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I am really puzzled by this statement and it has so many different versions in different places. Yesterday I did a homework to prove that a finite function $f$ is continuous if and only if ${f^{ - 1}}(G)$ is open for any open set $G$. Today I see an application of this statement without the restriction of "finite". And I found in a lecture note which says

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I am really confused. I want to know if this statement "a map $f$ is continuous if and only if for any open set $G$, ${f^{ - 1}}(G)$ is still open" is true in general? There is no need to impose a restriction like "finite map"? Can you provide a proof or a reference of proof? Thank you!

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    $\begingroup$ The standard definition of a continuous function (for topological spaces) is that the inverse image of an open set is open. It seems that you want to prove that the standard definition is equivalent to some other definition that you've been given. Perhaps it would be helpful to give more background. $\endgroup$ – Michael Burr Jul 2 '15 at 16:47
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First, the function $f$ is finite because it's codomain is $R$. If it's codomain was $R \cup \{\infty\}$, then it wouldn't necessarily be finite. See this question for what it means for a function to be finite.

Second, remember what it means for a set to be open relative to another set. In $R^n$, there is an induced topology on a subset $D$ of $R^n$ given by the sets $U \cap D$ where $U$ is open in $R^n$.

In principle, $D$ could be closed. So if you had an open set $U$ and looked at $f^{-1}(U)$, this set could also be closed, yet the function $f$ could still be continuous. The way to reconcile this is to consider the openness of $f^{-1}(U)$ relative to the induced topology on $D$.

For instance, consider $f:[0,1] \to R$ given by $f(x)=x$ and consider $f^{-1}((-2,2))=[0,1]$. The set is closed in $R$, but it is actually open in $[0,1]$. This is how it should be since $f(x)=x$ is still continuous on $[0,1]$.

So the point of the exercise is demonstrate how this reconciliation works. Makes sense?

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  • $\begingroup$ I would guess that the OP meant that the function $f$ is finite in the sense that the domain and/or codomain are finite. $\endgroup$ – Michael Burr Jul 2 '15 at 16:56
  • $\begingroup$ Thank you so much for the reply. Yes I am also confused by "$f$ is finite". If we say a map $f:D\rightarrow R$, then it already indicates $f$ is finite? $\endgroup$ – Tony Jul 2 '15 at 17:06
  • $\begingroup$ So the statement is not true in general? It has to be open relative to $D$ but not simply open in $R$? $\endgroup$ – Tony Jul 2 '15 at 17:09
  • $\begingroup$ The statement $f$ is continuous means $f^{-1}(U)$ is open iff $U$ is not true in general if you are talking about $f:D \to R$ with $D \subset R^n$. If $D$ is open, then the above is true. Otherwise, you have to correct it slightly as in the statement in your question. $\endgroup$ – abnry Jul 2 '15 at 17:42
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Let $V$ be relatively open/open in $f(D)$, i.e. $V = G \cap f(D)$ where $G$ is open in $\mathbb R^n$. In particular $f^{-1}(G) = f^{-1}(V)$.

Let $y\in V$, then by definition $\exists x_y \in D, f(x_y) = y$.

Since $f$ is continuous, we know there exists an open ball $B_\delta(x)$ where $f(B_\delta(x)) \subseteq B_\varepsilon(y) \subseteq V$.

It follows that $f^{-1}(V) = (\bigcup_{y \in V} B_\delta(x) ) \cap D$ (If you don't see it let me know). Union of open balls is open, and hence, $f^{-1}(G) = f^{-1}(V)$ is open.

The other direction is exactly the same except back wards. Continuity gives us $\varepsilon$ for each $\delta$ for the open balls, pre-open mapping gives $\varepsilon$ for each $\delta$ for continuity via pre-mapping of open balls.

The reason I avoided saying that "it's true because topogically that's how we define it", is because the reason we define it in topology is BECAUSE of this property of $\mathbb R^n$. We modelled topology after our intuition of $\mathbb R^n$.

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