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I have the following statements

At first, theorem 1:

Let
$A$ be a complete metric space,
$B$ a separable metric space,
$M$ in $Atimes B$ open and dense set, then

the room $A$ contains a dense set $N$ with the property, that for every point $a$ of $N$ in $B$ exists a dense set of points $y$, such that all points $(a,y)$ belongs to $M$. End theorem 1.

Let $F_n(X):={f:Xrightarrow mathbb{R}^n ;|;f;continuous }$ the space of continuous functions with mapping to $mathbb{R}^n$ on a compact metric space $X$ provided with sup-metric and
$G_n(X):={f:Xrightarrow mathbb{R}^n ;|; f; continuous,; f(r)neq 0, ; forall ; xin X}$ the space of continuous functions with mapping to $mathbb{R}^n$ without functions with zeros, on a compact metric space $X$ provided with sup-metric.

So $F_n(X)$ is a complete, separable metric space, $G_n(X)$ is open in $F_n(X)$ and we can write $F_n(X)$ as $F_1(X)times F_{n-1}(X)$.

So if $G_n(X)$ is dense in $F_n(X)$ we can use theorem 1.

Therefore there exists a everywhere dense of functions $fin F_1(X)$ such that:

a) For all functions $varphiin F_{n-1}$ who combined with $f$ yield an element $(f,varphi)in G_n(X)$, form a dense set in $F_{n-1}$

If we define for $fin F_1(X)$ with $E(f)$ as set of points $xin X$ with $f(x)=0$, then for $varphiin F_{n-1}$ is the relation $(f,varphi)in G_n(X)$ equivalent, that no point of $E(f)$ will be mapped through $varphi$ in the point $(0,0,ldots, 0)in mathbb{R}^{n-1}$.

I don't understand why $varphi(x)$ with $xin E(f)$ can't be mapped to the zero vector.
Because, let us start with $F_1(X)times F_1(X)$. Have a look a my picture (sorry for ugly writing). Visualisation of my imagination

so if we take one constant $f$ in $G_2(X)$ e.g. $f(x)=1$, then $varphi$ can be mapped to $0$ in $G_2$ (because the constraint that functions with zeros in $G_2$ are not allowed are not broken) and therefore it is allowed to use $varphi(x)$ with $xin E(f)$ if $f$ is not a function with zero.

Can anyone help me to find my logic mistake? Thanks

Therefore there exists a everywhere dense of functions $fin F_1(X)$ such that:

a) For all functions $varphiin F_{n-1}$ who combined with $f$ yield an element $(f,varphi)in G_n(X)$, form a dense set in $F_{n-1}$

That is there exists a dense set $N$ of $F_1(x)$ such that for each $fin N$ the set $G_f={varphiin F_{n-1}(X):(f,varphi)in G_{n}(X)}$ is dense. Note that the set $G_f$ depends on the function $f$ and it may be not the same for distinct $fin F_1(X)$.

If we define for $fin F_1(X)$ with $E(f)$ as set of points $xin X$ with $f(x)=0$, then for $varphiin F_{n-1}$ is the relation $(f,varphi)in G_n(X)$ equivalent, that no point of $E(f)$ will be mapped through $varphi$ in the point $(0,0,ldots, 0)in mathbb{R}^{n-1}$.

Right.

I don't understand why $varphi(x)$ with $xin E(f)$ can't be mapped to the zero vector.

Note, that we still speak about pairs $(f,varphi)in G_n(X)$. And if $xin E(f)$ and $varphi(x)=0$ then $(f(x),varphi(x))=0$, which contradicts $(f,varphi)in G_n(X)$.

so if we take one constant $f$ in $G_2(X)$ e.g. $f(x)=1$, then $varphi$ can be mapped to $0$ in $G_2$ (because the constraint that functions with zeros in $G_2$ are not allowed are not broken) and therefore it is allowed to use $varphi(x)$ with $xin E(f)$ if $f$ is not a function with zero.

We are allowed to use $varphi$ means $(f,varphi)in G_n(X)$ for a given fixed $f$. This doesn’t imply that we are allowed to use $varphi$ with a different function, say $f’in F_1(X)$. For a particular choice of $f$, when $f(x)=1$ for each $xin X$, a set $E(f)={xin X :f(x)=0}$ is empty. Then $G_f={varphiin F_{n-1}(X):(f,varphi)in G_{n}(X)}=F_{n-1}(X)$. But this does not mean that $G_{f’}=F_{n-1}(X)$ for any function $f’in F_1(X).$

Remark that in this case also $E(f)={xin X :f(x)=0}$ is empty, so we cannot "use $varphi(x)$ with $xin E(f)$", because there is no such $x$. But if $f’in F_1(X)$ is an other function then we are allowed to use a function $varphiin F_{n-1}$ with the function $f’$, that is $(f,varphi)in G_n(X)$ iff $varphi(x)ne 0$ for each $xin E(f’)$.

PS. It seems that the picture is not needed.