How does the author concludes that $f^{-1} (V)$ is open?

Question

In the book of Topology by Munkres, at page 128, it is given that

However how can the author conclude that

if $B(x,\delta)$ is a neighbourhood of x contained in $f^{-1} (V)$, then $f^{-1} (V)$ is open ? I mean $x$ is arbitrary, and we have found a ball around $x$ contained in $f^{-1} (V)$, but I can't see how does it mean that $f^{-1} (V)$ is open.

Answer 1

Each $B(x,\delta)$ is open and $f^{-1}(V) = \cup_{x \in f^{-1}(V)} B(x,\delta)$.

Answer 2

Recall the first definition at the beginning of Chapter 20 - The Metric Topology:

Definition: If $d$ is a metric on a set $X$, then the collection of all $\epsilon$-balls $B_d(x, \epsilon)$, for $x \in X$ and $\epsilon > 0$, is a basis for a topology on $X$, called the metric topology induced by $d$.

Using the definition of a basis, this means that:

A set $U$ is open in the metric topology induced by $d$ iff for each $x \in U$, there is some $\epsilon> 0$ such that $B_d(x, \epsilon) \subseteq U$.