On page 42, Spivak proves the Implicit Function Theorem. 
As shown in the picture, $k$ is a function defined on $W$, but how I can know that $\forall x\in A, (x,0)\in W$? This is the only part of the theorem that I do not understand.
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Check this out, I recently wrote notes about exactly this question. I believe it's a genuine error in Spivak's proof. My notes include what I believe is a correct proof that avoids this error. – littleO Sep 01 '22 at 07:55
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@littleO In your notes, you says that $A={x\in\mathbb R^n|(x,0)\in W}$ is an open set. May you explain this more clearly? – Rmal Sep 01 '22 at 08:15
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1I actually address that question in the comments after the proof. The set $A$ is the preimage of $W$ under the continuous mapping $x \mapsto \begin{bmatrix} x \ 0 \end{bmatrix}$. And it's a fundamental fact about continuous mappings that the preimage of an open set is guaranteed to be open. – littleO Sep 01 '22 at 08:33
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I believe it's a genuine error in Spivak's proof of the implicit function theorem. Here is an example to drive the point home.
In Spivak's proof, it could be the case that $f:\mathbb R^2 \to \mathbb R$ is defined by $$ f(x,y) = x^2 + y^2 - 25 $$ and $F:\mathbb R^2 \to \mathbb R^2$ is given by $$ F(x,y) = \begin{bmatrix} x \\ x^2 + y^2 - 25 \end{bmatrix}. $$ The point $\begin{bmatrix} a \\ b \end{bmatrix}$ could be $\begin{bmatrix} 3 \\ 4 \end{bmatrix}$, and the sets $A$ and $B$ introduced in Spivak's proof could be $A = (2,5)$ and $B = (3,5)$. The set $W = \{ \begin{bmatrix} x \\ f(x,y) \end{bmatrix} \mid x \in (2,5), y \in (3,5) \}$ is visualized in the figure below. We can see that if $x > 4$ then $\begin{bmatrix} x \\ 0 \end{bmatrix} \notin W$.
Also note that if $x > 4$ then there is no point $y \in B = (3,5)$ such that $f(x,y) = 0$. So in this example it is not true that if $x \in A$ then there exists $y \in B$ such that $f(x,y) = 0$.
littleO
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