer, but not too difficult to verify:

\begin{aligned} \mathbb I_{K \cap (L \cap M)} &= \mathbb I_K \cdot \mathbb I_{L \cap M} \\ &= \mathbb I_K \cdot (\mathbb I_L \cdot \mathbb I_M) \\ &= (\mathbb I_K \cdot \mathbb I_L) \cdot \mathbb I_M \\ &= \mathbb I_{K \cap L} \cdot \mathbb I_M = \mathbb I_{(K \cap L) \cap M}. \end{aligned}

Problem 4. Prove the following distributivity property

K \cap (L \cup M) = (K \cap L) \cup (K \cap M).

(Click for Solution)

Solution. We first observe that

(K \cap L) \cap (K \cap M) = (K \cap K) \cap (L \cap M) = K \cap (L \cap M).

Applying Problem 1 and Problem 2,

\begin{aligned} \mathbb I_{K \cap (L \cup M)} &= \mathbb I_K \cdot \mathbb I_{L \cup M} \\ &= \mathbb I_K \cdot ( \mathbb I_L + \mathbb I_M - \mathbb I_{L \cap M} ) \\ &= \mathbb I_K \cdot \mathbb I_L + \mathbb I_K \cdot \mathbb I_M - \mathbb I_K \cdot \mathbb I_{L \cap M} \\ &= \mathbb I_K \cdot \mathbb I_L + \mathbb I_K \cdot \mathbb I_M - \mathbb I_{K \cap (L \cap M)} \\ &= \mathbb I_{K \cap L} + \mathbb I_{K \cap M} - \mathbb I_{(K \cap L) \cap (K \cap M)} \\ &= \mathbb I_{(K \cap L) \cup (K \cap M)}. \end{aligned}

Problem 5. Prove the following propositions:

  • K = L \iff \overline K = \overline L,
  • \overline{\overline K} = K,
  • \overline{K \cup L} = \overline K \cap \overline L,
  • \overline{K \cap L} = \overline K \cup \overline L,
  • K \backslash L = K \backslash (K \cap L),
  • K \cap \overline K = \emptyset.
(Click for Solution)

Solution. By Problem 1,

\begin{aligned}K = L \quad &\iff \quad \mathbb I_K = \mathbb I_L \\ &\iff \quad 1 - \mathbb I_K = 1 - \mathbb I_L \\ &\iff \quad \mathbb I_{\overline K} = \mathbb I_{\overline L} \\ &\iff \quad \overline K = \overline L. \end{aligned}

For the second result,

\mathbb I_{ \overline{\overline K} } = 1 - \mathbb I_{ \overline K } = 1 - (1 - \mathbb I_K) = \mathbb I_K.

For the third result,

\begin{aligned} \mathbb I_{ \overline{ K \cup L } } &= 1 - ( \mathbb I_K + \mathbb I_L - \mathbb I_{K \cap L} ) \\ &= 1 - \mathbb I_K - \mathbb I_L + \mathbb I_K \cdot \mathbb I_L \\ &= (1 - \mathbb I_K) \cdot (1 - \mathbb I_L) \\ &= \mathbb I_{ \overline K } \cdot \mathbb I_{ \overline L } = \mathbb I_{ \overline K \cap \overline L }. \end{aligned}

For the fourth result,

\overline{ \overline K \cup \overline L } = \overline{ \overline K } \cap \overline{ \overline L } = K \cap L.

Therefore,

\overline{ K \cap L } = \overline{ \overline{ \overline K \cup \overline L } } = \overline K \cup \overline L.

For the fifth result,

\begin{aligned} \mathbb I_{K \cap \overline K} &= \mathbb I_{K \backslash K} \\ &= \mathbb I_K - \mathbb I_{K \cap K} \\ &= \mathbb I_K - \mathbb I_K = 0 = \mathbb I_\emptyset. \end{aligned}

For the sixth result,

\begin{aligned} K \backslash (K \cap L) &= K \cap \overline{(K \cap L)} \\ &= K \cap (\overline K \cup \overline L) \\ &= (K \cap \overline K) \cup (K \cap \overline L)\\ &= \emptyset \cup (K \cap \overline L) \\ &= K \cap \overline L = K \backslash L. \end{aligned}

Problem 6. Use the previous problems to deduce the following set identities.

  • K \cup \emptyset = K,
  • K \cup K = K,
  • K \cup \Omega = \Omega,
  • K \cup L = L \cup K,
  • K \cup (L \cup M) = (K \cup L) \cup M.
(Click for Solution)

Solution. We use Problem 1, Problem 2, and Problem 5.

  • \mathbb I_{ K \cup \emptyset } = \mathbb I_K + \mathbb I_{ \emptyset } - \mathbb I_{ K \cap \emptyset } = \mathbb I_K + \mathbb I_{ \emptyset } - \mathbb I_{ \emptyset } = \mathbb I_K,
  • \mathbb I_{ K \cup K } = \mathbb I_{ K } + \mathbb I_{ K } - \mathbb I_{ K \cap K } = \mathbb I_{ K } + \mathbb I_{ K } - \mathbb I_{ K } = \mathbb I_K,
  • \mathbb I_{ K \cup \Omega } = \mathbb I_{ K } + \mathbb I_{ \Omega } - \mathbb I_{ K \cap \Omega } = \mathbb I_{ K } + \mathbb I_{ \Omega } - \mathbb I_{ K } = \mathbb I_\Omega,
  • \mathbb I_{ \overline{K \cup L} } = \mathbb I_{ \overline K \cap \overline L } = \mathbb I_{ \overline L \cap \overline K } = \mathbb I_{ \overline{L \cup K} }.

