I am confused with a statement Enderton made in his text, Elements of Set Theory on page 202, Chapter 7. There were two Lemmas,
Lemma 7Q: For any ordinal number $\delta$ there is a function $F_{\delta} $ with domain $\delta$ such that $$ F_{\delta} ( \alpha )= \bigcup \{ \mathcal{P} F_{ \delta} ( \beta) \, | \, \beta \in \alpha \} $$
Proof: Apply transfinite recursion...
Lemma 7R: Let $\delta$ and $\epsilon$ be ordinal numbers; let $F_{\delta}$ and $F_\varepsilon$ be function of form in Lemma 7Q. Then $$F_\delta (\alpha) = F _ \varepsilon (\alpha)$$ for all $\alpha \in \delta \cap \varepsilon$.
And we have one statement following Lemma 7R,
The statement: In particular (by taking $\delta = \varepsilon$) we see that the function $F_{\delta}$ from lemma 7Q is unique. We can now...
My question is, the proof of 7Q uses the Transfinite Recursion Theorem, which already establishes the uniqueness of $F_\delta$. What is the point of this statement? Am I misunderstanding something here?
For reference, the recursion theorem is included below. Please tell me if not enough information is provided, or more clarification is needed. Thanks!
Transfinite Recursion Theorem Schema For any formula $\gamma(x,y)$ the following is a theorem. Assume $<$ is a well ordering on $A$. Assume that for any $f$ there is a unique $y$ such that $\gamma (f,y)$. Then there exists a unique function $F$ with domain $A$ such that $$ \gamma (F \upharpoonright \text{seg }t , F(t)) $$ for all $t \in A$.
I don't think you are misunderstanding anything here. The transfinite recursion theorem schema does indeed give uniqueness of the recursively defined function, and so your $F_δ$ is indeed uniquely defined.