Suppose $f : \mathbb{R}^n \rightarrow \mathbb{R}^m$ is continuous on $\mathbb{R}^n$, and suppose $U \subset \mathbb{R}^m$ is open. Show that: $$f^{-1}(U)$$ is open.
My attempt here was:
$f$ continuous on $\mathbb{R}^n \rightarrow$ $\forall a \in \mathbb{R}^n, \epsilon > 0$ $\exists \delta > 0$ s.t. $x \in B_\delta(a) \rightarrow f(x) \in B_\epsilon(f(a))$
$U$ open $\rightarrow \forall x \in U$ $\exists \epsilon$ s.t. $B_\epsilon(x) \subset U$
So, $x \in B_\delta(a) \rightarrow f(x) \in B_\epsilon(f(a))$
therefore $f(B_\delta(a)) \subset B_\epsilon(f(a))$
Now, choose any $a$ s.t. $f(a)$ inside $U$ and the $\epsilon$ for which the open condition of U is met in $f(a)$ therefore:
$$f(B_\delta(a)) \subset B_\epsilon(f(a)) \subset U$$
Apply $f^{-1}$ to every set, giving:
$$f^{-1}(f(B_\delta(a))) \subset f^{-1}(U)$$
And, since $B_\delta(a) \subset f^{-1}(f(B_\delta(a)))$
$$B_\delta(a) \subset f^{-1}(U)$$
Is this correct? It doesn't look like other proofs I've seen thus the question.
For simplicity I’ll work with $n=m=1$.
Theorem: Let $f:\Bbb{R}\to\Bbb{R}$ be a map. Then $f$ is continuous in the epsilon delta sense if and only if its continuous in the open set definition. (fun fact, this generalizes to the epsilon-delta definition coinciding with the open set definition when were dealing with any metric space)
Proof:
$(\Rightarrow$) Let $V \subset \Bbb{R}$ be open and let $x \in f^{-1}(V)$, then as $f(x) \in V$ and $V$ is open, we have that there exists some $\epsilon >0$ such that
$$(f(x)-\epsilon,f(x)+\epsilon) \subset V.$$
By epsilon-delta continuity, there exists $\delta>0$ such that
$$(x-\delta,x+\delta) \subset f^{-1}(V),$$
that is, $x \in (f^{-1}(V))^\circ$ is an interior point and $x$ was arbitrary, $f^{-1}(V)$ equals its interior and is thus open. ADDED: It is clear that $(f^{-1}(V))^\circ \subset f^{-1}(V)$, I just showed the reverse inclusion forcing the preimage to be open.
($\Leftarrow$) Let $\epsilon>0$ be given, and let $x \in \Bbb{R}$, then $f(x) \in \Bbb{R}$ and thus
$$V:=(f(x)-\epsilon,f(x)+\epsilon) \subset \Bbb{R}$$
is an open set, thus
$$f^{-1}(f(x)-\epsilon,f(x)+\epsilon) \subset \Bbb{R}$$
is open, and so there exists a $\delta>0$ such that
$$(x-\delta,x+\delta) \subset f^{-1}(V)$$
i.e.,
$$f(x-\delta,x+\delta) \subset V$$
which is precisely the $\epsilon-\delta$ definition of continuity.
$\blacksquare$
I'll leave it to you to extend this to the $n,m$ dimensional case.
Also please upvote if you deem my proof valid, I’m just a young mathematician trying to earn his stripes :)