The fifth identity takes a bit more effort, and is useful to analyse in the context of complements.

\begin{aligned} \mathbb I_{\overline{ K \cup (L \cup M) }} &= \mathbb I_{ \overline K \cap \overline{ L \cup M } } \\ &= \mathbb I_{ \overline K \cap ( \overline L \cap \overline M ) } \\ &= \mathbb I_{ ( \overline K \cap \overline L ) \cap \overline M } \\ &= \mathbb I_{ \overline { K \cup L } \cap \overline M } = \mathbb I_{ \overline { (K \cup L ) \cup M } }. \end{aligned}

Problem 7. Prove the distributivity properties:

\begin{aligned} K \cap (L \cup M) &= (K \cap L) \cup (K \cap M), \\ K \cup (L \cap M) &= (K \cup L) \cap (K \cup M). \end{aligned}

(Click for Solution)

Solution. The first identity is the content of Problem 4.

Applying Problem 5,

\begin{aligned} \overline{ K \cup (L \cap M) } &= \overline K \cap \overline{L \cap M} \\ &= \overline K \cap (\overline L \cup \overline M) \\ &= (\overline K \cap \overline L) \cup (\overline K \cap \overline M) \\ &= ( \overline{K \cup L} ) \cup ( \overline{K \cup M} ). \end{aligned}

Taking complements,

\begin{aligned} K \cup (L \cap M) &= \overline{ ( \overline{K \cup L} ) \cup ( \overline{K \cup M} ) } \\ &= \overline{ ( \overline{K \cup L} ) } \cap \overline{ ( \overline{K \cup M} ) } \\ &= ( K \cup L ) \cap ( K \cup M ). \end{aligned}

Problem 8. Prove that the following propositions are equivalent:

  • K \subseteq L,
  • K \cap \overline L = \emptyset,
  • K \cap L = K,
  • K \cup L = L.

Furthermore, in this case, prove that \mathbb I_{L \backslash K} = \mathbb I_L - \mathbb I_K.

(Click for Solution)

Solution. We first note that by propositional logic K \subseteq L is equivalent to

\overline K \cup L = \Omega.

Taking complements, the first proposition implies the second via

K \cap \overline L = \overline{ \overline K \cup L } = \overline{\Omega} = \emptyset.

Recalling that \Omega = L \cup \overline L, the second proposition implies the third via

\begin{aligned} K &= K \cap \Omega \\ &= K \cap (L \cup \overline L) \\ &= (K \cap L) \cup (K \cap \overline L) \\ &= ( K \cap L ) \cup \emptyset \\ &= K \cap L. \end{aligned}

Applying Problem 1 and Problem 2, the third proposition implies the fourth via

\begin{aligned} \mathbb I_{K \cup L} &= \mathbb I_K + \mathbb I_L - \mathbb I_{K \cap L} \\ &= \mathbb I_K + \mathbb I_L - \mathbb I_K = \mathbb I_L. \end{aligned}

The fourth proposition implies the first via K \subseteq K \cup L = L, since \subseteq holds unconditionally via propositional logic.

Finally, by Problem 5, if K \subseteq L, then

\begin{aligned} \mathbb I_{L \backslash K} &= \mathbb I_L - \mathbb I_{L \cap K} \\ &= \mathbb I_L - \mathbb I_{K}. \end{aligned}

Problem 9. Define the symmetric difference by

K \oplus L := (K \backslash L) \cup (L \backslash K).

Prove that K \oplus L = (K \cup L) \backslash (K \cap L).

(Click for Solution)

Solution. We first observe that

\begin{aligned} \mathbb I_{(K \backslash L) \cap (L \backslash K)} &= \mathbb I_{K \backslash L} \cdot \mathbb I_{L \backslash K}\\ &= \mathbb I_{K \cap \overline L} \cdot \mathbb I_{L \cap \overline K} \\ &= \mathbb I_{K \cap \overline L \cap L \cap \overline K} \\ &= \mathbb I_{K \cap \overline K \cap L \cap \overline L} \\ &= \mathbb I_{\emptyset \cap \emptyset } = \mathbb I_{\emptyset } = 0.\end{aligned}

Furthermore, (K \cap L) \subseteq (K \cup L) implies

\mathbb I_{(K \cup L) \backslash ( K \cap L ) } = \mathbb I_{K \cup L} - \mathbb I_{K \cap L}.

Therefore,

\begin{aligned}\mathbb I_{K \oplus L} &= \mathbb I_{(K \backslash L) \cup (L \backslash K)} \\ &= \mathbb I_{K \backslash L} + \mathbb I_{L \backslash K} - \mathbb I_{(K \backslash L) \cap (L \backslash K)} \\ &= \mathbb I_{K \cap \overline L} + \mathbb I_{L \cap \overline K} - 0 \\ &= \mathbb I_{K} \cdot \mathbb I_{\overline L} + \mathbb I_L \cdot \mathbb I_{\overline K} \\ &= \mathbb I_{K} \cdot (1 - \mathbb I_L) + \mathbb I_L \cdot(1 - \mathbb I_K) \\ &= \mathbb I_{K} - \mathbb I_{K} \cdot \mathbb I_{L} + \mathbb I_{L} - \mathbb I_{L} \cdot \mathbb I_{K} \\&= (\mathbb I_{K} + \mathbb I_{L} - \mathbb I_{K \cap L}) - \mathbb I_{K \cap L}\\ &= \mathbb I_{K \cup L} - \mathbb I_{K \cap L} \\ &= \mathbb I_{(K \cup L) \backslash ( K \cap L ) }. \end{aligned}

—Joel Kindiak, 7 Feb 25, 1729H

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