How to properly define a set that is a union of pairwise disjoint sets?

Question

I understand that the term "disjoint" is ambiguous, so I want to explicitly define the set $A$ in a clear and, especially, concise way:

$A = B \cup C \cup D \cup \{e,f\}$
$B = \{b_{1}, b_{2}, \dots, b_{n}\}$
$C = \{c_{1}, c_{2}, \dots, c_{m}\}$
$D = \{d_{1}, d_{2}, \dots, d_{k}\}$

It should be clear that the sets $B, C, D, \{e, f\}$ are pairwise disjoint. What would be the correct and most concise way to explicitly state this property? Probably $B \cap C \cap D \cap \{e, f\} = \emptyset$?

No; pairwise disjoint will mean $A \cap B = \emptyset$ and $B \cap C= \emptyset$ and... — Mauro ALLEGRANZA, Sep 15 '21 at 10:42
To the best of my knowledge, this is usually done as follows. Instead of $B,C,D,{e,f}$ use the notation $A_1 \triangleq B$, $A_2 \triangleq C$, $A_3 \triangleq D$, $A_4 \triangleq {e,f}$. Then, we can say $A_i \cap A_j = \emptyset$ for all $i,j \in {1,2,3,4}$, $i\neq j$. — Andreas Lenz, Sep 15 '21 at 10:46
@MauroALLEGRANZA Thank you, yes. Are you aware of a "shorter" form of this expression? — John Doe, Sep 15 '21 at 10:47
@Andreas Great! Thank you! But would this replace the union definition $A = B \cup C \cup D \cup {e,f}$? — John Doe, Sep 15 '21 at 10:49
One way: see Eric Wofsey's answer to What is the difference between disjoint union and union? and here. Another way: see What is the definition of the $\uplus$ symbol? and "Disjoint unions are also sometimes written ..." here. FYI, in blackboard lectures I sat in (late 1970s through the 1980s), the version with the dot was almost always used. — Dave L. Renfro, Sep 15 '21 at 13:24

M. Logic · Answer 1 · 2021-09-15T13:40:32.147

For any $A_0,\cdots,A_{n-1}$, usually we say $A_0,\cdots,A_{n-1}$ are pairwise disjoint if

\begin{equation} \text{for any }i\neq j<n\text{ we have }A_i\cap A_j=\varnothing;\qquad(*) \end{equation}

and we say $A_0,\cdots,A_{n-1}$ are disjoint if

\begin{equation} A_0\cap\cdots\cap A_n=\varnothing.\qquad(\dagger) \end{equation}

It's not hard to see that $(*)$ implies $(\dagger)$ while the converse doesn't hold. To see the former, without loss of generality suppose $A_0\cap A_1=\varnothing$, then $A_0\cap\cdots\cap A_n=\varnothing$, and to see the latter, setting $A_0=\{0,1\},A_1=\{1,2\},A_2=\{6\}$ we have $A_0\cap A_1\cap A_2=\varnothing$ but $A_0\cap A_1=\{1\}\neq\varnothing$.

How to properly define a set that is a union of pairwise disjoint sets?

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