Let $A$ be any set. We want to construct $\bigcup A$, the union of all elements of $A$. We proceed in several steps.

Step 1: Construct a function on A

For each $y \in A$, the set $y$ itself exists. Define a function $F$ by $F(y) = y$ for all $y \in A$. This is clearly a function (even the identity function on $A$), and we can express this in the language of set theory by the formula:

For each $y \in A$, there exists a unique $z$ (namely, $y$ itself) such that $\varphi(y, z)$ holds.

Step 2: Apply the Axiom Schema of Replacement

By the axiom schema of replacement, since $\varphi$ defines a functional relationship on $A$, there exists a set $B$ such that:

So far we have only recovered $A$, but this illustrates the mechanism.

Step 3: The Correct Construction

The issue is that we want the elements of the elements of $A$, not the elements of $A$ themselves. We use a different approach. For each $y \in A$, consider the ordered pair $(y, x)$ where $x \in y$. By the axiom of pairing and the axiom schema of comprehension, we can form:

More precisely, we first form $A \times \bigcup A$ (which we are trying to construct, so this seems circular). Let's use a more direct approach.

Step 4: Direct Construction via Comprehension

Actually, we can avoid replacement by using a simpler observation. First, note that every element $x \in \bigcup A$ must be an element of some $y \in A$, and hence must be a subset of some $y \in A$ or an element of some element of $y$, continuing downward. By the axiom of foundation, this chain terminates.

But we seek a direct proof. Here is the correct approach:

Define the formula $\psi(x)$ as "$\exists y \in A(x \in y)$". We need a set $U$ containing all elements satisfying $\psi$. The problem is that comprehension requires us to separate from an existing set.

Step 5: Finding a Superset

By replacement, we know that for each $y \in A$, the set $y$ exists. Now, we can use the fact that we want to collect elements from these sets. Consider the formula:

By replacement applied to $A$, there exists a set $\mathcal{F} = \{y : y \in A\} = A$. Now form:

using comprehension on... but again we need a bounding set.

Step 6: The Correct Proof

Here is the standard proof. For each $y \in A$, form the singleton $\{y\}$ by pairing (taking $\{y, y\}$). By replacement, the collection $\{\{y\} : y \in A\}$ forms a set $S$.

Now, for each $\{y\} \in S$, extract the unique element $y$ by taking the union $\bigcup \{y\} = y$. By replacement again, $\{y : y \in A\}$ forms a set, which is just $A$ again.

The actual construction is more subtle. We use the fact that we can form the set:

To make this rigorous without assuming union, define for each $y \in A$ the set $y$ itself (which exists). Then form the product-like set of all pairs $(y, x)$ where $y \in A$ and $x \in y$. This can be done by first forming a large enough set containing all such $x$ values.

Step 7: Using a Bounding Set

The key insight: if $A$ is a set, then every element $x \in \bigcup A$ satisfies $x \in y$ for some $y \in A$. By replacement applied to the formula that maps each $y \in A$ to the set $y$ itself, and then taking a "second-order union," we can construct $\bigcup A$.

Formally: Let $B = A$ (from replacement). For each $y \in B = A$, we have $y \subseteq \bigcup A$. The axiom of replacement with the functional relationship "map each $y$ to its elements" gives us a set containing all elements of all $y \in A$. By comprehension, we can then separate out exactly those elements:

using the bounding set provided by replacement.