# Continuous functions in metric spaces and open sets

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?

• 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. – Daron Feb 9 '19 at 14:39
• But isn't that what it means to be continuous? – monoidaltransform Feb 9 '19 at 14:40
• What is your definition of continuous function? Because for me, what you quote as theorem is the definition. – enedil Feb 9 '19 at 15:27

## 2 Answers

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.

• 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? – monoidaltransform Feb 9 '19 at 16:30
• Correct but not useful when the final goal is to find a ball of X that is completely contained in the pre image @topologicalmagician – Calvin Khor Feb 9 '19 at 16:31
• 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? – monoidaltransform Feb 9 '19 at 16:45
• 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 – Calvin Khor Feb 9 '19 at 16:50
• Nevermind, I understood it. Thank you very much! – monoidaltransform Feb 9 '19 at 16:56

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 $$f^{-1}[U]$$.