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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.

criddell
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    This is exactly the condition that the elements of S are pairwise disjoint: two elements are either equal or their intersection is empty. – Wuestenfux Apr 09 '22 at 15:03

2 Answers2

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(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?

ryang
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  • So in sets, there's no distinction between equivalence and identity? If I say set X= set Y I'm saying that not only is X equivalent to Y, but that X is Y. Is that right? But then (2) says "for every two sets X, Y". If they are equal, then they aren't two sets - it's a single set. I'm confused... – criddell Apr 09 '22 at 18:04
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    @criddell Equivalence, identity and equality are generally not synonyms. Two sets can be equivalent (which commonly means they have the same cardinality) without being equal. And when I hear "set identity", I scratch my head then think this, which osntensibly isn't what you mean. – ryang Apr 09 '22 at 18:18
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    @criddell But then (2) says "for every two sets X, Y". If they are equal, then they aren't two sets - it's a single set. Please first slowly re-read my first two sentences. $\quad\quad$ Done? OK, now, Given the identity $(x+y)^2\equiv x^2+y^2+2xy,$ is it legitimate to substitute the exact same value into both $x$ and $y;?$ If the answer is Yes, then similarly, "every pair of sets X and Y are disjoint" $\implies$ "every pair of sets X and X are disjoint" $\implies$ "every set set X is disjoint with itself". – ryang Apr 09 '22 at 18:30
  • The Wikipedia page on disjoint sets says "the empty set is the only set that is disjoint from itself". Doesn't that contradict your last statement? Is clause (2) basically saying: if you pick any two sets from S, they will either be disjoint or else you picked the same set twice – criddell Apr 09 '22 at 18:52
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    (Continuing my previous comment) $⟹$ "every set X is disjoint with itself" $⟹$ "every set X is empty". Note that I've merely been paraphrasing my Answer above; and to be super clear, this last conclusion is the consequence of using the alternative (and wrong) Property #2 that you wanted, that is, the original Property #2 but with "either X = Y or" struck out. – ryang Apr 09 '22 at 19:01
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    @criddell Doesn't that contradict your last statement? No. $\quad\quad$ if you pick any two sets from S, they will either be disjoint or else you picked the same set twice? Yes. – ryang Apr 09 '22 at 19:02
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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.

Frank William Hammond
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  • so {{1}, {1}, {1}, {2, 3}} is a partition of {1, 2, 3}? If so, what's the value of allowing {1} to be there more than once? The sets {1} and {1} aren't disjoint so doesn't that violate the pairwise disjoint requirement of a partition? – criddell Apr 09 '22 at 15:41
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    In set theory you can't have the same element more than once in a single set. In this case, the set {{1},{1},{1},{2,3}} is equal to the set {{1},{2,3}}. – Frank William Hammond Apr 10 '22 at 08:18