Mathematical-Reasoning · Unit 28 · Video 5 · Interactive Practice

Partitions and Equivalence Relations: Same Information

IKey Formulas

Condition	For a collection $\mathcal{C}$ of subsets of $A$
(1) nonempty pieces	$V \ne \emptyset$ for each $V \in \mathcal{C}$
(2) every element covered	for each $x \in A$ there is $V \in \mathcal{C}$ with $x \in V$
(3) pairwise disjoint	for all $V, W \in \mathcal{C}$ : $V = W$ or $V \cap W = \emptyset$
Classes $\Rightarrow$ partition	$\mathcal{C} = \{ [x] \mid x \in A \}$ is a partition of $A$
Partition $\Rightarrow$ relation	$x \sim y$ when $x$ and $y$ lie in the same piece; its classes are the pieces

Key Insight: Conditions (2) and (3) together say that each element of $A$ lies in exactly one piece; an equivalence relation and a partition carry the same information.

IITesting the Three Conditions

Each row is one piece of a collection of subsets of $A = \{1, 2, 3, 4, 5, 6\}$ ; a partition must pass all three conditions.

💡 Challenge: turn the empty-piece collection into a partition with three nonempty pieces.

IIIFrom a Partition to “Same Piece”

Declaring $x \sim y$ when $x$ and $y$ share a piece fills one square block of pairs per piece.

Partition Element x 4

IVQuiz Questions

Problem 1 · Spot the Partition

Which collection is a partition of $A = \{1, 2, 3, 4, 5, 6\}$ ?

✅ Correct! Three nonempty pieces, every element appears, and no element appears twice.

❌ Condition (3) fails.

\{1, 2\} \cap \{2, 3, 4\} = \{2\}

, but the pieces are not equal.

❌ Condition (2) fails.

3

lies in no piece.

❌ Condition (1) fails. A partition never contains the empty set.

Show solution

(a) contains $2$ in two different pieces, so condition (3) fails. (b) misses $3$ , so condition (2) fails. (d) contains $\emptyset$ , so condition (1) fails. (c) has the nonempty pieces $\{1, 6\}$ , $\{2, 3\}$ , $\{4, 5\}$ , whose union is $A$ and which are pairwise disjoint: a partition.

Problem 2 · Which Condition Fails?

On $A = \{1, 2, 3, 4, 5, 6\}$ , which condition of a partition does $\mathcal{C} = \{\{1, 2, 3\}, \{3, 4\}, \{5, 6\}\}$ fail?

✅ Correct!

\{1, 2, 3\} \cap \{3, 4\} = \{3\}

while the pieces differ; everything else holds.

❌ Not quite. The union is

\{1, 2, 3, 4, 5, 6\} = A

, so every element is covered.

❌ Not quite. Condition (2) holds:

1, \ldots, 6

all appear.

❌ Not quite. The element

3

lies in two different pieces.

Show solution

Condition (1): all three pieces are nonempty. Condition (2): $\{1, 2, 3\} \cup \{3, 4\} \cup \{5, 6\} = A$ . Condition (3): $\{1, 2, 3\} \ne \{3, 4\}$ , yet $\{1, 2, 3\} \cap \{3, 4\} = \{3\} \ne \emptyset$ . So only condition (3) fails.

Problem 3 · Counting the Pairs

Let $\sim$ be the "same piece" relation of the partition $\{\{1, 4\}, \{2, 5\}, \{3, 6\}\}$ of $A = \{1, 2, 3, 4, 5, 6\}$ . How many ordered pairs $(x, y) \in A \times A$ satisfy $x \sim y$ ?

✅ Correct! Each piece of size

2

gives

2 \cdot 2 = 4

pairs, and there are three pieces.

❌ Not quite. That counts pieces; each piece gives several ordered pairs.

❌ Not quite. Remember the pairs

(x, x)

: the relation is reflexive.

❌ Not quite. List the pairs inside one piece.

Show solution

Inside the piece $\{1, 4\}$ the pairs are $(1, 1), (1, 4), (4, 1), (4, 4)$ ; a piece with $k$ elements gives $k^2$ ordered pairs, and pairs from different pieces are not related. So the relation has

2^2 + 2^2 + 2^2 = 12

ordered pairs, and its classes are $[1] = \{1, 4\}$ , $[2] = \{2, 5\}$ , $[3] = \{3, 6\}$ : the pieces.

Problem 4 · Pieces of the Real Line

On $\mathbb{R}$ define $x \sim y$ when $\lfloor x \rfloor = \lfloor y \rfloor$ . Its classes form a partition of $\mathbb{R}$ . Which piece contains $-0.5$ ?

✅ Correct!

\lfloor -0.5 \rfloor = -1

, and

\lfloor z \rfloor = -1

exactly when

-1 \le z \lt 0

❌ Check the endpoints.

\lfloor 0 \rfloor = 0 \ne -1

and

\lfloor -1 \rfloor = -1

❌ Not quite. That groups by rounding;

\lfloor 0.25 \rfloor = 0 \ne -1

❌ Not quite.

\lfloor -0.5 \rfloor = -1

, not

0

: the floor rounds down.

Show solution

$\lfloor -0.5 \rfloor = -1$ , so $[-0.5] = \{ z \in \mathbb{R} \mid \lfloor z \rfloor = -1 \} = [-1, 0)$ . In general the pieces are the intervals $[k, k + 1)$ for $k \in \mathbb{Z}$ : infinitely many nonempty pieces, which cover $\mathbb{R}$ and never overlap.

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