Continuous functions in metric spaces and open sets

Question

WTS: For two metric spaces $X,Y$ If $f:X \rightarrow Y$ is continuous then for every open set $U\subset Y$, $f^{-1}(U)$ is open. May someone please verify if this proof is correct?

Proof: Assume $f$ is continuous. Let $U$ be an open set in $Y$. Let $x\in f^{-1}(U)$. Since $U$ is open, there exists an $\epsilon>0$ such that $N_{\epsilon}(f(x)) \subset U$. Since the function is continuous, $\exists \delta >0$ such that whenever $p\in f^{-1}(U)$ and $p\in N_{\delta}(x)$ then $f(p)\in N_{\epsilon}(f(x))$. Since $f(p)\in N_{\epsilon}(f(x))$ then $p\in f^{-1}(U)$ and so $N_{\delta}(x) \subset U$.

The question I have is that I think the proof is correct, but I don't see why the $p$ cannot be in the intersection of the $N_{\delta}(x) $. and $f^{-1}(U)$.

May someone please clarify and tell me what I should do to improve the proof? Please?

Answer 1

You must have meant to finish with $N_\delta (x) \subset f^{-1}(U)$ instead of $\subset U$. But also you shouldn't assume that $p\in f^{-1}(U)$ and $p \in N_\delta(x)$ when using continuity of $f$, this is like using continuity of $f$ as a function on the induced metric subspace, $f:f^{-1}(U)\to Y$. Instead, you want to get your small ball $N_\delta(x)\subset X$ and prove that it is a subset of $f^{-1}(U)$.

The inclusion $A \subset f^{-1}(f(A))$ and the implication $B\subset C \implies f^{-1}(B) \subset f^{-1}(C)$, true for any function $f$ and sets $A,B,C$ will be useful.

Answer 2

Proof: Assume $f$ is continuous. Let $U$ be an open set in $Y$. Let $x\in f^{-1}(U)$. Since $U$ is open,

and $f(x) \in U$

there exists an $\epsilon>0$ such that $N_{\epsilon}(f(x)) \subset U$. Since the function is continuous,

using the $\epsilon$-$\delta$ continuity definition at $x$ and our $\epsilon$

$\exists \delta >0$ such that whenever $p\in N_{\delta}(x)$ then $f(p)\in N_{\epsilon}(f(x))$.

(omit $p \in f^{-1}[U]$ because we have to show it, not assume it)

Now, if $p$ is such that $p \in N_\delta(x)$ then $f(p)\in N_{\epsilon}(f(x)) \subseteq U$ and so $p\in f^{-1}(U)$ and as $p$ was arbitrary, $N_{\delta}(x) \subset f^{-1}[U]$, and so $x$ is an interior point of $f^{-1}[U]$.