如何证明不交并的分配律？求证$(A\cup B) \uplus C \subseteq (A\uplus C)\cup (B \uplus C)$

Hey there! Your initial approach is perfectly valid—leaning into the explicit set-theoretic definition of the disjoint union is exactly the right way to go. Let's walk through this step by step, and then extend it to prove the full distributive law for disjoint unions.

First, Recap the Disjoint Union Definition #

As you wrote, the standard construction for the disjoint union $X \uplus Y$ is:

X ⊎ Y = {(x, 0) | x ∈ X} ∪ {(y, 1) | y ∈ Y}

The tags 0 and 1 just let us distinguish elements that might be in both $X$ and $Y$ (since even if $x \in X \cap Y$, $(x,0)$ and $(x,1)$ are distinct ordered pairs).

Proving $(A\cup B) \uplus C \subseteq (A\uplus C)\cup (B \uplus C)$ #

Let's take any arbitrary element $k \in (A\cup B) \uplus C$. By definition, $k$ must fall into one of two categories:

Case 1: $k = (x, 0)$ where $x \in A \cup B$ #

By the definition of union, $x \in A$ or $x \in B$:

• If $x \in A$, then $(x, 0) \in A \uplus C$ (since $A \uplus C$ includes all $(a,0)$ for $a \in A$). Since $A \uplus C$ is a subset of $(A\uplus C)\cup(B\uplus C)$, this means $k \in (A\uplus C)\cup(B\uplus C)$.
• If $x \in B$, similarly, $(x,0) \in B \uplus C$, so $k$ is also in the right-hand union.

Case 2: $k = (x, 1)$ where $x \in C$ #

Here, $(x,1) \in A \uplus C$ (since $A \uplus C$ includes all $(c,1)$ for $c \in C$). Again, this puts $k$ inside $(A\uplus C)\cup(B\uplus C)$.

In both cases, $k$ belongs to the right-hand set. Since $k$ was arbitrary, we've proven the subset relation.

Proving the Full Distributive Law: $(A\cup B) \uplus C = (A\uplus C)\cup (B \uplus C)$ #

To confirm this is an equality (not just a subset), we need to show the reverse direction: $(A\uplus C)\cup (B \uplus C) \subseteq (A\cup B) \uplus C$.

Take any $k \in (A\uplus C)\cup (B \uplus C)$. Then $k$ is in $A\uplus C$ or $B\uplus C$:

Case 1: $k \in A\uplus C$ #

• If $k = (x,0)$ where $x \in A$, then $x \in A \cup B$, so $(x,0) \in (A\cup B) \uplus C$.
• If $k = (x,1)$ where $x \in C$, then directly, $(x,1) \in (A\cup B) \uplus C$.

Case 2: $k \in B\uplus C$ #

• If $k = (x,0)$ where $x \in B$, then $x \in A \cup B$, so $(x,0) \in (A\cup B) \uplus C$.
• If $k = (x,1)$ where $x \in C$, again, $(x,1) \in (A\cup B) \uplus C$.

Every element in the right-hand set is in the left-hand set, so the reverse subset holds. Combining both directions, we've proven the distributive law.

内容的提问来源于stack exchange，提问作者oldselflearner1959