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The question is:

Suppose $f\colon X\to Y$ is continuous on a compact metric space $X$, $Y$ is a metric space and $C\subset Y$ is closed. Show that for any open neighborhood $U$ of $f^{-1}(C)$ in $X$, there exists an open neighborhood $V$ of $C$ in $Y$ such that $f^{-1}(V)⊂U$.

I have tried to argue that $f^{-1}(C)$ is closed (and hence compact) by continuity of $f$ and compactness of $X$, then $C=f(f^{-1}(C))$ is also compact by continuity of $f$ again.

But the compactness of the two sets seems don't help me very much. Can anyone help me? Thx!

Lawrence
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2 Answers2

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Let $U$ be an open neighborhood of $f^{-1}(C) \Longrightarrow U^c\cap f^{-1}(C)=\emptyset$.

Then $U^c$ is compact $\Longrightarrow f(U^c)$ is compact $\Longrightarrow$ closed and $C\cap f(U^c)= \emptyset$
(Proof: If $x\in C\cap f(U^c)$ then $x=f(a)$ for some $a\in U^c \Longrightarrow$ $a\in U^c\cap f^{-1}(C)=\emptyset$ ↯).

Now we find $V$ open with $C\subset V $ and $V\cap f(U^c)=\emptyset$.

Show that $f^{-1}(V) \cap U^c=\emptyset$. Therefore $f^{-1}(V)\subseteq U$ .

P..
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  • Sorry that I don't even know why $C∩f(U^c)=∅$. Thank for your help and kindly response. – Lawrence Dec 12 '12 at 20:21
  • If $x\in C\cap f(U^c)$ then $x=f(a)$ for some $a\in U^c$. Then $a\in U^c$ and $a\in f^{-1}(C)$ ↯. – P.. Dec 12 '12 at 20:25
  • After establishing $f(U^c)$ is closed, then $Y\f(U^c)$ is open in $Y$ and $C⊂Y\f(U^c)$ as $C∩f(Uc)=∅$.

    So let $V=Y\f(U^c)$, then $f^{-1}(V)⊂U$?

    My Claim:

    $x∈f^{-1}(V) => f(x)∈Y\f(U^c) => f(x)∉f(U^c)$

    Suppose $x∉U$ then $x∈U^c$, $f(x)∈f(U^c) ↯$.

     – Lawrence Dec 12 '12 at 22:36
  • @Lawrence: That's right. – P.. Dec 13 '12 at 05:43
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This does not seem to be true as it is currently stated. (Edit: Or before it was added that $Y$ is also a metric space).

Take $Y=\{0,1\}$ and $\tau=\{\emptyset,\{0\},\{0,1\}\}$, and choose $f:[0,1]\to (\{0,1\},\tau\,)$ by setting $f(\frac{1}{2})=1$ and $f(x)=0$ otherwise. The preimage of every open set is open so this function is continuous, and $[0,1]$ is a compact metric space. Take $C=\{1\}$, which is a closed subset of $\{0,1\}$ since its complement is open, and $f^{-1}(C)=\{\frac{1}{2}\}$. Now by taking $U=]\frac{1}{4},\frac{3}{4}[$, for example, as an open neighborhood of $f^{-1}(C)$ in $[0,1]$, then $f^{-1}(\{0,1\})=[0,1]\not\subset ]\frac{1}{4},\frac{3}{4}[$, and $\{0,1\}$ is the only open set containing $C$. So for any open neighbourhood $V$ of $C$ we have $f^{-1}(V)\not\subset U$.

T. Eskin
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  • I don't understand what ({0,1},τ)is but I wonder how can $f$ be continuous. Why can you state that the preimage of every open set is open in this case? Since I think f^-1(1)={1/2} is not open in [0,1]. – Lawrence Dec 12 '12 at 19:38
  • @Lawrence: By $({0,1},\tau)$ I mean the two-point set ${0,1}$ equipped with the topology $\tau$. In this case, $\frac{1}{2}\notin {0,1}$, so we can't take preimage of $\frac{1}{2}$ (or to be precise, it is the empty set). On the other hand, $f$ is continuous because $f^{-1}(\emptyset)=\emptyset$, $f^{-1}({0})=[0,\frac{1}{2}[\cup]\frac{1}{2},1]$ and $f^{-1}({0,1})=[0,1]$. So the preimage of every open set is open. Note also that ${1}$ is not an open set since it is not a member of $\tau$. – T. Eskin Dec 12 '12 at 19:38
  • I have not be introduced with the concept of ({0,1},τ). So I am not quite understand why {1} is not an open set here. Or can you suggest me what is the metric associated with Y? In fact this question is in the past exam paper of my university in Analysis course so if it is incorrect, then I should ask my professor the details. – Lawrence Dec 12 '12 at 19:49
  • @ThomasE.: Probably here $Y$ is a metric space. – P.. Dec 12 '12 at 19:51
  • @Lawrence: In this case, there does not exist a metric associated to $Y$, since it is not even Hausdorff. $(Y,\tau)$ is just a topological pair here. – T. Eskin Dec 12 '12 at 19:51
  • And I am looking on the definition of topological space so I can understand why {1} is not open here. – Lawrence Dec 12 '12 at 19:51
  • Hah, since I don't know so much on topology and I don't even know there is such thing you have described. Now I find there is a sentence "In this paper, unless otherwise specified, letters X, Y, Z, etc., will stand for general metric spaces." at the begining of my paper.

    So, Y is a metric space. Sorry for causing inconvenience.

     – Lawrence Dec 12 '12 at 19:54
  • Pambos is correct. Sorry that I have told you there is no description about Y but actually there is, which is, Y is a metric space. – Lawrence Dec 12 '12 at 19:56
  • @Lawrence. Okey, that explains it :-) – T. Eskin Dec 12 '12 at 19:56
  • I have corrected the problem by adding Y is a metric space. – Lawrence Dec 12 '12 at 20:00