About Implicit Function Theorem in “Principles of Mathematical Analysis” by Walter Rudin. I think $U$ is not necessary.

Question

I am reading "Principles of Mathematical Analysis" by Walter Rudin.

I have a question about the implicit function theorem in this book.

Is the open set $U$ in the following statement so important?

Then there exist open sets $U\subset \mathbb{R}^{n+m}$ and $W\subset \mathbb{R}^m$, with $(a, b)\in U$ and $b \in W$, having the following property:
To every $y \in W$ corresponds a unique $x$ such that
$$(x,y) \in U$$
and
$$f(x,y)=0.$$

Isn't the following statement without $U$ enough?

Then there exists an open set $W\subset \mathbb{R}^m$, with $b \in W$, having the following property:
To every $y \in W$ corresponds a unique $x$ such that
$$x \in \mathbb{R}^n$$
and
$$f(x,y)=0.$$

Answer 1

Your variant is too strong. More precisely you should say

Then there exists an open set $W\subset \mathbb{R}^m$, with $b \in W$, having the following property:
To every $y \in W$ corresponds a unique $x$ such that
$$(x,y) \in E$$ and $$f(x,y)=0.$$

This means that you have a globally unique $x$, whereas Rudin states it is locally unique ($x$ is assumed to satisfy $(x,y) \in U$). This does not exclude that there is $x' \ne x$ such that $(x',y) \in E \setminus U$ with $f(x',y) = 0$. In fact, this phenomenon will occur in general, and it prevents your variant from being useful because it is not true for all functions satisfing Rudin's prerequisites.