Set Partition: Subsets must be disjoint or equal?

Question

I'm reading a Mathematical Proofs by Polimeni, Chartrand, and Zhang and their definition of a set partition is confusing me:

A partition of A can be defined as a collection S of subsets of A satisfying the three properties:

(1) $ X \neq \emptyset $ for every set $ X \in S $;

(2) for every two sets $X,Y \in S$, either $X = Y$ or $X \cap Y = \emptyset$;

(3) $ \cup_{X \in S} X = A $.

How is the $X = Y$ part of (2) compatible with the idea that the subsets have to be pairwise disjoint? If X and Y are equivalent, then they aren't disjoint and I don't think they can be part of a partition.

Clearly, there's something very basic I'm not understanding. Please clue me in.

Answer 1

(2) for every two sets $X,Y \in S$, either $X = Y$ or $X \cap Y = \emptyset$

How is the $X = Y$ part of (2) compatible with the idea that the subsets have to be pairwise disjoint? If X and Y are equivalent, then they aren't disjoint and I don't think they can be part of a partition.

If property #2 was “for every two sets $X,Y \in S,$ either X = Y or $X \cap Y = \emptyset$” instead, then $X \cap X = \emptyset$ for every set $X$ in $S,$ that is, every set $X$ in $S$ equals $\emptyset.$

To be clear, property #2 does not say “for every two distinct sets $X,Y\in S$, either $X = Y$ or $X \cap Y = \emptyset$”.

so $\{\{1\}, \{1\}, \{1\}, \{2, 3\}\}$ is a partition of $\{1, 2, 3\}\;?$

Yes, because $\{\{1\}, \{1\}, \{1\}, \{2, 3\}\}=\{\{1\},\{2, 3\}\}.$

If so, what's the value of allowing $\{1\}$ to be there more than once?

Can elements in a set be duplicated?

Answer 2

It's simple. Condition (2) can be shown, using basic logic, to be equivalent to saying that if $X\neq Y$, then $X\cap Y=\emptyset$. This means that two sets from $S$ are either the same one or their intersection is empty.