Prove that countable intersection/union of zero/cozero sets is a zero/cozero set. 
Definition
A subset $Z$ of a topological space $X$ is a zero-set if there exist a continuous real valued function $f:X\rightarrow\Bbb R$ such that
$$
Z=f^{-1}[0]
$$
So we say that a subset $Z^C$ of $X$ is cozero-set if $X\setminus Z^C$ is zero-set.

So to follow I tried to prove that the countable intersection/union of zero/cozero sets is a zero/cozero sets so that on this purpose let be
$$
\mathcal Z:=\big\{Z_n:n\in\Bbb N\big\}\quad\text{and}\quad\mathcal Z^C:=\big\{Z_n^C:n\in\Bbb N\big\}
$$
two countable collection of zero and cozero sets resectively.

First of all if $f_n$ is for each $n\in\Bbb N$ a continuous real valued function in $X$ such that
$$
f_n(x)=0
$$
for any $x\in X$ then we observe that without loss of generality we can suppose that the inequality
$$
\begin{equation}\tag{1}\label{eq:simple}
|f_n(x)|<\frac 12
\end{equation}
$$
holds for any $x\in X$ because the function
$$
\varphi:\Bbb R\ni y\to \frac{|y|}{2\big(|y|+1\big)}\in\Bbb R
$$
is continuous and such that
$$
\big|\varphi(y)\big|\le\frac 12
$$
for any $y\in\Bbb R$ so that finally $\varphi\circ f_n$ is a continuous real valued function in $X$ satisfying the inequality \eqref{eq:simple} and such that
$$
\big(\varphi\circ f_n\big)(x)=0
$$
for any $x\in Z_n$. So we observe that the inequality
$$
\Big|\big(f_n(x)\big)^n\Big|=\big|f(x)\big|^n\le\frac 1{2^n}
$$
holds for each $n\in\Bbb N$ and for each $x\in X$ so that the serie
$$
f:=\sum_{n\in\omega}f_n^n
$$
totally converges to a continuous function (remember that $f_n$ is for any $n\in\omega$ cotinuous!) $f$ so that we let to prove that
$$
Z_f=\bigcap_{n\in\omega}Z_n
$$
So we observe that the identity
$$
f(x)=\sum_{n\in\omega}\big(f_n(x)\big)^n=\sum_{n\in\omega}0=0
$$
holds clearly for any $n\in\omega$ when $x\in\bigcap_{n\in\omega}Z_n$ so that
$$
\bigcap_{n\in\omega}Z_n\subseteq Z_f
$$
After all, if $x_0\notin\bigcap_{n\in\omega}Z_n$ then the set
$$
Z_x:=\{n\in\omega:x\notin Z_n\}
$$
is nonempty and thus the infimum of the set
$$
F_0:=\big\{f_n(x_0):n\in\omega\big\}
$$
is positive (remember that $f_n$ is for each $n\in\omega$ positive definite!) and strictly less then $1$, that is there exists $\epsilon\in(0,1)$ such that
$$
\epsilon^n\le\big(f_n(x)\big)^n
$$
for any $n\in\omega$ and thus such that
$$
0<\frac 1{1-\epsilon}=\sum_{n\in\omega}\epsilon^n\le\sum_{n\in\omega}\big(f_n(x)\big)^n=f(x)
$$
so that we conclude that
$$
X\setminus\bigcap_{n\in\omega}Z_n\subseteq X\setminus Z_f
$$
that is
$$
Z_f\subseteq\bigcap_{n\in\omega}Z_n
$$
So we proved that the countable intersection of zero sets is a zero set. Now $X\setminus Z_n^C$ is for each $n\in\Bbb N$ a zero set so that also the intersecion $\bigcap_{n\in\omega}\big(X\setminus Z_n^C\big)$ is a zero set and thus observing that
$$
\bigcup_{n\in\omega}Z^C_n=X\setminus\Biggl(X\setminus\bigcup_{n\in\omega}Z^C_n\Biggl)=X\setminus\bigcap_{n\in\omega}\big(X\setminus Z_n^C\big)
$$
we conclude that the countable union of cozero sets is a cozero set.

So I ask if the proof I gave are correct and in particular I ask if the totally convergent and infimum argumens are well applied: could someone help me, please?
 A: You are correct. The topological generalization of the classical $\epsilon$-$\delta$ statement that $f:\Bbb R\to\Bbb R$ is continuous at a point $x\in\Bbb R$ is:
The function $f:X\to Y$ is continuous at $x\in X$ iff for every open $V$ of $Y$ such that $f(x)\in V,$ there exists an open $U$ of $X$ with $x\in U,$ such that the image $f[U]\subset V.$
Then we have: $f:X\to Y$ is continuous iff $f$ is continuous at each $x\in X.$
Now in your work, for $m\in\omega$ let $g_m(x)=\sum_{n\le m}(f_n(x))^n.$ Then each  $g_m$ is continuous, and $g_m$ converges uniformly to $f.$ Therefore $f$ is continuous.
Proof: Let $x\in X$ and let $V$ be an open subset of $\Bbb R$ with $f(x)\in V.$ Take $r>0$ such that $(-r+f(x),r+f(x))\subset V.$ Take $m\in\omega$ large enough that $\forall x'\in X\,(|f(x')-g_m(x')|<r/3).$ Take an open $U$ in $X$ with $x\in U,$ such that $g_m[U]\subset (-r/3+f(x),r/3+f(x)).$ Now if $x'\in U$ then $$|f(x')-f(x)|\le |f(x')-g_m(x')|+|g_m(x')-g_m(x)|+|g_m(x)-f(x)|<$$ $$<r/3+r/3+r/3=r$$ implying $f(x')\in V.$ So $f$ is continuous at $x.$
