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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?

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  • $\begingroup$ First you assume "Such that whenever $p \in f^{-1}(U)$. . . " which is exactly what you are trying to prove. Try to drop that assumption at the earlier stage of the proof. $\endgroup$
    – Daron
    Feb 9, 2019 at 14:39
  • $\begingroup$ But isn't that what it means to be continuous? $\endgroup$
    – user643073
    Feb 9, 2019 at 14:40
  • $\begingroup$ What is your definition of continuous function? Because for me, what you quote as theorem is the definition. $\endgroup$
    – enedil
    Feb 9, 2019 at 15:27

2 Answers 2

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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.

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  • $\begingroup$ For a given function f: X $\rightarrow$ Y is it not the case that if f is continuous for all x in X then f is continuous at any subset of X? $\endgroup$
    – user643073
    Feb 9, 2019 at 16:30
  • $\begingroup$ Correct but not useful when the final goal is to find a ball of X that is completely contained in the pre image @topologicalmagician $\endgroup$
    – Calvin Khor
    Feb 9, 2019 at 16:31
  • $\begingroup$ so if x $\in$ X and x$\in$ $N_{\delta}(p)$ then $f(x) \in N_{\epsilon}(f(p))$ , but the issue im having is that I know that x is in the intersection of a ball(nbhd) and X is a metric space but does tha mean that the neighbourhood is a subset of X? $\endgroup$
    – user643073
    Feb 9, 2019 at 16:45
  • $\begingroup$ Sorry can you write that again, x is in the intersection of a ball and...? Remember the point of the exercise is to get that the pre image is open in X @topologicalmagician $\endgroup$
    – Calvin Khor
    Feb 9, 2019 at 16:50
  • $\begingroup$ Nevermind, I understood it. Thank you very much! $\endgroup$
    – user643073
    Feb 9, 2019 at 16:56
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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 $f^{-1}[U]$.

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