First, we have the following definition.
Def: Let $V \subseteq D \subset \mathbf R^n$. We'll say that the subset $V$ is open in $D$ if for every point $P\in V$ there exists $\epsilon>0$, such that $B(P, \epsilon) \cap D \subset V$, where $B(P, \epsilon)$ is a neibhorhood of P.
I have the following theorem:
The function $f(P)$ with $dom(f)=D\subset \mathbf R^n$ and with values in $\mathbf R^m$ is continious iff for every open subset $U \subset \mathbf R^m$ we have that $f^{-1}(U)$ is open in D.
OK, so the proof goes like this (I'll write only the first part since there's the thing I don't understand):
Let f be continuous in all the points of it's domain D and let $U \subset \mathbf R^m$ is an open set. We'll prove that $f^{-1}(U)$ is an open set too. Let $P_0 \in f^{-1}(U)$. If $U$ is an open set, there exists $\epsilon>0$ such that $B(f(P_0), \epsilon) \subset U$. By the Cauchy definition of continuity, there exists $\delta > 0$ s.t. for every point $P$ of D when $d(P_0, P) < \delta$ we have that $d(f(P_0), f(P)) < \epsilon$. Expressing this in terms of sets, we have that $B(P_0, \delta) \cap D \subset f^{-1}(U)$, which is what we were trying to prove.
What I don't get is the last part: Isn't the last part an implication $A \implies B$ where we have that $B$ is true, so how do we make sure that A is true out of that? I mean, how did it happen that for every point $P$ of D $d(P_0, P) < \delta$ is true?
Thanks in advance!
The proof seems to be wrong. In you proof $P_0\in U\subset \mathbb{R}^m$ (codomain) but $P_0$ also in $D\subset \mathbb{R}^n$ (domain). Some of your "$P_0$" needs to be changed to "$f(P_0)$" for your proof to be correct.
You have to prove that $f^{-1}(U)$ is open if $U$ is open. So you take $P_0\in f^{-1}(U)$ (not in $U$). Then $f(P_0)\in U.$ Since $U$ is open, then for some $\epsilon>0$ we have $B(f(P_0),\epsilon)\subset U.$ By Cauchy definition of continuity there exists $\delta>0$ such that if $d(P_0,P)<\delta$ then $d(f(P_0),f(P))<\epsilon.$ (Not all $P\in D$ need to satisfy the condition $d(P_0,P)<\delta,$ but the statement is that if it does, then the condition $d(f(P_0),f(P))<\epsilon$ is satisfied too.) Thus, if $P\in B(P,\delta)\cap D,$ then $f(P)\in B(f(P_0),\epsilon)\subset U,$ which means that $B(P,\delta)\cap D \subset f^{-1}(U)$ and hence $f^{-1}(U)$ is open.