When I read Chapter II, Section 31: "B-measurable Functions" in Topology by Kuratowski, I have a problem in understanding the proof of Theorem 3.
Here is the image of that theorem.
I can not understand the left direction proof ($\Leftarrow$), which I marked with a blue line, especially the last sentence: "Secondly, the condition ...".
There is written "for sufficiently large $k$, if $|y-f(x)|<\frac{1}{k}$, then $y\in G$ (since $G$ open)."
I can not connect it with the definition of open set, i.e. the set G is open if for every $z\in G$, there exists $r>0$ such that $B(z,r)\subset G$.
So, can anyone explain, why if $G$ is open, implies that, for sufficiently large $k$, if $|y-f(x)|<\frac{1}{k}$, then $y\in G$?
Thanks for any explanation.
Note: Definition of B-measurable function
A mapping $f:X\to Y$ is said to be B-measurable of class $a$ (or, briefly, of class $a$) if, for every closed subset $F\subset Y$, the set $f^{-1}(F)$ is borelian of multiplicative class $a$.
And $\delta[A]$ denotes diameter of set $A$.
He wants to show
$$f^{-1}(G) = Z := \bigcup\{Z^k_n \mid f(Z^k_n) \subset G \} .$$ By definition $Z \subset f^{-1}(G)$ since $f(Z^k_n) \subset G$ is equivalent to $Z^k_n \subset f^{-1}(G)$.
Next he considers $x \in f^{-1}(G)$ which is equivalent to $f(x) \in G$. Since $G$ is open, there exists $r > 0$ such that $B(f(x),r) \subset G$. This means that if $\lvert y - f(x) \rvert < r$, then $y \in G$. Now we may take w.l.o.g. $r = 1/k$ for some sufficiently large $k$. We know that $x \in Z^k_n$ for some $n$. Since $\delta(f(Z^k_n)) < 1/k$, we get $\lvert f(x) - f(x') \rvert < 1/k$ for all $x' \in Z^k_n$. Thus $f(x') \in G$ for all $x' \in Z^k_n$, i.e. $f(Z^k_n) \subset G$. This shows that $x \in Z^k_n \subset Z$.