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Let $f$ be a continuous function from a metric space $X$ into $Y$. If $V\subset Y$ and $V$ is open, then show that $f^{-1}(V)$ is open.

The proofs I've seen of the fact that open sets have open preimages either use the fact that continuous functions map limit points to limit points, or they use a completely topological proof.

Is there a more basic metric feeling proof? Something that just uses the basic definition of open sets, and the basic definition of continuity? Or are these sequential/topological arguments the only arguments to make?

hrkrshnn
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    Actually thats the definition of a continuous function $f$. – PenasRaul Sep 06 '14 at 08:18
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    @PenasRaul That is one definition of continuity, used in general topology. In metric spaces, the ε–δ style definition is (perhaps) more commonly used. It is, of course, equivalent, but that equivalence is what the question is about, no? – Harald Hanche-Olsen Sep 06 '14 at 08:21
  • Bottom of https://www.dpmms.cam.ac.uk/~wtg10/easyanalysis1.html feels metrical. – isomorphismes Nov 11 '14 at 03:02

2 Answers2

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Let $f$ be a continuous function from a metric space $X$ into $Y$. If $V\subset Y$ and $V$ is open, then we shall prove that $f^{-1}(V)$ is open.

Suppose that $p\in X$ and $f(p)\in V$. Since $V$ is open, there exists $\varepsilon>0$ such that $y=f(x) \in V$ if $d_{Y}(f(x),f(p))<\varepsilon$, and since $f$ is continuous at $p$, there exists $\delta>0$ such that $d_{Y}(f(x),f(p))<\varepsilon$ if $d_{X}(x,p)<\delta$. Thus $x\in f^{-1}{(V)}$ as soon as $d_X(x,p)<\delta$.

Aqqqq
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hrkrshnn
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  • I find the logic in the proof not solid enough: because if e.g. $d_{X}(x,p)\geq\delta$ it is still possible that $d_{Y}(f(x),f(p))<\varepsilon$. – Aqqqq Oct 01 '18 at 08:22
  • That is not an issue. We only need that the membership hold true whenever $d_X(x,p)<\delta$. We do not care what happens otherwise. – hlcrypto123 Sep 29 '19 at 15:08
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    My question though is, what if the domain is not all of $X$? How can we guarantee that $\delta$ can be small enough such that the entire ball around $x$ is contained in the domain? – hlcrypto123 Sep 29 '19 at 15:11
  • @hicrypto123. Not sure if the domain "not all of $X$" matters here. The metric is defined on the domain. – hrkrshnn Sep 29 '19 at 15:25
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    Great, even though at first I was confused by part >0 such that y =() ∈ ....

    In that epsilon neighborhood of f(p), all of its elements $z \neq f(p)$ might have a preimage in X, but also just some of them or none (just $f(p)$). The important part is there exists the delta neighborhood around arbitrary $p$ from the preimage that maps into V, more specifically into the neighborhood of $f(p)$, but not necessarily to every its point.

     – Adam May 12 '22 at 11:03
  • Who guarantees that $f^{-1}$ exist? $f$ is not injective in general. – ric.san Jul 11 '23 at 15:14
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    @ric.san $f$ doesn't have to be injective. $f^{-1}$ can be a one-to-many function. The proof still works. – hrkrshnn Jul 11 '23 at 22:52
  • I also think that, as written, although correct, it's not immediately clear that there aren't additional points $q \in X$ but not in the $\delta$ neighbourhood of $p$, that also map to the $\epsilon$-neighbourhood of $f(p)$. Instead of starting with points $p \in X$, I'd begin with an open set $V \subset Y$ and look at the pre-image $p$ of every point in $V$. Then say that the pre-image of every open set $V \subset Y$ is the union of open sets in X. – Dion Silverman Nov 26 '24 at 20:42
  • It's not incorrect. But reading it, I had the same question, and had to work through why it worked. So perhaps the proof could be arranged to avoid other people having the same question.

    Alternatively, perhaps another line could just be appended stating that since this is true for every $x = f^{-1}(V)$...

     – Dion Silverman Nov 26 '24 at 20:45
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Let $f$ be a continuous function from a metric space $X$ into $Y$ and $V \subset Y$ be an open set.

Suppose $x \in f^{-1}(V)$. To prove $f^{-1}(V)$ is an open set, we need to show by the definition, that there exists $r > 0$ such that $B_X(x, r) \subset f^{-1}(V)$.

Firstly, notice that $f(x) \in V$ and since $V$ is the open set in $Y$, there exists $\epsilon > 0$, such that

$$B_Y(f(x), \epsilon) \subset V.$$

Then, by the definition of the continuity of the function $f$, there exists $\delta > 0$ such that

$$f(B_X(x, \delta)) \subset B_Y(f(x), \epsilon) \subset V.$$

Therefore, $f(B_X(x, \delta)) \subset V$, which implies $B_X(x, \delta) \subset f^{-1}(V)$.

Jan E.
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