集合并运算结合性的证明方法问询

Hey there! I totally get your frustration—sometimes textbook proofs can feel way more convoluted than they need to be, especially when a straightforward line of reasoning seems right there. Let's walk through your approach to proving the associativity of set unions, using the axiom you referenced.

First, let's restate the union axiom clearly for context:

For two sets $A$ and $B$, there exists a set $A\cup B$ such that $x\in A\cup B\iff x\in A\lor x\in B$.

Now, your proposed proof is actually spot-on, and it's a great example of using the definition directly to cut through unnecessary complexity. Here's how it plays out step by step:

Starting with $(A\cup B)\cup C$, by the definition of set union, we have:
$$x\in (A\cup B)\cup C\iff x\in A\cup B\lor x\in C$$
Using the union definition again on $A\cup B$, this simplifies to:
$$x\in (A\cup B)\cup C\iff x\in A\lor x\in B\lor x\in C$$

Now let's do the same for $A\cup (B\cup C)$. Applying the union definition first to the outer union:
$$x\in A\cup (B\cup C)\iff x\in A\lor x\in B\cup C$$
Then applying the union definition to $B\cup C$, we get:
$$x\in A\cup (B\cup C)\iff x\in A\lor x\in B\lor x\in C$$

Since both $(A\cup B)\cup C$ and $A\cup (B\cup C)$ have exactly the same elements (any $x$ is in one if and only if it's in at least one of $A$, $B$, or $C$), by the extensionality axiom (which states that two sets are equal if they have exactly the same elements), we can conclude that:
$$(A\cup B)\cup C = A\cup (B\cup C)$$

Your core reasoning is perfect—you're using the definition of union repeatedly to reduce both expressions to the same logical condition, which directly gives you equality. Tao's longer proof might be covering more foundational steps (like explicitly invoking extensionality, or breaking down every logical equivalence in more detail for strict rigor), but your approach is concise and correct for anyone already comfortable with the basic axioms.

备注：内容来源于stack exchange，提问作者Aryaan